The number of different true permutation polynomial based interleavers under Zhao and Fan sufficient conditions

Abstract

Interleavers are important blocks of turbo codes, their type and dimensions having a significant influence on the performances of the mentioned codes. If appropriately chosen, permutation polynomial (PP) based interleavers lead to remarkable performances of these codes. Necessary and sufficient conditions for a polynomial to be PP are known only for degrees up to three. Sufficient conditions for the coefficients of a polynomial of any degree to be PP were presented by Zhao and Fan (Electron Lett 43(8):449–451, 2007). In this paper, we determine the number of true different PPs of degrees 3, 4 and 5 that meet the sufficient conditions given by Zhao and Fan. This is of particular interest when we need to find higher than second degree PP based interleavers for turbo codes, because the Zhao and Fan sufficient conditions avoid brute force searching of the PP coefficients. Some remarks are made for the Long Term Evolution standard interleaver lengths.

Appendices

Appendix 1

1.1 a. Proof [Theorem 4.1]

From the case ZF2) in Sect. 4, it follows that the number of possible combinations for the coefficient $q_1$ is equal to $\prod \nolimits _{j=1}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$ and the number of coefficients $q_2$ and $q_3$, respectively, is equal to $\prod \nolimits _{j=1}^s {p_j^{n_{N,p_j } -1} }$. The value $q_3 =0$ results only when $q_{3,j} =0,\forall j=\overline{1,s}$, that is, for a single combination of coefficients $q_{3,j} ,j=\overline{1,s}$, that has to be removed. The number of CPPs will be equal to that in (12).

1.2 b. Proof [Theorem 4.2]

From the analysis for the cases ZF1.a) and ZF2) in Sect. 4, the number of possible combinations for the coefficient $q_1$ is equal to $1\cdot \Phi \left( {N/2} \right) =\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, and the number of coefficients $q_2$ and $q_3$, respectively, is equal to $1\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, from which one value is zero. By removing the value $q_3 =0$ the number of CPPs will be equal to:

$$\begin{aligned} C_{N,CPPs}= & {} \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\= & {} \left( {\prod _{j=2}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$

(51)

1.3 c. Proof [Theorem 4.3]

From the cases ZF1.b) and ZF2) in Sect. 4, it follows that the number of possible combinations for the coefficient $q_1$ is equal to $\Phi \left( N \right) =2^{n_{N,2} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, and the total number of coefficients $q_2$ and $q_3$, respectively, is equal to $2^{n_{N,2} -2}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, from which one value is zero. By removing the value $q_3 =0$, the number of CPPs will be equal to:

$$\begin{aligned} C_{N,CPPs}= & {} \left( {2^{n_{N,2} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\= & {} 2^{2\cdot n_{N,2} -3}\cdot \prod _{j=2}^s p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$

(52)

1.4 d. Proof [Theorem 4.4]

The number of possible combinations for the coefficient $q_1$ is equal to $2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of coefficients $q_2$ is equal to $3^{n_{N,3} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1}}$ and the number of coefficients $q_3$ is equal to $3^{n_{N,3} -2}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,3} >1$, and equal to $\left\lceil {\frac{1}{3}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$, if $n_{N,3} =1$, from which one is zero. We note that, if $n_{N,3} =1$, the single value for $q_3 \left( {\hbox {mod } 3} \right)$ is 0, so that from the $\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1}}$ total possible values for $q_3$, we will retain only those for which $q_3 <N/3$, that is $\left\lceil {\frac{1}{3}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ values. By removing the value $q_3 =0$, the number of CPPs will be equal to:

$$\begin{aligned} C_{N,CPPs}= & {} \left( {2\cdot 3^{n_{N,3} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} \left( {2\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot \prod _{j=2}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \end{aligned}$$

(53)

1.5 e. Proof [Theorem 4.5]

The proof for Theorem 4.5 is similar to that of Theorem 4.4, by replacing N by N / 2. Therefore, the number of CPPs will be equal to:

$$\begin{aligned} C_{N,CPPs}= & {} \left( {2\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} \left( {2\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot \prod _{j=3}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \end{aligned}$$

(54)

1.6 f. Proof [Theorem 4.6]

The number of possible combinations for the coefficient $q_1$ is equal to $\Phi \left( N \right) =2^{n_{N,2} -1}\cdot 2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right)$, the number of coefficients $q_2$ is equal to $2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}$ and the number of coefficients $q_3$ is equal to $\left\lceil {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$, from which one is zero. By removing the value $q_3 =0$, the number of CPPs will be equal to:

$$\begin{aligned} C_{N,CPPs}= & {} \left( {2^{n_{N,2} -1}\cdot 2\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\left\lceil {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} 2^{2\cdot \left( {n_{N,2} -1} \right) }\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) } \nonumber \\&\cdot \prod _{j=3}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \cdot } \nonumber \\&\cdot \left( {\left\lceil {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1}\right) \nonumber \\ \end{aligned}$$

(55)

Appendix 2

1.1 a. Proof [Theorem 4.7]

From the case ZF2) in Sect. 4, the number of possible combinations for the coefficient $q_1$ is equal to $\prod \nolimits _{j=1}^s p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right)$ and the number of coefficients $q_2$, $q_3$, and $q_4$, respectively, is equal to $\prod \nolimits _{j=1}^s {p_j^{n_{N,p_j } -1} }$. By removing the value $q_4 =0$, the number of 4-PPs will be equal to that in (20).

1.2 b. Proof [Theorem 4.8]

From the analysis for the cases ZF1.a) and ZF2) in Sect. 4, the number of possible combinations for the coefficient $q_1$ is equal to $1\cdot \Phi \left( {N/2} \right) =\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of coefficients $q_2$, $q_3$, and $q_4$, respectively, is equal to $1\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, from which one value is zero. By removing the value $q_4 =0$, the number of 4-PPs will be equal to:

$$\begin{aligned} C_{N,4{\hbox {-}}PPs}= & {} \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \cdot \left( {\prod _{j=2}^s{p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\= & {} \left( {\prod _{j=2}^s {p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$

(56)

1.3 c. Proof [Theorem 4.9]

The number of possible combinations for the coefficient $q_1$ is equal to $2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of coefficients $q_2$ is equal to $3^{n_{N,3} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$ and the number of coefficients $q_3$ and $q_4$, respectively, is equal to $3^{n_{N,3} -2}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,3} \ge 2$, and equal to $\left\lceil {\frac{1}{3}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$, if $n_{N,3} =1$, from which one is zero. By removing the value $q_4 =0$, the number of 4-PPs will be equal to:

$$\begin{aligned} C_{N,4{\hbox {-}}PPs}= & {} \left( {2\cdot 3^{n_{N,3} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}}\right) \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} \left( {2\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot \prod _{j=2}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \end{aligned}$$

(57)

1.4 d. Proof [Theorem 4.10]

From the cases ZF1.b) and ZF2) in Sect. 4, the number of possible combinations for the coefficient $q_1$ is equal to $\Phi \left( N \right) =2\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of coefficients $q_2$, $q_3$ and $q_4$, respectively, is equal to $\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, from which one is zero. By removing the value $q_4 =0$, the number of 4-PPs will be equal to:

$$\begin{aligned} C_{N,4{\hbox {-}}PPs}= & {} \left( {2\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) {\cdot } \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\= & {} 2\cdot \prod _{j=2}^s p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$

(58)

1.5 e. Proof [Theorem 4.11]

The proof for Theorem 4.11 is similar to that of Theorem 4.9, by replacing N with N / 2, thus, the number of 4-PPs will be equal to:

$$\begin{aligned} C_{N,4-PPs}= & {} \left( {2\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} \left( {2\cdot 3^{2{\cdot } \left( {n_{N,3} -1} \right) }\cdot \prod _{j=3}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \end{aligned}$$

(59)

1.6 f. Proof [Theorem 4.12]

From the cases ZF1.b) and ZF2) in Sect. 4, the number of possible combinations for the coefficient $q_1$ is equal to $\Phi \left( N \right) =2^{n_{N,2} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of coefficients $q_3$ is equal to $2^{n_{N,2} -2}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, the number of even or odd coefficients $q_2$ is equal to $2^{n_{N,2} -2}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, and the number of even or odd coefficients $q_4$ is equal to $2^{n_{N,2} -4}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,2} \ge 4$, from which one is zero. If $n_{N,2} =3$, there are $\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_4$ (all even), and $2\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_2$, (all even). Therefore, if $n_{N,2} \ge 4$, the number of 4-PPs will be equal to:

$$\begin{aligned} C_{N,4{\hbox {-}}PPs}= & {} \left( {2^{n_{N,2} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -4}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&+\left( {2^{n_{N,2} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j }-1}}} \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -4}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\= & {} 2^{3\cdot n_{N,2} -5}\cdot \prod _{j=2}^s p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -3}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$

(60)

We mention that, if $n_{N,2} =3$, the number of 4-PPs will be equal to:

$$\begin{aligned} C_{N,4-PPs}= & {} \left( {2^{3-1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \cdot \left( {2^{3-2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\= & {} 16\cdot \prod _{j=2}^s p_j^{3\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) , \end{aligned}$$

(61)

so that, the relation (60) is also true.

1.7 g. Proof [Theorem 4.13]

From the cases ZF1.b) and ZF2) in Sect. 4, the number of possible combinations for the coefficients $q_1$ is equal to $\Phi \left( N \right) =2\cdot 2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of coefficients $q_3$ is equal to $3^{n_{N,3} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, the number of coefficients $q_2$ is equal to $3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, and the number of coefficients $q_4$ is equal to $3^{n_{N,3} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,3} \ge 2$, from which one is zero. If $n_{N,3} =1$, there are $\prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_2$, and $\left\lceil {\frac{1}{3}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ coefficients $q_3$ and $q_4$, respectively. Therefore, the number of 4-PPs will be equal to:

$$\begin{aligned} C_{N,4{\hbox {-}}PPs}= & {} \left( {4\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} 4\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot \prod _{j=3}^s p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \end{aligned}$$

(62)

1.8 h. Proof [Theorem 4.14]

From the cases ZF1.b) and ZF2) in Sect. 4, the number of possible combinations for the coefficient $q_1$ is equal to $\Phi \left( N \right) =2^{n_{N,2} -1}\cdot 2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of coefficients $q_3$ is equal to $2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,3} \ge 2$, the number of even or odd coefficients $q_2$ is equal to $2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, and the number of even or odd coefficients $q_4$ is equal to $2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,2} \ge 4$ and $n_{N,3} \ge 2$, from which one is zero. If $n_{N,2} =3$ and $n_{N,3} \ge 2$, there are $2\cdot 3^{n_{N,3} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_3$, $3^{n_{N,3} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_4$ (all even), and $2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_2$ (all even). If $n_{N,2} \ge 4$ and $n_{N,3} =1$, there are $\left\lceil {\frac{1}{3}\cdot 2^{n_{N,2} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ coefficients $q_3$, $\left\lceil {\frac{1}{3}\cdot 2^{n_{N,2} -4}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ even coefficients $q_4$, $\llbracket {\frac{1}{3}\cdot 2^{n_{N,2} -4}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket$ odd coefficients $q_4$, and $2^{n_{N,2} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ even or odd coefficients $q_2$. If $n_{N,2} =3$ and $n_{N,3} =1$, there are $\left\lceil {\frac{1}{3}\cdot 2\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ coefficients $q_3$, $\left\lceil {\frac{1}{3}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ coefficients $q_4$ (all even), and $2\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_2$ (all even). Therefore, if $n_{N,2} \ge 4$, the number of 4-PPs will be equal to:

(63)

In the last equation we used (11).

We mention that, if $n_{N,2} =3$, the number of 4-PPs will be equal to:

$$\begin{aligned} C_{N,4{\hbox {-}}PPs}= & {} \left( {2^{3-1}\cdot 2\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left\lceil {2\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} 16\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot \prod _{j=3}^s p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left\lceil {2\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) , \end{aligned}$$

(64)

so that the relation (63) is also true.

Appendix 3

1.1 a. Proof [Theorem 4.15]

From the case ZF2) in Sect. 4, the number of possible combinations for the coefficients $q_1$ is equal to $\prod \nolimits _{j=1}^s p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right)$ and the number of coefficients $q_2$, $q_3$, $q_4$, and $q_5$, respectively, is equal to $\prod \nolimits _{j=1}^s {p_j^{n_{N,p_j } -1} }$. By removing the value $q_5 =0$, the number of 5-PPs will be equal to that in (30).

1.2 b. Proof [Theorem 4.16]

From the analysis for the cases ZF1.a) and ZF2) in Sect. 4, the number of possible combinations for the coefficients $q_1$ is equal to $1\cdot \Phi \left( {N/2} \right) =\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$ and the number of coefficients $q_2$, $q_3$, $q_4$, and $q_5$, respectively, is equal to $1\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, from which one is zero. By removing the value $q_5 =0$, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\= & {} \left( {\prod _{j=2}^s {p_j^{4\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$

(65)

1.3 c. Proof [Theorem 4.17]

The number of possible combinations for the coefficient $q_1$ is equal to $2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of coefficients $q_2$ is equal to $3^{n_{N,3} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$ and the number of coefficients $q_3$, $q_4$, and $q_5$, respectively, is equal to $3^{n_{N,3} -2}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,3} \ge 2$, and equal to $\left\lceil {\frac{1}{3}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$, if $n_{N,3} =1$, from which one is zero. By removing the value $q_5 =0$, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {2\cdot 3^{n_{N,3} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} \left( {2\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot \prod _{j=2}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1}\right) } } \right) \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil } \right) ^{2} \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \end{aligned}$$

(66)

1.4 e. Proof [Theorem 4.18]

From the cases ZF1.b) and ZF2) in Sect. 4, the number of possible combinations for the coefficient $q_1$ is equal to $\Phi \left( N \right) =2\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$ and the number of coefficients $q_2$, $q_3$, $q_4$ and $q_5$, respectively, is equal to $\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, from which one is zero. By removing the value $q_5 =0$, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {2{\cdot } \prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) {\cdot } \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\= & {} 2\cdot \prod _{j=2}^s p_j^{4\cdot \left( {n_{N,p_j } -1} \right) }\cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$

(67)

1.5 f. Proof [Theorem 4.19]

The number of possible combinations for the coefficient $q_1$ is equal to $4\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of coefficients $q_2$, $q_3$, and $q_4$, respectively, is equal to $5^{n_{N,5} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$ and the number of coefficients $q_5$ is equal to $5^{n_{N,5} -2}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,5} \ge 2$, and equal to $\left\lceil {\frac{1}{5}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$, if $n_{N,5} =1$, from which one is zero. By removing the value $q_5 =0$, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {4\cdot 5^{n_{N,5} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {5^{n_{N,5} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {5^{n_{N,5} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {5^{n_{N,5} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\left\lceil {5^{n_{N,5} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} \left( {4\cdot 5^{4\cdot \left( {n_{N,5} -1} \right) }\cdot \prod _{j=2}^s {p_j^{4\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {\left\lceil {5^{n_{N,5} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \end{aligned}$$

(68)

1.6 g. Proof [Theorem 4.20]

The proof for Theorem 4.20 is similar to that of Theorem 4.17, by replacing N with N / 2. Therefore, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {2\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} \left( {2\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot \prod _{j=3}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil } \right) ^{2} \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \end{aligned}$$

(69)

1.7 h. Proof [Theorem 4.21]

From the cases ZF1.b) and ZF2) in Sect. 4, the number of possible combinations for the coefficient $q_1$ is equal to $\Phi \left( N \right) =2^{n_{N,2} -1}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of even or odd coefficients $q_2$ and $q_3$, respectively, is equal to $2^{n_{N,2} -2}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, the number of even or odd coefficients $q_4$ and $q_5$, respectively, is equal to $2^{n_{N,2} -4}\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,2} \ge 4$, from which one is zero. If $n_{N,2} =3$, there are $\prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_4$ and $q_5$, respectively, all of them even, from which one is zero, and $2\cdot \prod \nolimits _{j=2}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_2$ and $q_3$, respectively, all of them even. Therefore, the number of 5-PPs, if $n_{N,2} \ge 4$, will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {2^{n_{N,2} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -4}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left[ \left( {2^{n_{N,2} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \right. \nonumber \\&\quad \cdot \left( {2^{n_{N,2} -4}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\quad +\left( {2^{n_{N,2} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\quad \left. \cdot \left( {2^{n_{N,2} -4}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \right] \nonumber \\&+\left( {2^{n_{N,2} -1}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}\cdot \left( {p_j -1} \right) } } \right) \nonumber \\ \end{aligned}$$

$$\begin{aligned}&\cdot \left( {2^{n_{N,2} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -4}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left[ \left( {2^{n_{N,2} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \right. \nonumber \\&\quad \cdot \left( {2^{n_{N,2} -4}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\&\quad +\left( {2^{n_{N,2} -2}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\quad \left. \cdot \left( {2^{n_{N,2} -4}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \right] \nonumber \\= & {} 2^{4\cdot \left( {n_{N,2} -2} \right) }\cdot \prod _{j=2}^s p_j^{4\cdot \left( {n_{N,p_j } -1} \right) }\cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -3}\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$

(70)

and, if $n_{N,2} =3$, this number will be:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {4\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \cdot \left( {2\cdot \prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1}}} \right) \cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\= & {} 16\cdot \prod _{j=2}^s p_j^{4\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left( {\prod _{j=2}^s {p_j^{n_{N,p_j } -1} } -1} \right) , \end{aligned}$$

(71)

so that, the relation (70) is also true.

1.8 i. Proof [Theorem 4.22]

The proof for Theorem 4.22 is similar to that of Theorem 4.19, by replacing N with N / 2. Therefore, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {4\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\left\lceil {5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} \left( {4\cdot 5^{4\cdot \left( {n_{N,5} -1} \right) }\cdot \prod _{j=3}^s {p_j^{4\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {\left\lceil {5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \end{aligned}$$

(72)

1.9 j. Proof [Theorem 4.23]

From the cases ZF1.b) and ZF2) in Sect. 4, the number of possible combinations for the coefficient $q_1$ is equal to $\Phi \left( N \right) =2\cdot 2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of coefficients $q_2$ is equal to $3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ and the number of coefficients $q_3$, $q_4$ and $q_5$, respectively, is equal to $3^{n_{N,3} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,3} \ge 2$, from which one is zero. If $n_{N,3} =1$, there are $\prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_2$ and $\left\lceil {\frac{1}{3}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ coefficients $q_3$, $q_4$, and $q_5$, respectively. Therefore, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {4\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} 4\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot \prod _{j=3}^s p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1}\right) \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil } \right) ^{2} \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \end{aligned}$$

(73)

1.10 k. Proof [Theorem 4.24]

The number of possible combinations for the coefficient $q_1$ is equal to $2\cdot 4\cdot 3^{n_{N,3} -1}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of coefficients $q_2$ is equal to $3^{n_{N,3} -1}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, the number of coefficients $q_3$ and $q_4$, respectively, is equal to $3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,3} \ge 2$, and equal to $\left\lceil {\frac{1}{3}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$, if $n_{N,3} =1$. The number of coefficients $q_5$ is equal to $3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,3} \ge 2$ and $n_{N,5} \ge 2$, equal to $\left\lceil {\frac{1}{3}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$, if $n_{N,3} =1$ and $n_{N,5} \ge 2$, equal to $\left\lceil {\frac{1}{5}\cdot 3^{n_{N,3} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$, if $n_{N,5} =1$ and $n_{N,3} \ge 2$, and equal to $\left\lceil {\frac{1}{15}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$, if $n_{N,3} =1$ and $n_{N,5} =1$, from which one is zero. By removing the value $q_5 =0$, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {8{\cdot } 3^{n_{N,3} -1}\cdot 5^{n_{N,5} -1}{\cdot } \prod _{j=3}^s {p_j^{n_{N,p_j } -1} {\cdot } \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} \left( 8\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot 5^{2\cdot \left( {n_{N,5} -1} \right) } \right. \nonumber \\&\quad \left. \cdot \prod _{j=3}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil } \right) ^{2} \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\ \end{aligned}$$

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1.11 l. Proof [Theorem 4.25]

From the cases ZF1.b) and ZF2) in Sect. 4, the number of possible combinations for the coefficient $q_1$ is equal to $\Phi \left( N \right) =2\cdot 4\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of coefficients $q_2$, $q_3$, and $q_4$, respectively, is equal to $5^{n_{N,5} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ and the number of coefficients $q_5$ is equal to $5^{n_{N,5} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,5} \ge 2$, from which one is zero. If $n_{N,5} =1$, there are $\prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_2$, $q_3$, and $q_4$, respectively, and $\left\lceil {\frac{1}{5}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ coefficients $q_5$. Therefore, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {8\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\left\lceil {5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} 8\cdot 5^{4\cdot \left( {n_{N,5} -1} \right) }\cdot \prod _{j=3}^s p_j^{4\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left( {\left\lceil {5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \end{aligned}$$

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1.12 m. Proof [Theorem 4.26]

From the cases ZF1.b) and ZF2) in Sect. 4, the number of possible combinations for the coefficient $q_1$ is equal to $\Phi \left( N \right) =2^{n_{N,2} -1}\cdot 2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of even or odd coefficients $q_2$ is equal to $2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, the number of even or odd coefficients $q_3$ is equal to $2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,3} \ge 2$, and the number of even or odd coefficients $q_4$ and $q_5$, respectively, is equal to $2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,2} \ge 4$ and $n_{N,3} \ge 2$, from which one is zero. If $n_{N,2} =3$ and $n_{N,3} \ge 2$, there are $2\cdot 3^{n_{N,3} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_2$, all of them even, $2\cdot 3^{n_{N,3} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_3$, all of them even, and $3^{n_{N,3} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_4$, and $q_5$, respectively, all of them even. If $n_{N,2} \ge 4$ and $n_{N,3} =1$, there are $2^{n_{N,2} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ even or odd coefficients $q_2$, $\left\lceil {\frac{1}{3}\cdot 2^{n_{N,2} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ even coefficients $q_3$, $\llbracket {\frac{1}{3}\cdot 2^{n_{N,2} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket$ odd coefficients $q_3$, and $\left\lceil {\frac{1}{3}\cdot 2^{n_{N,2} -4}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ even coefficients $q_4$ and $q_5$, respectively, and $\llbracket {\frac{1}{3}\cdot 2^{n_{N,2} -4}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket$ odd coefficients $q_4$ and $q_5$, respectively. If $n_{N,2} =3$ and $n_{N,3} =1$, there are $2\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_2$, all of them even, $\left\lceil {2\cdot \frac{1}{3}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ coefficients $q_3$, all of them even, and $\left\lceil {\frac{1}{3}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ coefficients $q_4$ and $q_5$, respectively, all of them even. Therefore, if $n_{N,2} \ge 4$, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {2^{n_{N,2} -1}\cdot 2\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left\lceil {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left\{ \left\lceil {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \right. \nonumber \\&\quad \cdot \left( {\left\lceil {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\&\quad +\,\llbracket {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\quad \left. \cdot \llbracket {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \right\} \nonumber \\&+\,\left( {2^{n_{N,2} -1}\cdot 2\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left[ {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right] \nonumber \\&\cdot \left\{ \left\lceil {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \right. \nonumber \\&\quad \cdot \left( {\left\lceil {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\&\quad +\llbracket {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\quad \left. \cdot \llbracket {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \right\} \nonumber \end{aligned}$$

$$\begin{aligned}&\qquad \qquad =2^{2\cdot \left( {n_{N,2} -1} \right) }{\cdot } 3^{2\cdot \left( {n_{N,3} -1} \right) }{\cdot } \prod _{j=3}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } {\cdot } \left( {p_j -1} \right) } \nonumber \\&\qquad \qquad \qquad \cdot \left\{ \left\lceil {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \right. \nonumber \\&\qquad \qquad \qquad \cdot \left\lceil {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\qquad \qquad \qquad \cdot \left( {\left\lceil {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\&\qquad \qquad \qquad +\left\lceil {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\qquad \qquad \qquad \cdot \llbracket {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\qquad \qquad \qquad \cdot \llbracket {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\qquad \qquad \qquad +\llbracket {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\qquad \qquad \qquad \cdot \left\lceil {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\qquad \qquad \qquad \cdot \left( {\left\lceil {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\&\qquad \qquad \qquad +\llbracket {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\qquad \qquad \qquad \cdot \llbracket {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\qquad \qquad \qquad \left. \cdot \llbracket {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \right\} \end{aligned}$$

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Thus, if $n_{N,2} \ge 4$ and $n_{N,3} \ge 2$, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} 2^{4\cdot n_{N,2} -7}\cdot 3^{4\cdot n_{N,3} -6} \nonumber \\&\cdot \prod _{j=3}^s p_j^{4\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -3}\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1} } -1} \right) \end{aligned}$$

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and if $n_{N,2} \ge 4$ and $n_{N,3} =1$, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} 2^{2\cdot \left( {n_{N,2} -1} \right) }\cdot \prod _{j=3}^s p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left\{ \left\lceil {2^{n_{N,2} -4}\cdot \frac{1}{3}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \right. \nonumber \\&\quad \cdot \left\lceil {2^{n_{N,2} -2}\cdot \frac{1}{3}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\quad \cdot \left( {\left\lceil {2^{n_{N,2} -4}\cdot \frac{1}{3}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\&\quad +\left\lceil {2^{n_{N,2} -4}\cdot \frac{1}{3}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\quad \cdot \llbracket {2^{n_{N,2} -2}\cdot \frac{1}{3}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\quad \cdot \llbracket {2^{n_{N,2} -4}\cdot \frac{1}{3}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\quad +\llbracket {2^{n_{N,2} -4}\cdot \frac{1}{3}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\quad \cdot \left\lceil {2^{n_{N,2} -2}\cdot \frac{1}{3}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\quad \cdot \left( {\left\lceil {2^{n_{N,2} -4}\cdot \frac{1}{3}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\&\quad +\llbracket {2^{n_{N,2} -4}\cdot \frac{1}{3}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\quad \llbracket {2^{n_{N,2} -2}\cdot \frac{1}{3}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\quad \left. \cdot \llbracket {2^{n_{N,2} -4}\cdot \frac{1}{3}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \right\} \end{aligned}$$

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If $n_{N,2} =3$, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {4\cdot 2\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2\cdot 3^{n_{N,3} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left\lceil {2\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} 16\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot \prod _{j=3}^s p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left\lceil {2\cdot 3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \end{aligned}$$

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1.13 n. Proof [Theorem 4.27]

The proof for Theorem 4.27 is similar to that of Theorem 4.24, by replacing N with N / 2. The number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {8{\cdot } 3^{n_{N,3} -1}{\cdot } 5^{n_{N,5} -1}{\cdot } \prod _{j=4}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} \left( 8\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot 5^{2\cdot \left( {n_{N,5} -1} \right) } \right. \nonumber \\&\quad \left. \cdot \prod _{j=4}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil } \right) ^{2} \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\ \end{aligned}$$

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1.14 o. Proof [Theorem 4.28]

From the cases ZF1.b) and ZF2) in Sect. 4, the number of possible combinations for the coefficient $q_1$ is equal to $\Phi \left( N \right) =2^{n_{N,2} -1}\cdot 4\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of even or odd coefficients $q_2$ and $q_3$, respectively, is equal to $2^{n_{N,2} -2}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, the number of even or odd coefficients $q_4$, is equal to $2^{n_{N,2} -4}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ and the number of even or odd coefficients $q_5$, is equal to $2^{n_{N,2} -4}\cdot 5^{n_{N,5} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,2} \ge 4$ and $n_{N,5} \ge 2$, from which one is zero. If $n_{N,2} =3$ and $n_{N,5} \ge 2$, there are $2\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_2$ and $q_3$, respectively, all of them even, $5^{n_{N,5} -1}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_4$, all of them even, and $5^{n_{N,5} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_5$, all of them even, from which one is zero. If $n_{N,2} \ge 4$ and $n_{N,5} =1$, there are $2^{n_{N,2} -2}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ even or odd coefficients $q_2$ and $q_3$, respectively, $2^{n_{N,2} -4}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ even or odd coefficients $q_4$ and $\left\lceil {\frac{1}{5}\cdot 2^{n_{N,2} -4}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ even coefficients $q_5$, from which one is zero and $\llbracket {\frac{1}{5}\cdot 2^{n_{N,2} -4}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket$ odd coefficients $q_5$. If $n_{N,2} =3$ and $n_{N,5} =1$, there are $2\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_2$ and $q_3$, respectively, all of them even, $\prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1} }$ coefficients $q_4$, all of them even, and $\left\lceil {\frac{1}{5}\cdot \prod \nolimits _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ coefficients $q_5$, all of them even, from which one is zero. Therefore, if $n_{N,2} \ge 4$, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {2^{n_{N,2} -1}\cdot 4{\cdot } 5^{n_{N,5} -1}{\cdot } \prod _{j=3}^s {p_j^{n_{N,p_j } -1} {\cdot } \left( {p_j -1} \right) } } \right) \\&\cdot \left( {2^{n_{N,2} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \\&\cdot \left( {2^{n_{N,2} -4}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \\&\cdot \left\{ \left( {2^{n_{N,2} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \right. \\&\quad \cdot \left( {\left\lceil {2^{n_{N,2} -4}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \\&\quad +\left( {2^{n_{N,2} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \\&\quad \left. \cdot \llbracket {2^{n_{N,2} -4}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \right\} \\&\quad +\left( {2^{n_{N,2} -1}{\cdot } 4{\cdot } 5^{n_{N,5} -1}{\cdot } \prod _{j=3}^s {p_j^{n_{N,p_j } -1} {\cdot } \left( {p_j -1} \right) } } \right) \\&\cdot \left( {2^{n_{N,2} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \\&\cdot \left( {2^{n_{N,2} -4}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \\&\cdot \left\{ \left( {2^{n_{N,2} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \right. \\&\quad \cdot \left( {\left\lceil {2^{n_{N,2} -4}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \end{aligned}$$

$$\begin{aligned}&\qquad \qquad \quad \qquad +\left( {2^{n_{N,2} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\qquad \qquad \qquad \left. \cdot \llbracket {2^{n_{N,2} -4}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \right\} \nonumber \\&\qquad \quad =2^{4\cdot n_{N,2} -6}\cdot 5^{4\cdot \left( {n_{N,5} -1} \right) } \cdot \prod _{j=3}^s {p_j^{4\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \nonumber \\&\qquad \qquad \quad \cdot \left( \left\lceil {2^{n_{N,2} -4}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \right. \nonumber \\&\qquad \qquad \qquad \left. +\llbracket {2^{n_{N,2} -4}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \rrbracket -1 \right) \nonumber \\&\qquad \quad =2^{4\cdot n_{N,2} -6}\cdot 5^{4\cdot \left( {n_{N,5} -1} \right) }\cdot \prod _{j=3}^s p_j^{4\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \nonumber \\&\qquad \qquad \quad \cdot \left( {\left\lceil {2^{n_{N,2} -3}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\ \end{aligned}$$

(81)

In the last equation we used (11).

If $n_{N,2} =3$, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5{\hbox {-}}PPs}= & {} \left( {4\cdot 4\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {2\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {2\cdot 5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {5^{n_{N,5} -1}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left( {\left\lceil {5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} 64\cdot 5^{4\cdot \left( {n_{N,5} -1} \right) }\cdot \prod _{j=3}^s p_j^{4\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) \nonumber \\&\cdot \left( {\left\lceil {5^{n_{N,5} -2}\cdot \prod _{j=3}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \end{aligned}$$

(82)

so that, the relation (81) is also true.

1.15 p. Proof [Theorem 4.29]

The proof for Theorem 4.29 is similar to that of Theorem 4.24, by replacing N with N / 4 and multiplying the number of coefficients $q_1$ by 2. The number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5-PPs}= & {} \left( {16{\cdot } 3^{n_{N,3} -1}{\cdot } 5^{n_{N,5} -1}{\cdot } \prod _{j=4}^s {p_j^{n_{N,p_j } -1} {\cdot } \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left( {3^{n_{N,3} -1}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} \left( 16\cdot 3^{2\cdot \left( {n_{N,5} -1} \right) }\cdot 5^{2\cdot \left( {n_{N,5} -1} \right) } \right. \nonumber \\&\quad \left. \cdot \prod _{j=4}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil } \right) ^{2} \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\ \end{aligned}$$

(83)

1.16 q. Proof [Theorem 4.30]

From the cases ZF1.b) and ZF2) in Sect. 4, the number of possible combinations for the coefficient $q_1$ is equal to $2^{n_{N,2} -1}\cdot 2\cdot 3^{n_{N,3} -1}\cdot 4\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=4}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) }$, the number of even or odd coefficients $q_2$ is equal to $2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=4}^s {p_j^{n_{N,p_j } -1} }$ and the number of even or odd coefficients $q_3$ is equal to $2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=4}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,3} \ge 2$. If $n_{N,3} =1$, the number of even coefficients $q_3$ is equal to $\left\lceil {\frac{1}{3}\cdot 2^{n_{N,2} -2}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ and the number of odd coefficients $q_3$ is equal to $\llbracket \frac{1}{3}\cdot 2^{n_{N,2} -2}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=4}^s {p_j^{n_{N,p_j } -1}} \rrbracket$. The number of even or odd coefficients $q_4$ is equal to $2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=4}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,2} \ge 4$ and $n_{N,3} \ge 2$. If $n_{N,2} \ge 4$ and $n_{N,3} =1$, the number of even coefficients $q_4$ is equal to $\left\lceil {\frac{1}{3}\cdot 2^{n_{N,2} -4}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$ and the number of odd coefficients $q_4$ is equal to $\llbracket \frac{1}{3}\cdot 2^{n_{N,2} -4}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=4}^s {p_j^{n_{N,p_j } -1}} \rrbracket$. If $n_{N,2} =3$, the number of coefficients $q_4$ is equal to $\left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod \nolimits _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$, all of them even. The number of even or odd coefficients $q_5$ is equal to $2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod \nolimits _{j=4}^s {p_j^{n_{N,p_j } -1} }$, if $n_{N,2} \ge 4$, $n_{N,3} \ge 2$ and $n_{N,5} \ge 2$. If $n_{N,2} \ge 4$ and $n_{N,3} =1$ and/or $n_{N,5} =1$, the number of even coefficients $q_5$ is equal to $\left\lceil {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod \nolimits _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$, from which one is zero, and the number of odd coefficients $q_5$ is equal to $\llbracket {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod \nolimits _{j=4}^s {p_j^{n_{N,p_j } -1}}} \rrbracket$. If $n_{N,2} =3$, the number of coefficients $q_5$ is equal to $\left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod \nolimits _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil$, all of them even, from which one is zero. By removing the value $q_5 =0$, the number of 5-PPs, if $n_{N,2} \ge 4$, will be equal to:

$$\begin{aligned}&C_{N,5{\hbox {-}}PPs} \\&\quad =\left( {8\cdot 2^{n_{N,2} -1}\cdot 3^{n_{N,3} -1}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \\&\qquad \cdot \left( {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j} -1}}} \right) \\&\qquad \cdot \left\lceil {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \\&\qquad \cdot \left\{ \left\lceil {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \right. \\&\qquad \cdot \left( {\left\lceil {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \\&\qquad +\llbracket {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \\&\qquad \left. {\cdot \llbracket {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \rrbracket } \right\} \\&\qquad +\,\left( {8\cdot 2^{n_{N,2} -1}\cdot 3^{n_{N,3} -1}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } } \right) \\&\qquad \cdot \left( {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -1}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}}\right) \\&\qquad \cdot \llbracket {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \end{aligned}$$

$$\begin{aligned}&\cdot \left\{ \left\lceil {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \right. \nonumber \\&\quad \cdot \left( {\left\lceil {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\&\quad +\llbracket {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\quad \left. {\cdot \llbracket {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \rrbracket } \right\} \nonumber \\= & {} \left( {2^{2\cdot n_{N,2} }\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot 5^{2\cdot \left( {n_{N,5} -1} \right) }\cdot \prod _{j=4}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } } \right) \nonumber \\&\cdot \left\{ {\left\lceil {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil } \right. \nonumber \\&\quad \cdot \left\{ \left\lceil {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \right. \nonumber \\&\quad \cdot \left( {\left\lceil {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\&\quad +\llbracket {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\quad \left. {\cdot \llbracket {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \rrbracket } \right\} \nonumber \\&\quad +\llbracket {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\quad \cdot \left\{ \left\lceil {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \right. \nonumber \\&\quad \cdot \left( {\left\lceil {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\&\quad +\llbracket {2^{n_{N,2} -2}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \rrbracket \nonumber \\&\quad \left. \left. {\cdot \llbracket {2^{n_{N,2} -4}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \rrbracket } \right\} \right\} \end{aligned}$$

(84)

Thus, if $n_{N,2} \ge 4$, $n_{N,3} \ge 2$ and $n_{N,5} \ge 2$, the number of 5-PPs will be equal to:

$$\begin{aligned}&C_{N,5-PPs} =\left( 2^{4\cdot n_{N,2} -5}\cdot 3^{4\cdot n_{N,3} -6}\cdot 5^{4\cdot \left( {n_{N,5} -1} \right) } \right. \nonumber \\&\quad \left. \cdot \prod _{j=4}^s {p_j^{4\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\cdot \left( {2^{n_{N,2} -3}\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1} } -1} \right) \nonumber \\ \end{aligned}$$

(85)

If $n_{N,2} =3$, the number of 5-PPs will be equal to:

$$\begin{aligned} C_{N,5-PPs}= & {} \left( 8\cdot 4\cdot 3^{n_{N,3} -1}\cdot 5^{n_{N,5} -1} \right. \nonumber \\&\quad \left. \cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1} \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\cdot \left( {2\cdot 3^{n_{N,3} -1}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right) \nonumber \\&\cdot \left\lceil {2\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\= & {} \left( 64\cdot 3^{2\cdot \left( {n_{N,3} -1} \right) }\cdot 5^{2\cdot \left( {n_{N,5} -1} \right) } \right. \nonumber \\&\quad \left. \cdot \prod _{j=4}^s {p_j^{2\cdot \left( {n_{N,p_j } -1} \right) } \cdot \left( {p_j -1} \right) } \right) \nonumber \\&\cdot \left\lceil {2\cdot 3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -1}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil \nonumber \\&\cdot \left( {\left\lceil {3^{n_{N,3} -2}\cdot 5^{n_{N,5} -2}\cdot \prod _{j=4}^s {p_j^{n_{N,p_j } -1}}} \right\rceil -1} \right) \nonumber \\ \end{aligned}$$

(86)

Appendix 4

In this appendix we give some examples of checking some formulas proved in the previous appendices, for more comprehensive cases, i.e., when the prime decomposition of interleaver length contains all types of prime factors involved in the theorems for a certain degree of PPs.

1.1 a. Example for Theorem 4.6

Let $N=2100=2^{2}\cdot 3\cdot 5^{2}\cdot 7$. According to (17), the number of CPPs will be equal to:

$$\begin{aligned} C_{2100,CPPs}= & {} 2^{2\cdot \left( {2-1} \right) }\cdot 3^{2\cdot \left( {1-1} \right) }\cdot 5^{2\cdot \left( {2-1} \right) }\cdot \left( {5-1} \right) \cdot 7^{2\cdot \left( {1-1} \right) } \\&\cdot \left( {7-1} \right) \cdot \left( {\left\lceil {2^{2-2}\cdot 3^{1-2}\cdot 5^{2-1}\cdot 7^{1-1}} \right\rceil -1} \right) = \\= & {} 4\cdot 1\cdot 25\cdot 4\cdot 1\cdot 6\cdot \left( {\left\lceil {1\cdot 3^{-1}\cdot 5\cdot 1} \right\rceil -1} \right) \\= & {} 2400\cdot \left( {2-1} \right) =2400. \end{aligned}$$

Further, we detail the coefficients of the 2400 true different CPPs, under Zhao and Fan sufficient conditions. Because $6\left| N \right.$, it requires that $q_3 <N/6=350$ and $q_2 <N/2=1050$. The coefficients $q_3$ and $q_2$ have to be multiples of $2\cdot 3\cdot 5\cdot 7=210$, and the coefficient $q_1$ has to be relatively prime with 2100. Therefore, the coefficient $q_3$ can take only the value 210, the coefficient $q_2$ can take the values 0, 210, 420, 630 and 840 (5 values), and the coefficient $q_1$ can take $\Phi \left( {2100} \right) =2100\cdot \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{4}{5}\cdot \frac{6}{7}=480$ values. The total number of CPPs is equal to $1\cdot 5\cdot 480=2400$, just the number found by the formula we determined.

1.2 b. Example for Theorem 4.14

Let $N=1080=2^{3}\cdot 3^{3}\cdot 5$. According to (27), the number of 4-PPs will be equal to:

$$\begin{aligned} C_{1080,4{\hbox {-}}PPs}= & {} 2^{2\cdot \left( {3-1} \right) }\cdot 3^{2\cdot \left( {3-1} \right) }\cdot 5^{2\cdot \left( {1-1} \right) }\cdot \left( {5-1} \right) \\&\cdot \left\lceil {2^{3-2}\cdot 3^{3-2}\cdot 5^{1-1}} \right\rceil \\&\cdot \left( {\left\lceil {2^{3-3}\cdot 3^{3-2}\cdot 5^{1-1}} \right\rceil -1} \right) \\= & {} 16\cdot 81\cdot 1\cdot 4\cdot 6\cdot \left( {3-1} \right) \\= & {} 62{,}208. \end{aligned}$$

Since $24\left| N \right.$, it requires that $q_4 <N/{24}=45$, $q_3 <N/6=180$ and $q_2 <N/2=540$. The coefficients $q_4$ and $q_2$ have to be multiples of $3\cdot 5=15$, the coefficient $q_3$ has to be multiple of $2\cdot 3\cdot 5=30$, and the coefficient $q_1$ has to be relatively prime with 1080. Moreover, the coefficients $q_4$ and $q_2$ have to meet the condition that the sum $q_4 +q_2$ is even. Therefore, the coefficient $q_3$ can take the values 0, 30, 60, 90, 120, 150 (6 values), the pair of coefficients $\left( {q_4 ,q_2 } \right)$ can take the values (15, 15), (15, 45), (15, 75), ..., (15, 525) (18 pairs with both coefficients odd) and (30, 0), (30, 30), (30, 60), ..., (30, 510) (18 pairs with both coefficients even), i.e. a total of 36 pairs. The coefficient $q_1$ can take $\Phi \left( {1080} \right) =1080\cdot \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{4}{5}=288$ values. The total number of 4-PPs is equal to $36\cdot 6\cdot 288=62{,}208$, just the number found by the fomula we determined.

1.3 c. Examples for Theorem 4.30

For this theorem we give two examples to use both the simpler formula (85) and the general one given by (46).

Firstly, let $N=25{,}200=2^{4}\cdot 3^{2}\cdot 5^{2}\cdot 7$. According to (85) the number of 5-PPs will be equal to:

$$\begin{aligned} C_{25200,5{\hbox {-}}PPs}= & {} \left( {2^{4\cdot 4-5}\cdot 3^{4\cdot 2-6}\cdot 5^{4\cdot \left( {2-1} \right) }\cdot 7^{4\cdot \left( {1-1} \right) }\cdot \left( {7-1} \right) } \right) \\&\cdot \left( {2^{4-3}\cdot 3^{2-2}\cdot 5^{2-2}\cdot 7^{1-1}-1} \right) \\= & {} \left( {2048\cdot 9\cdot 625\cdot 1\cdot 6} \right) \cdot \left( {2\cdot 1\cdot 1\cdot 1-1} \right) \\= & {} 69{,}120{,}000. \end{aligned}$$

Since 3|N, 5|N and 8|N, it requires that $q_5 <N/{120}=210$, $q_4 <N/{24}=1050$, $q_3 <N/6=4200$ and $q_2 <N/2=12{,}600$. The coefficients $q_5$, $q_4$, $q_3$ and $q_2$ have to be multiples of $3\cdot 5\cdot 7=105$, and the coefficient $q_1$ has to be relatively prime with 25,200. Moreover, the coefficients $q_5$ and $q_3$ have to meet the condition that the sum $q_5 +q_3$ is even and the coefficients $q_4$ and $q_2$ have to meet the condition that the sum $q_4 +q_2$ is even. Therefore, the pair of coefficients $\left( {q_5 ,q_3 } \right)$ can take the values (105, 105), (105, 315), (105, 525), ..., (105, 4095) (20 pairs with both coefficients odd), and the pair of coefficients $\left( {q_4 ,q_2 } \right)$ can take the values (105, 105), (105, 315), (105, 425), ..., (105, 12495), (315, 105), (315, 315), (315, 425), ..., (315, 12495), (525, 105), (525, 315), (525, 425), ..., (525, 12495), (735, 105), (735, 315), (735, 425), ..., (735, 12495), (945, 105), (945, 315), (945, 425), ..., (945, 12495) ($5\cdot 60=300$ pairs with both coefficients odd) and (0, 0), (0, 210), (0, 420), ..., (0, 12390), (210, 0), (210, 210), (210, 420), ..., (210, 12390), (420, 0), (420, 210), (420, 420), ..., (420, 12390), (630, 0), (630, 210), (630, 420), ..., (630, 12390), (840, 0), (840, 210), (840, 420), ..., (840, 12390), ($5\cdot 60=300$ pairs with both coefficients even), i.e. a total of 600 pairs. The coefficient $q_1$ can take $\Phi \left( {25200} \right) =25200\cdot \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{4}{5}\cdot \frac{6}{7}=5760$ values. The total number of 5-PPs is equal to $20\cdot 600\cdot 5760=69120000$, just the number we found by the formula we determined.

Now, let $N=2400=2^{5}\cdot 3\cdot 5^{2}$. According to (46), the number of 5-PPs will be equal to:

$$\begin{aligned} C_{2400,5-PPs}= & {} \left( {2^{2\cdot 5}\cdot 3^{2\cdot \left( {1-1} \right) }\cdot 5^{2\cdot \left( {2-1} \right) }} \right) \\&\cdot \left\{ \left\lceil {2^{5-4}\cdot 3^{1-2}\cdot 5^{2-1}} \right\rceil \right. \\&\quad \cdot \left\{ \left\lceil {2^{5-2}\cdot 3^{1-2}\cdot 5^{2-1}} \right\rceil \right. \\&\qquad \cdot \left( {\left\lceil {2^{5-4}\cdot 3^{1-2}\cdot 5^{2-2}} \right\rceil -1} \right) \\&\qquad +\llbracket {2^{5-2}\cdot 3^{1-2}\cdot 5^{2-1}} \rrbracket \\&\qquad \left. \cdot \llbracket {2^{5-4}\cdot 3^{1-2}\cdot 5^{2-2}} \rrbracket \right\} \\&\quad +\llbracket {2^{5-4}\cdot 3^{1-2}\cdot 5^{2-1}} \rrbracket \\&\quad \cdot \left\{ \left\lceil {2^{5-2}\cdot 3^{1-2}\cdot 5^{2-1}} \right\rceil \right. \\&\qquad \cdot \left( {\left\lceil {2^{5-4}\cdot 3^{1-2}\cdot 5^{2-2}} \right\rceil -1} \right) \\&\qquad +\llbracket {2^{5-2}\cdot 3^{1-2}\cdot 5^{2-1}} \rrbracket \\&\qquad \left. \left. {\cdot \llbracket {2^{5-4}\cdot 3^{1-2}\cdot 5^{2-2}} \rrbracket } \right\} \right\} \\= & {} \left( {1024\cdot 1\cdot 25} \right) \\&\cdot \left\{ \left\lceil {\frac{10}{3}} \right\rceil \cdot \left\{ \left\lceil {\frac{40}{3}} \right\rceil \cdot \left( {\left\lceil {\frac{2}{3}} \right\rceil -1} \right) \right. \right. \\&\qquad +\llbracket {\frac{40}{3}} \rrbracket \left. {\cdot \llbracket {\frac{2}{3}} \rrbracket } \right\} +\llbracket {\frac{10}{3}} \rrbracket \\&\quad \cdot \left\{ \left\lceil {\frac{40}{3}} \right\rceil \cdot \left( {\left\lceil {\frac{2}{3}} \right\rceil -1} \right) \left. +\llbracket {\frac{40}{3}} \rrbracket \left. \cdot \llbracket {\frac{2}{3}} \rrbracket \right\} \right\} \right. \\= & {} 25{,}600\cdot \left\{ 4\cdot \left\{ 14\cdot \left( {1-1} \right) +13\left. {\cdot 1} \right\} \right. \right. \\&\left. +3\cdot \left\{ 14\cdot \left( {1-1} \right) +13 {\cdot 1} \right\} \right\} \\= & {} 25{,}600\cdot 7\cdot 13=2{,}329{,}600 \end{aligned}$$

In this case it requires that $q_5 <N/{120}=20$, $q_4 <N/{24}=100$, $q_3 <N/6=400$ and $q_2 <N/2=1200$. The coefficients $q_5$, $q_4$, $q_3$ şi $q_2$ have to be multiples of $3\cdot 5=15$. The other conditions the coefficients $q_5$, $q_4$, $q_3$ and $q_2$ have to meet are the same as those for $N=25{,}200$, and the coefficient $q_1$ has to be relatively prime with 2400. Therefore, the pair of coefficients $\left( {q_5 ,q_3 } \right)$ can take the values (15, 15), (15, 45), (15, 75), ..., (15, 375) (13 pairs with both coefficients odd). The pair of coefficients $\left( {q_4 ,q_2 } \right)$ can take the values (15, 15), (15, 45), (15, 75), ..., (15, 1185), (45, 15), (45, 45), (45, 75), ..., (45, 1185), (75, 15), (75, 45), (75, 75), ..., (75, 1185) ($3\cdot 40=120$ pairs with both coefficients odd) and (0, 0), (0, 30), (0, 60), ..., (0, 1170), (30, 0), (30, 30), (30, 60), ..., (30, 1170), (60, 0), (60, 30), (60, 60), ..., (60, 1170), (90, 0), (90, 30), (90, 60), ..., (90, 1170) ($4\cdot 40=160$ pairs with both coefficients even), i.e. a total of 280 pairs. The coefficient $q_1$ can take $\Phi \left( {2400} \right) =2400\cdot \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{4}{5}=640$ values. The total number of 5-PPs is equal to $13\cdot 280\cdot 640=2{,}329{,}600$, just the number we found by the formula we determined.