Annals of Combinatorics

, Volume 23, Issue 3–4, pp 659–682

# Partitions into Distinct Parts Modulo Powers of 5

Article

## Abstract

If $$p_D(n)$$ denotes the number of partitions of n into distinct parts, it is known that for $$\alpha \ge 1$$ and $$n\ge 0$$,
\begin{aligned} p_D\left( 5^{2\alpha +1}n+\frac{5^{2\alpha +2}-1}{24}\right) \equiv 0\pmod {5^\alpha }. \end{aligned}
We give a completely elementary proof of this fact.

## Keywords

Distinct parts Congruences Powers of 5

## Mathematics Subject Classification

Primary 11P83 Secondary 05A17

## 1 Introduction

Let $$p_D(n)$$ denote the number of partitions of n into distinct parts. Then
\begin{aligned} \sum _{n\ge 0}p_D(n)q^n=(-q;q)_\infty =\frac{E(q^2)}{E(q)}, \end{aligned}
where
\begin{aligned} E(q)=(q;q)_\infty . \end{aligned}
Baruah and Begum  proved the following results:
\begin{aligned}&\displaystyle \sum _{n\ge 0}p_D(5n+1)q^n=\frac{E(q^2)^2E(q^5)^3}{E(q)^4E(q^{10})},\end{aligned}
(1.1)
\begin{aligned}&\displaystyle \quad \sum _{n\ge 0}p_D(25n+1)q^n=\frac{E(q^2)^3E(q^5)^4}{E(q)^5E(q^{10})^2}\nonumber \\&\displaystyle \quad \times \left( 1+160q\left( \frac{E(q^2)E(q^{10})^3}{E(q)^3E(q^5)}\right) +2800q^2\left( \frac{E(q^2)E(q^{10})^3}{E(q)^3E(q^5)}\right) ^2\right. \nonumber \\&\displaystyle \left. \quad +16000q^3\left( \frac{E(q^2)E(q^{10})^3}{E(q)^3E(q^5)}\right) ^3+32000q^4\left( \frac{E(q^2)E(q^{10})^3}{E(q)^3E(q^5)}\right) ^4\right) , \end{aligned}
(1.2)
as well as the corresponding result for $$\sum \nolimits _{n\ge 0}p_D(125n+26)q^n$$.

Inspired by their work, we prove the following general result.

### Theorem 1.1

For$$\alpha \ge 1$$,
\begin{aligned}&\sum _{n\ge 0}p_D\left( 5^{2\alpha -1}n+\frac{5^{2\alpha }-1}{24}\right) q^n =\gamma \sum _{i=1}^{(5^{2\alpha }-1)/24}x_{2\alpha -1,i}\zeta ^{i-1}, \end{aligned}
(1.3)
\begin{aligned}&\sum _{n\ge 0}p_D\left( 5^{2\alpha }n+\frac{5^{2\alpha }-1}{24}\right) q^n =\delta \sum _{i=1}^{(5^{2\alpha +1}-5)/24}x_{2\alpha ,i}\zeta ^{i-1}, \end{aligned}
(1.4)
where
\begin{aligned} \gamma =\frac{E(q^2)^2E(q^5)^3}{E(q)^4E(q^{10})},\quad \delta =\frac{E(q^2)^3E(q^5)^4}{E(q)^5E(q^{10})^2},\quad \zeta =q\frac{E(q^2)E(q^{10})^3}{E(q)^3E(q^5)} \end{aligned}
and where the coefficient vectors$${{\mathbf {x}}}_\alpha =(x_{\alpha ,1},\,x_{\alpha ,2},\ \ldots \ )$$ are given recursively by
\begin{aligned} {{\mathbf {x}}}_1=(1,\,0,\ \ldots \ ), \end{aligned}
and for$$\alpha \ge 1$$,
\begin{aligned} {{\mathbf {x}}}_{2\alpha }={{\mathbf {x}}}_{2\alpha -1}A \end{aligned}
and
\begin{aligned} {{\mathbf {x}}}_{2\alpha +1}={{\mathbf {x}}}_{2\alpha }B, \end{aligned}
whereAis the matrix$$(\alpha _{i,j})_{i,j\ge 1}$$andBis the matrix$$(\beta _{i,j})_{i,j\ge 1}$$, where the$$\alpha _{i,j}$$and$$\beta _{i,j}$$are given by
\begin{aligned} \sum _{i,j\ge 1}\alpha _{i,j}x^iy^j=\frac{N_\alpha }{D'} \end{aligned}
and
\begin{aligned} \sum _{i,j\ge 1}\beta _{i,j}x^iy^j=\frac{N_\beta }{D'}, \end{aligned}
where
\begin{aligned} N_\alpha&=(y+160y^2+2800y^3+16000y^4+32000y^5)x\nonumber \\&\quad +(180y^2+3000y^3+16800y^4 +32000y^5)x^2\nonumber \\&\quad +(75y^2+1215y^3+6600y^4+12000y^5)x^3\nonumber \\&\quad +(14y^2+220y^3+1150y^4+2000y^5)x^4\nonumber \\&\quad +(y^2+15y^3+75y^4+125y^5)x^5,\\ N_\beta&=(5y+660y^2+14400y^3+120000y^4+448000y^5+640000y^6)x\nonumber \\&\quad +(y+680y^2+14900y^3+123200y^4+456000y^5+640000y^6)x^2\nonumber \\&\quad +(265y^2+5785y^3+47500y^4+174000y^5+240000y^6)x^3\nonumber \\&\quad +(46y^2+1000y^3+8150y^4+29500y^5+40000y^6)x^4\nonumber \\&\quad +(3y^2+65y^3+525y^4+1875y^5+2500y^6)x^5 \end{aligned}
and
\begin{aligned} D'&=1-(205y+4300y^2+34000y^3+120000y^4+160000y^5)x\nonumber \\&\quad -(215y+4475y^2+35000y^3+122000y^4+160000y^5)x^2\nonumber \\&\quad -(85y+1750y^2+13525y^3+46000y^4+60000y^5)x^3\nonumber \\&\quad -(15y+305y^2+2325y^3+7875y^4+10000y^5)x^4\nonumber \\&\quad -(y+20y^2+150y^3+500y^4+625y^5)x^5. \end{aligned}
Furthermore, for$$\alpha \ge 1$$,
\begin{aligned} x_{2\alpha +1,i}&\equiv 0\pmod {5^{\alpha }}, \\ x_{2\alpha +2,i}&\equiv 0\pmod {5^{\alpha }}, \end{aligned}
from which it follows that for$$\alpha \ge 1$$,
\begin{aligned} p_D\left( 5^{2\alpha +1}n+\frac{5^{2\alpha +2}-1}{24}\right)&\equiv 0\pmod {5^{\alpha }}, \end{aligned}
(1.5)
\begin{aligned} p_D\left( 5^{2\alpha +2}n+\frac{5^{2\alpha +2}-1}{24}\right)&\equiv 0\pmod {5^{\alpha }}. \end{aligned}
(1.6)
(Of course, (1.6) is a special case of (1.5).)

This result is due to Rødseth  and independently to Gordon and Hughes . See also Lovejoy .

## 2 Preliminaries

Let
\begin{aligned} R(q)=\left( \begin{matrix} q,q^4\\ q^2,q^3\end{matrix};q^5\right) _\infty ,\quad \chi (-q)=(q;q^2)_\infty =\frac{E(q)}{E(q^2)}. \end{aligned}
Then ([4, (8.1.1)])
\begin{aligned} E(q)=E(q^{25})\left( \frac{1}{R(q^5)}-q-q^2R(q^5)\right) , \end{aligned}
([4, (8.4.4)])
\begin{aligned} \frac{1}{E(q)}&=\frac{E(q^{25})^5}{E(q^5)^6} \left( \frac{1}{R(q^5)^4}+\frac{q}{R(q^5)^3}+\frac{2q^2}{R(q^5)^2}+\frac{3q^3}{R(q^5)}+5q^4\right. \\&\quad \left. -3q^5R(q^5)+2q^6R(q^5)^2-q^7R(q^5)^3+q^8R(q^5)^4 \right) , \end{aligned}
([4, (40.2.3)])
\begin{aligned} R(q^2)-R(q)^2=2q\left( \begin{matrix} q,\,q,\, q^9,q^9\\ q^3,q^5,q^5,q^7\end{matrix};q^{10}\right) _\infty , \end{aligned}
(2.1)
([4, (40.2.4)])
\begin{aligned} R(q^2)+R(q)^2=2\left( \begin{matrix} q,\,q^4,q^6,q^9\\ q^2,q^5,q^5,q^8\end{matrix};q^{10}\right) _\infty , \end{aligned}
(2.2)
([4, (41.1.3)])
\begin{aligned} 1-qR(q)R(q^2)^2=\left( \begin{matrix} q,\,q^4,q^5,q^5,q^6,q^9\\ q^2,q^3,q^3,q^7,q^7,q^8\end{matrix};q^{10}\right) _\infty , \end{aligned}
(2.3)
([4, (41.1.2)])
\begin{aligned} 1+qR(q)R(q^2)^2=\left( \begin{matrix} q^2,q^2,q^5,q^5,q^8,q^8\\ q,\,q^4,q^4,q^6,q^6,q^9\end{matrix};q^{10}\right) _\infty , \end{aligned}
(2.4)
([4, (34.8.4)])
\begin{aligned} \frac{E(q^2)^4}{E(q)^2}-q\frac{E(q^{10})^4}{E(q^5)^2}=\frac{E(q^2)E(q^5)^3}{E(q)E(q^{10})} \end{aligned}
and ([4, (34.8.3)])
\begin{aligned} \frac{E(q^5)^4}{E(q^{10})^2}-\frac{E(q)^4}{E(q^2)^2}=4q\frac{E(q)E(q^{10})^3}{E(q^2)E(q^5)}. \end{aligned}
We require the following results.

### Lemma 2.1

\begin{aligned}&\displaystyle \frac{R(q^2)}{R(q)^2}-\frac{R(q)^2}{R(q^2)} =4q\frac{\chi (-q)}{\chi (-q^5)^5}, \end{aligned}
(2.5)
\begin{aligned}&\displaystyle \frac{R(q^2)-R(q)^2}{R(q^2)+R(q)^2} =qR(q)R(q^2)^2, \end{aligned}
(2.6)
\begin{aligned}&\displaystyle \frac{1}{R(q)R(q^2)^2}-q^2R(q)R(q^2)^2 =\frac{\chi (-q^5)^5}{\chi (-q)}, \end{aligned}
(2.7)
\begin{aligned}&\displaystyle \frac{1-qR(q)R(q^2)^2}{1+qR(q)R(q^2)^2} =\frac{R(q)^2}{R(q^2)}, \end{aligned}
(2.8)
\begin{aligned}&\displaystyle \frac{R(q)}{R(q^2)^3}+q^2\frac{R(q^2)^3}{R(q)} =\frac{\chi (-q^5)^5}{\chi (-q)}-2q+4q^2\frac{\chi (-q)}{\chi (-q^5)^5}, \end{aligned}
(2.9)
\begin{aligned}&\displaystyle \frac{1}{R(q)^3R(q^2)}+q^2R(q)^3R(q^2)=\frac{\chi (-q^5)^5}{\chi (-q)}+2q+4q^2\frac{\chi (-q)}{\chi (-q^5)^5}, \end{aligned}
(2.10)
\begin{aligned}&\displaystyle \frac{\chi (-q^5)^5}{\chi (-q)}+q =\frac{E(q^2)^4E(q^5)^2}{E(q)^2E(q^{10})^4} \end{aligned}
(2.11)
and
\begin{aligned} 1-4q\frac{\chi (-q)}{\chi (-q^5)^5}=\frac{E(q)^4E(q^{10})^2}{E(q^2)^2E(q^5)^4}. \end{aligned}
(2.12)

### Proof of (2.5)

If we multiply (2.1) by (2.2) and divide by $$R(q)^2R(q^2)$$, we find that
\begin{aligned} \frac{R(q^2)}{R(q)^2}-\frac{R(q)^2}{R(q^2)}&=\frac{(R(q^2)-R(q)^2)(R(q^2)+R(q)^2)}{R(q)^2R(q^2)}\\&=\frac{2q\left( \begin{matrix} q,\,q,\,q^9,q^9\\ q^3,q^5,q^5,q^7\end{matrix};q^{10}\right) _\infty \cdot 2\left( \begin{matrix} q,\,q^4,q^6,q^9\\ q^2,q^5,q^5,q^8\end{matrix};q^{10}\right) _\infty }{\left( \begin{matrix} q,\,q,\,q^4,q^4,q^6,q^6,q^9,q^9,q^2,q^8\\ q^2,q^2,q^3,q^3,q^7,q^7,q^8,q^8,q^4,q^6\end{matrix};q^{10}\right) _\infty }\\&=4q\left( \begin{matrix} q,\,q^3,q^5,q^7,q^9\\ q^5,q^5,q^5,q^5,q^5\end{matrix};q^{10}\right) _\infty \\&=4q\frac{(q;q^2)_\infty }{(q^5;q^{10})_\infty ^5}\\&=4q\frac{\chi (-q)}{\chi (-q^5)^5}. \end{aligned}
$$\square$$

### Proof of (2.6)

If we divide (2.1) by (2.2), we obtain
\begin{aligned} \frac{R(q^2)-R(q)^2}{R(q^2)+R(q)^2}&=q\frac{\left( \begin{matrix} q,\,q,\,q^9,q^9\\ q^3,q^5,q^5,q^7\end{matrix};q^{10}\right) _\infty }{\left( \begin{matrix} q,\,q^4,q^6,q^9\\ q^2,q^5,q^5,q^8\end{matrix};q^{10}\right) _\infty }\\&=q\left( \begin{matrix} q,\,q^2,q^8,q^9\\ q^3,q^4,q^6,q^7\end{matrix};q^{10}\right) _\infty \\&=q\left( \begin{matrix} q,\,q^4,q^6,q^9,q^2,q^2,q^8,q^8\\ q^2,q^3,q^7,q^8,q^4,q^4,q^6,q^6\end{matrix};q^{10}\right) _\infty \\&=qR(q)R(q^2)^2. \end{aligned}
$$\square$$

### Proof of (2.7)

If we multiply (2.3) by (2.4) and divide by $$R(q)R(q^2)^2$$, we find
\begin{aligned}&\frac{1}{R(q)R(q^2)^2}-q^2R(q)R(q^2)^2 =\frac{(1-qR(q)R(q^2)^2)(1+qR(q)R(q^2)^2)}{R(q)R(q^2)^2}\\&\quad =\frac{\left( \begin{matrix} q,\,q^4,q^5,q^5,q^6,q^9\\ q^2,q^3,q^3,q^7,q^7,q^8\end{matrix};q^{10}\right) _\infty \left( \begin{matrix} q^2,q^2,q^5,q^5,q^8,q^8\\ q,\,q^4,q^4,q^6,q^6,q^9\end{matrix};q^{10}\right) _\infty }{\left( \begin{matrix} q,\,q^4,q^6,q^9\\ q^2,q^3,q^7,q^8\end{matrix};q^{10}\right) _\infty \left( \begin{matrix} q^2,q^2,q^8,q^8\\ q^4,q^4,q^6,q^6\end{matrix};q^{10}\right) _\infty }\\&\quad =\left( \begin{matrix} q^5,q^5,q^5,q^5,q^5\\ q,\,q^3,q^5,q^7,q^9\end{matrix};q^{10}\right) _\infty \\&\quad =\frac{\chi (-q^5)^5}{\chi (-q)}. \end{aligned}
$$\square$$

### Proof of (2.8)

If we divide (2.3) by (2.4), we obtain
\begin{aligned} \frac{1-qR(q)R(q^2)^2}{1+qR(q)R(q^2)^2}&=\frac{\left( \begin{matrix} q,\,q^4,q^5,q^5,q^6,q^9\\ q^2,q^3,q^3,q^7,q^7,q^9\end{matrix};q^{10}\right) _\infty }{\left( \begin{matrix} q^2,q^2,q^5,q^5,q^8,q^8\\ q,\,q^4,q^4,q^6,q^6,q^9\end{matrix};q^{10}\right) _\infty }\\&=\left( \begin{matrix} q,\,q,\,q^4,q^4,q^6,q^6,q^9,q^9,q^4,q^6\\ q^2,q^2,q^3,q^3,q^7,q^7,q^8,q^8,q^2,q^8\end{matrix};q^{10}\right) _\infty \\&=\frac{R(q)^2}{R(q^2)}. \end{aligned}
$$\square$$

### Proof of (2.9)

Note that (2.6) is equivalent to (2.8), because they are both equivalent to
\begin{aligned} R(q^2)-R(q)^2=qR(q)^3R(q^2)^2+qR(q)R(q^2)^3. \end{aligned}
(2.13)
If we divide (2.13) by $$R(q)R(q^2)^3$$ and rearrange, we find that
\begin{aligned} \frac{R(q)}{R(q^2)^3}=\frac{1}{R(q)R(q^2)^2}-q\frac{R(q)^2}{R(q^2)}-q, \end{aligned}
(2.14)
while if we divide (2.13) by $$R(q)^2$$, rearrange and multiply by q, we obtain
\begin{aligned} q^2\frac{R(q^2)^3}{R(q)}=-q^2R(q)R(q^2)^2+q\frac{R(q^2)}{R(q)^2}-q. \end{aligned}
(2.15)
If we add (2.14) and (2.15), we obtain
\begin{aligned} \frac{R(q)}{R(q^2)^3}+q^2\frac{R(q^2)^3}{R(q)}&=\left( \frac{1}{R(q)R(q^2)^2}-q^2R(q)R(q^2)^2\right) -2q\\&\quad +q\left( \frac{R(q^2)}{R(q)^2}-\frac{R(q)^2}{R(q^2)}\right) \\&=\frac{\chi (-q^5)^5}{\chi (-q)}-2q+4q^2\frac{\chi (-q)}{\chi (-q^5)^5}. \end{aligned}
$$\square$$

### Proof of (2.10)

If we multiply (2.5) by (2.7) and add (2.9), we find that
\begin{aligned}&\frac{1}{R(q)^3R(q^2)}+q^2R(q)^3R(q^2)\\&\quad =\left( \frac{R(q^2)}{R(q)^2}-\frac{R(q)^2}{R(q^2)}\right) \left( \frac{1}{R(q)R(q^2)^2}-q^2R(q)R(q^2)^2\right) \\&\qquad +\left( \frac{R(q)}{R(q^2)^3}+q^2\frac{R(q^2)^3}{R(q)}\right) \\&\quad =4q\frac{\chi (-q)}{\chi (-q^5)^5}\cdot \frac{\chi (-q^5)^5}{\chi (-q)}+\left( \frac{\chi (-q^5)^5}{\chi (-q)}-2q+4q^2\frac{\chi (-q)}{\chi (-q^5)^5}\right) \\&\quad =\frac{\chi (-q^5)^5}{\chi (-q)}+2q+4q^2\frac{\chi (-q)}{\chi (-q^5)^5}. \end{aligned}
$$\square$$

### Proof of (2.11)

\begin{aligned} \frac{\chi (-q^5)^5}{\chi (-q)}+q&=\frac{E(q^2)E(q^5)^5}{E(q)E(q^{10})^5}+q\\&=\frac{E(q^5)^2}{E(q^{10})^4}\left( \frac{E(q^2)E(q^5)^3}{E(q)E(q^{10})}+q\frac{E(q^{10})^4}{E(q^5)^2}\right) \\&=\frac{E(q^5)^2}{E(q^{10})^4}\cdot \frac{E(q^2)^4}{E(q)^2}. \end{aligned}
$$\square$$

### Proof of (2.12)

\begin{aligned} 1-4q\frac{\chi (-q)}{\chi (-q^5)^5}&=1-4q\frac{E(q)E(q^{10})^5}{E(q^2)E(q^5)^5}\\&=\frac{E(q^{10})^2}{E(q^5)^4}\left( \frac{E(q^5)^4}{E(q^{10})^2}-4q\frac{E(q)E(q^{10})^3}{E(q^2)E(q^5)}\right) \\&=\frac{E(q^{10})^2}{E(q^5)^4}\cdot \frac{E(q)^4}{E(q^2)^2}. \end{aligned}
$$\square$$

## 3 Proof of (1.1)

In this section, we effectively reproduce the proof of Baruah and Begum .

We have
\begin{aligned} \sum _{n\ge 0}p_D(n)q^n&=(-q;q)_\infty =\frac{E(q^2)}{E(q)}\\&=\frac{E(q^{25})^5}{E(q^5)^6}\left( \frac{1}{R(q^5)^4}+\frac{q}{R(q^5)^3}+\frac{2q^2}{R(q^5)^2}+\frac{3q^3}{R(q^5)}+5q^4\right. \\&\quad \left. -3q^5R(q^5)+2q^6R(q^5)^2-q^7R(q^5)^3+q^8R(q^5)^4\right) \\&\quad \times E(q^{50})\left( \frac{1}{R(q^{10})}-q^2-q^4R(q^{10})\right) . \end{aligned}
It follows that
\begin{aligned}&\sum _{n\ge 0}p_D(5n+1)q^n\\&\quad =\frac{E(q^5)^5E(q^{10})}{E(q)^6}\\&\qquad \times \left( \left( \frac{1}{R(q)^3R(q^2)}+q^2R(q)^3R(q^2)\right) -5q-2q\left( \frac{R(q^2)}{R(q)^2}-\frac{R(q)^2}{R(q^2)}\right) \right) \\&\quad =\frac{E(q^5)^5E(q^{10})}{E(q)^6}\\&\qquad \times \left( \left( \frac{\chi (-q^5)^5}{\chi (-q)}+2q+4q^2\frac{\chi (-q)}{\chi (-q^5)^5}\right) -5q-2q\cdot 4q\frac{\chi (-q)}{\chi (-q^5)^5}\right) \\&\quad =\frac{E(q^5)^5E(q^{10})}{E(q)^6}\left( \frac{\chi (-q^5)^5}{\chi (-q)}-3q-4q^2\frac{\chi (-q)}{\chi (-q^5)^5}\right) \\&\quad =\frac{E(q^5)^5E(q^{10})}{E(q)^6}\left( \frac{\chi (-q^5)^5}{\chi (-q)}+q\right) \left( 1-4q\frac{\chi (-q)}{\chi (-q^5)^5}\right) \\&\quad =\left( \frac{E(q^5)^5E(q^{10})}{E(q)^6}\right) \left( \frac{E(q^2)^4E(q^5)^2}{E(q)^2E(q^{10})^4}\right) \left( \frac{E(q)^4E(q^{10})^2}{E(q^2)^2E(q^5)^4}\right) \\&\quad =\frac{E(q^2)^2E(q^5)^3}{E(q)^4E(q^{10})}. \end{aligned}
$$\square$$

Note that (1.1) is the case $$\alpha =1$$ of (1.3).

## 4 The Modular Equation

We obtain the modular equation for $$\zeta$$.

Let $$\zeta (q^5)=Z$$.

### Theorem 4.1

\begin{aligned}&\zeta ^5-(205Z+4300Z^2+34000Z^3+120000Z^4+160000Z^5)\zeta ^4\nonumber \\&\quad -(215Z+4475Z^2+35000Z^3+122000Z^4+160000Z^5)\zeta ^3\nonumber \\&\quad -(85Z+1750Z^2+13525Z^3+46500Z^4+60000Z^5)\zeta ^2\nonumber \\&\quad -(15Z+305Z^2+2325Z^3+7875Z^4+10000Z^5)\zeta \nonumber \\&\quad -(Z+20Z^2+150Z^3+500Z^4+625Z^5)=0. \end{aligned}
(4.1)

### Proof

Let H be the huffing operator, given by
\begin{aligned} H\left( \sum _{n} a(n)q^n\right) =\sum _{n} a(5n)q^{5n}. \end{aligned}
We can show, using extremely lengthy but elementary calculations (see Sect. 9 “Appendix”), that
\begin{aligned} H(\zeta )&=41Z+860Z^2+6800Z^3+24000Z^4+32000Z^5, \end{aligned}
(4.2)
\begin{aligned} H(\zeta ^2)&=86Z+10195Z^2+366600Z^3+6534800Z^4+68384000Z^5\nonumber \\&\quad +450720000Z^6+1907200000Z^7+5056000000Z^8\nonumber \\&\quad +7680000000Z^9+5120000000Z^{10}, \end{aligned}
(4.3)
\begin{aligned} H(\zeta ^3)&=51Z+27495Z^2+2836265Z^3+128688900Z^4+3343692000Z^5\nonumber \\&\quad +56283680000Z^6 +656205600000Z^7+5502096000000Z^8\nonumber \\&\quad +33821312000000Z^9+153192960000000Z^{10}\nonumber \\&\quad +506956800000000Z^{11}+1195008000000000Z^{12}\nonumber \\&\quad +1904640000000000Z^{13}+1843200000000000Z^{14}\nonumber \\&\quad +819200000000000Z^{15}, \end{aligned}
(4.4)
\begin{aligned} H(\zeta ^4)&=12Z+32674Z^2+8579260Z^3+831492275Z^4+42958434000Z^5\nonumber \\&\quad +1396773180000Z^6+31314949600000Z^7+511802288800000Z^8\nonumber \\&\quad +6319880448000000Z^9+60349364480000000Z^{10}\nonumber \\&\quad +452174745600000000Z^{11}+2679038592000000000Z^{12}\nonumber \\&\quad +12574269440000000000Z^{13}+46561935360000000000Z^{14}\nonumber \\&\quad +134544588800000000000Z^{15}+297365504000000000000Z^{16}\nonumber \\&\quad +485949440000000000000Z^{17}+553779200000000000000Z^{18}\nonumber \\&\quad +393216000000000000000Z^{19}+131072000000000000000Z^{20} \end{aligned}
(4.5)
and
\begin{aligned} H(\zeta ^5)&=Z+21370Z^2+13932050Z^3+2684902125Z^4+251131688125Z^5\nonumber \\&\quad +14097638650000Z^6+532547945100000Z^7+14515766554000000Z^8\nonumber \\&\quad +298883447380000000Z^9+4797842366000000000Z^{10}\nonumber \\&\quad +61395781800000000000Z^{11}+636255683040000000000Z^{12}\nonumber \\&\quad +5398601306880000000000Z^{13}+37772239436800000000000Z^{14}\nonumber \\&\quad +21875584000000000000000Z^{15}+1049457704960000000000000Z^{16}\nonumber \\&\quad +4160657715200000000000000Z^{17}+13552680960000000000000000Z^{18}\nonumber \\&\quad +35909189632000000000000000Z^{19}+76195266560000000000000000Z^{20}\nonumber \\&\quad +126438604800000000000000000Z^{21}+158138368000000000000000000Z^{22}\nonumber \\&\quad +140247040000000000000000000Z^{23}+78643200000000000000000000Z^{24}\nonumber \\&\quad +20971520000000000000000000Z^{25}. \end{aligned}
(4.6)
Let $$\eta$$ be a fifth root of unity other than 1, and for $$i=0,\,1,\,2,\,3,\,4$$ define
\begin{aligned} \zeta _i=\zeta (\eta ^iq). \end{aligned}
Then the power sums $$\pi _1,\ \ldots \ ,\pi _5$$ of the $$\zeta _i$$ are given by
\begin{aligned} \pi _1&=\zeta _0+\ \cdots \ +\zeta _4=5H(\zeta ),\nonumber \\ \pi _2&=\zeta _0^2+\ \cdots \ +\zeta _4^2=5H(\zeta ^2),\nonumber \\ \cdots&\nonumber \\ \pi _5&=\zeta _0^5+\ \cdots \ +\zeta _4^5=5H(\zeta ^5). \end{aligned}
(4.7)
From (4.7) we obtain the symmetric functions $$\sigma _1,\ \ldots ,\sigma _5$$ of the $$\zeta _i$$,
\begin{aligned} \sigma _1&=\sum _{i}\zeta _i=\pi _1\nonumber \\&=205Z+4300Z^2+34000Z^3+120000Z^4+160000Z^5,\nonumber \\ \sigma _2&=\sum _{i<j}\zeta _i\zeta _j=\frac{1}{2}(\pi _1\sigma _1-\pi _2)\nonumber \\&=-215Z-4475Z^2-35000Z^3-122000Z^4-160000Z^5,\nonumber \\ \sigma _3&=\sum _{i<j<k}\zeta _i\zeta _j\zeta _k=\frac{1}{3}(\pi _1\sigma _2-\pi _2\sigma _1+\pi _3)\nonumber \\&=85Z+1750Z^2+13525Z^3+46000Z^4+60000Z^5,\nonumber \\ \sigma _4&=\sum _{i<j<k<l}\zeta _i\zeta _j\zeta _k\zeta _l=\frac{1}{4}(\pi _1\sigma _3-\pi _2\sigma _2+\pi _3\sigma _1-\pi _4)\nonumber \\&=-15Z-305Z^2-2325Z^3-7875Z^4-10000Z^5,\nonumber \\ \sigma _5&=\zeta _0\zeta _1\cdots \zeta _4 =\frac{1}{5}(\pi _1\sigma _4-\pi _2\sigma _3+\pi _3\sigma _2-\pi _4\sigma _1+\pi _5)\nonumber \\&=Z+20Z^2+150Z^3+500Z^4+625Z^5. \end{aligned}
Now, $$\zeta _0,\ \ldots \ ,\zeta _4$$ are the roots of
\begin{aligned}&(X-\zeta _0)(X-\zeta _1)(X-\zeta _2)(X-\zeta _3)(X-\zeta _4)\nonumber \\&\quad =X^5-\sigma _1X^4+\sigma _2X^3-\sigma _3X^2+\sigma _4X-\sigma _5=0, \end{aligned}
or,
\begin{aligned}&X^5-(205Z+4300Z^2+34000Z^3+120000Z^4+160000Z^5)X^4\nonumber \\&\quad -(215Z+4475Z^2+35000Z^3+122000Z^4+160000Z^5) X^3\nonumber \\&\quad -(85Z+1750Z^2+13525Z^3+46500Z^4+60000Z^5)X^2\nonumber \\&\quad -(15Z+305Z^2+2325Z^3+7875Z^4+10000Z^5)X\nonumber \\&\quad -(Z+20Z^2+150Z^3+500Z^4+625Z^5)=0. \end{aligned}
In particular, $$\zeta$$ is a root, and we obtain (4.1). $$\square$$

### Remark 4.2

It is truly remarkable, amazing even, that although $$\pi _1,\ \ldots \ ,\pi _5$$ are polynomials of degree up to 25, $$\sigma _1,\ \ldots \ ,\sigma _5$$ are of degree 5.

## 5 Some Important Recurrences and Generating Functions

Let U be the unitizing operator, given by
\begin{aligned} U\left( \sum _{n} a(n)q^n\right) =\sum _{n} a(5n)q^n. \end{aligned}
It follows from (4.1) that for $$i\ge 6$$, $$u_i=U(\zeta ^i)$$ satisfies the recurrence
\begin{aligned} u_i&=(205\zeta +4300\zeta ^2+34000\zeta ^3+120000\zeta ^4+160000\zeta ^5)u_{i-1}\nonumber \\&\quad +(215\zeta +4475\zeta ^2+35000\zeta ^3+122000\zeta ^4+160000\zeta ^5)u_{i-2}\nonumber \\&\quad +(85\zeta +1750\zeta ^2+13525\zeta ^3+46500\zeta ^4+60000\zeta ^5)u_{i-3}\nonumber \\&\quad +(15\zeta +305\zeta ^2+2325\zeta ^3+7875\zeta ^4+10000\zeta ^5)u_{i-4}\nonumber \\&\quad +(\zeta +20\zeta ^2+150\zeta ^3+500\zeta ^4+625\zeta ^5)u_{i-5}. \end{aligned}
(5.1)
The recurrence (5.1), together with the five initial values $$u_1,\,u_2,\ \ldots \ ,u_5$$, which can be read off from (4.2)–(4.6) by replacing Z by $$\zeta$$, gives
\begin{aligned} \sum _{i\ge 1}u_ix^i=\frac{N}{D}, \end{aligned}
(5.2)
where
\begin{aligned} N&=(41\zeta +860\zeta ^2+6800\zeta ^3+24000\zeta ^4+32000\zeta ^5)x\nonumber \\&\quad +(86\zeta +1790\zeta ^2+14000\zeta ^3+48800\zeta ^4+64000\zeta ^5)x^2\nonumber \\&\quad +(51\zeta +1050\zeta ^2+8115\zeta ^3+27900\zeta ^4+36000\zeta ^5)x^3\nonumber \\&\quad +(12\zeta +244\zeta ^2+1869\zeta ^3+6300\zeta ^4+8000\zeta ^5)x^4\nonumber \\&\quad +(\zeta +20\zeta ^2+150\zeta ^3+500\zeta ^4+625\zeta ^5)x^5 \end{aligned}
(5.3)
and
\begin{aligned} D&=1-(205\zeta +4300\zeta ^2+34000\zeta ^3+120000\zeta ^4+160000\zeta ^5)x\nonumber \\&\quad -(215\zeta +4475\zeta ^2+35000\zeta ^3+122000\zeta ^4+160000\zeta ^5)x^2\nonumber \\&\quad -(85\zeta +1750\zeta ^2+13525\zeta ^3+46500\zeta ^4+60000\zeta ^5)x^3\nonumber \\&\quad -(15\zeta +305\zeta ^2+2325\zeta ^3+7875\zeta ^4+10000\zeta ^5)x^4\nonumber \\&\quad -(\zeta +20\zeta ^2+150\zeta ^3+500\zeta ^4+625\zeta ^5)x^5. \end{aligned}
(5.4)
From (5.2)–(5.4), we deduce that for $$i\ge 1$$,
\begin{aligned} U(\zeta ^i)=u_i=\sum _{j=1}^{5i}\mu _{i,j}\zeta ^j, \end{aligned}
where the $$\mu _{i,j}$$ are given by
\begin{aligned} \sum _{i=1}^\infty \sum _{j=1}^{5i}\mu _{i,j}x^iy^j=\frac{N'}{D'}, \end{aligned}
where
\begin{aligned} N'&=(41y+860y^2+6800y^3+24000y^4+32000y^5)x\nonumber \\&\quad +(86y+1790y^2+14000y^3+48800y^4+64000y^5)x^2\nonumber \\&\quad +(51y+1050y^2+8115y^3+27900y^4+36000y^5)x^3\nonumber \\&\quad +(12y+244y^2+1869y^3+6300y^4+8000y^5)x^4\nonumber \\&\quad +(y+20y^2+150y^3+500y^4+625y^5)x^5 \end{aligned}
and
\begin{aligned} D'&=1-(205y+4300y^2+34000y^3+120000y^4+160000y^5)x\nonumber \\&\quad -(215y+4475y^2+35000y^3+122000y^4+160000y^5)x^2\nonumber \\&\quad -(85y+1750y^2+13525y^3+46500y^4+60000y^5)x^3\nonumber \\&\quad -(15y+305y^2+2325y^3+7875y^4+10000y^5)x^4\nonumber \\&\quad -(y+20y^2+150y^3+500y^4+625y^5)x^5. \end{aligned}
(5.5)
More importantly, if we multiply (4.1) by $$\gamma$$ and apply the operator U, we see that $$v_i=U(\gamma \zeta ^{i-1})$$ satisfy the recurrence (5.1) (with v for u).
Also, using the same sort of calculations as in Sect. 4 (see Sect. 9 “Appendix”),
\begin{aligned} v_1&=U(\gamma )=\delta (1+160\zeta +2800\zeta ^2+16000\zeta ^3+32000\zeta ^4), \end{aligned}
(5.6)
\begin{aligned} v_2&=U(\gamma \zeta )=\delta (385\zeta +40100\zeta ^2+1312800\zeta ^3+20912000\zeta ^4+189920000\zeta ^5\nonumber \\&\quad +1043200000\zeta ^6+3456000000\zeta ^7+6400000000\zeta ^8+5120000000\zeta ^9), \end{aligned}
(5.7)
\begin{aligned} v_3&=U(\gamma \zeta ^2)=\delta (290\zeta +119015\zeta ^2+11235600\zeta ^3+476348000\zeta ^4\nonumber \\&\quad +11537760000\zeta ^5+179434400000\zeta ^6+1908992000000\zeta ^7\nonumber \\&\quad +14377472000000\zeta ^8+77783040000000\zeta ^9+301644800000000\zeta ^{10}\nonumber \\&\quad +821248000000000\zeta ^{11}+1495040000000000\zeta ^{12}\nonumber \\&\quad +1638400000000000\zeta ^{13} +819200000000000\zeta ^{14}), \end{aligned}
(5.8)
\begin{aligned} v_4&=U(\gamma \zeta ^3)=\delta (99\zeta +157795\zeta ^2+36522125\zeta ^3+3308569500\zeta ^4\nonumber \\&\quad +161943150000\zeta ^5+4995603800000\zeta ^6+105933588800000\zeta ^7\nonumber \\&\quad +1628976896000000\zeta ^8+18797435520000000\zeta ^9\nonumber \\&\quad +166360908800000000\zeta ^{10}+1143762304000000000\zeta ^{11}\nonumber \\&\quad +6142300160000000000\zeta ^{12}+25729781760000000000\zeta ^{13}\nonumber \\&\quad +83330457600000000000\zeta ^{14}+204857344000000000000\zeta ^{15}\nonumber \\&\quad +370032640000000000000\zeta ^{16}+463667200000000000000\zeta ^{17}\nonumber \\&\quad +360448000000000000000\zeta ^{18}+131072000000000000000\zeta ^{19}) \end{aligned}
(5.9)
and
\begin{aligned} v_5&=U(\gamma \zeta ^4)=\delta (16\zeta +118090\zeta ^2+63835100\zeta ^3+11315760375\zeta ^4\nonumber \\&\quad +1002222145000\zeta ^5+53778439200000\zeta ^6+1946392973200000\zeta ^7\nonumber \\&\quad +50789296612000000\zeta ^8+998696483520000000\zeta ^9\nonumber \\&\quad +15256932894400000000\zeta ^{10}+185007570368000000000\zeta ^{11}\nonumber \\&\quad +1807671489280000000000\zeta ^{12}+14376293539840000000000\zeta ^{13}\nonumber \\&\quad +93630345523200000000000\zeta ^{14}+500636522496000000000000\zeta ^{15}\nonumber \\&\quad +2195582095360000000000000\zeta ^{16}+7860788428800000000000000\zeta ^{17}\nonumber \\&\quad +22768123904000000000000000\zeta ^{18}+52564656128000000000000000\zeta ^{19}\nonumber \\&\quad +94522572800000000000000000\zeta ^{20}+127664128000000000000000000\zeta ^{21}\nonumber \\&\quad +121896960000000000000000000\zeta ^{22}+73400320000000000000000000\zeta ^{23}\nonumber \\&\quad +209715200000000000000000000\zeta ^{24}). \end{aligned}
(5.10)
It follows that for $$i\ge 1$$,
\begin{aligned} U(\gamma \zeta ^{i-1})=\delta \sum _{j=1}^{5i}\alpha _{i,j}\zeta ^{j-1}, \end{aligned}
(5.11)
where
\begin{aligned} \sum _{i=1}^\infty \sum _{j=1}^{5i}\alpha _{i,j}x^iy^j=\frac{N_\alpha }{D'}, \end{aligned}
and where
\begin{aligned} N_\alpha&=(y+160y^2+2800y^3+16000y^4+32000y^5)x\nonumber \\&\quad +(180y^2+3000y^3+16800y^4 +32000y^5)x^2\nonumber \\&\quad +(75y^2+1215y^3+6600y^4+12000y^5)x^3\nonumber \\&\quad +(14y^2+220y^3+1150y^4+2000y^5)x^4\nonumber \\&\quad +(y^2+15y^3+75y^4+125y^5)x^5 \end{aligned}
and $$D'$$ is given in (5.5).

Similarly, if we multiply (4.1) by $$q^{-1}\delta$$ and apply the operator U, we see that $$w_i=U(q^{-1}\delta \zeta ^{i-1})$$ satisfy (5.1) (with w for u).

Also,
\begin{aligned} w_1&=U(q^{-1}\delta )=\gamma (5+660\zeta +14400\zeta ^2+120000\zeta ^3+448000\zeta ^4+640000\zeta ^5), \end{aligned}
(5.12)
\begin{aligned} w_2&=U(q^{-1}\delta \zeta )=\gamma (1+1705\zeta +171700\zeta ^2+6083200\zeta ^3+110016000\zeta ^4\nonumber \\&\quad +178080000\zeta ^5+797120000\zeta ^6+34688000000\zeta ^7+94720000000\zeta ^8\nonumber \\&\quad +148480000000\zeta ^9+1024000000000\zeta ^{10}), \end{aligned}
(5.13)
\begin{aligned} w_3&=U(q^{-1}\delta \zeta ^2)=\gamma (1545\zeta +523885\zeta ^2+48836000\zeta ^3+2157580000\zeta ^4\nonumber \\&\quad +55972480000\zeta ^5+950485600000\zeta ^6+11233328000000\zeta ^7\nonumber \\&\quad +95713408000000\zeta ^8 +598718720000000\zeta ^9+2762265600000000\zeta ^{10}\nonumber \\&\quad +9317888000000000\zeta ^{11} +22405120000000000\zeta ^{12}\nonumber \\&\quad +36454400000000000\zeta ^{13}+36044800000000000\zeta ^{14}\nonumber \\&\quad +16384000000000000\zeta ^{15}), \end{aligned}
(5.14)
\begin{aligned} w_4&=U(q^{-1}\delta \zeta ^3)=\gamma (686\zeta +753625\zeta ^2+161075075\zeta ^3+14497246500\zeta ^4\nonumber \\&\quad +727863490000\zeta ^5+23458401400000\zeta ^6+526452595200000\zeta ^7\nonumber \\&\quad +8658501792000000\zeta ^8 +107918950400000000\zeta ^9\nonumber \\&\quad +1042082905600000000\zeta ^{10}+7904596864000000000\zeta ^{11} \nonumber \\&\quad +47450048000000000000\zeta ^{12}+225774243840000000000\zeta ^{13} \nonumber \\&\quad +847926476800000000000\zeta ^{14}+2486042624000000000000\zeta ^{15}\nonumber \\&\quad +5577277440000000000000\zeta ^{16}+9255321600000000000000\zeta ^{17} \nonumber \\&\quad +107151360000000000000000\zeta ^{18}+7733248000000000000000\zeta ^{19}\nonumber \\&\quad +2621440000000000000000\zeta ^{20}) \end{aligned}
(5.15)
and
\begin{aligned} w_5&=U(q^{-1}\delta \zeta ^4) =\gamma (163\zeta +630970\zeta ^2+295013300\zeta ^3\nonumber \\&\quad +50030923625\zeta ^4+4413689785000\zeta ^5+240963519250000\zeta ^6\nonumber \\&\quad +8992052284600000\zeta ^7+244243690752000000\zeta ^8+5037514186320000000\zeta ^9\nonumber \\&\quad +81262009334400000000\zeta ^{10}+1047144506208000000000\zeta ^{11}\nonumber \\&\quad +10942698476160000000000\zeta ^{12}+93715045227520000000000\zeta ^{13}\nonumber \\&\quad +662259232256000000000000\zeta ^{14}+387577451008000000000000\zeta ^{15}\nonumber \\&\quad +18796453150720000000000000\zeta ^{16}+75357109452800000000000000\zeta ^{17}\nonumber \\&\quad +248290942976000000000000000\zeta ^{18}+665623035904000000000000000\zeta ^{19}\nonumber \\&\quad +1429384069120000000000000000\zeta ^{20}+2401107968000000000000000000\zeta ^{21}\nonumber \\&\quad +3040870400000000000000000000\zeta ^{22}+2731540480000000000000000000\zeta ^{23}\nonumber \\&\quad +1551892480000000000000000000\zeta ^{24}+419430400000000000000000000\zeta ^{25}). \end{aligned}
(5.16)
It follows that for $$i\ge 1$$,
\begin{aligned} U(q^{-1}\delta \zeta ^{i-1})=\gamma \sum _{j=1}^{5i+1}\beta _{i,j}\zeta ^{j-1}, \end{aligned}
(5.17)
where
\begin{aligned} \sum _{i=1}^\infty \sum _{j=1}^{5i+1}\beta _{i,j}x^iy^j=\frac{N_\beta }{D'}, \end{aligned}
and where
\begin{aligned} N_\beta&=(5y+660y^2+14400y^3+120000y^4+448000y^5+640000y^6)x\nonumber \\&\quad +(y+680y^2+14900y^3+123200y^4+456000y^5+640000y^6)x^2\nonumber \\&\quad +(265y^2+5785y^3+47500y^4+174000y^5+240000y^6)x^3\nonumber \\&\quad +(46y^2+1000y^3+8150y^4+29500y^5+40000y^6)x^4\nonumber \\&\quad +(3y^2+65y^3+525y^4+1875y^5+2500y^6)x^5 \end{aligned}
and $$D'$$ is given in (5.5).

## 6 Proof of the First Part of Theorem 1.1

The first part of Theorem 1.1 follows by a simple induction from (1.1), (5.11) and (5.17), as we now demonstrate.

We know that (1.3) is true for $$\alpha =1$$. Suppose (1.3) is true for some $$\alpha \ge 1$$. Then
\begin{aligned} \sum _{n\ge 0}p_D\left( 5^{2\alpha -1}n+\frac{5^{2\alpha }-1}{24}\right) q^n=\gamma \sum _{i=1}^{(5^{2\alpha }-1)/24}x_{2\alpha -1,i}\zeta ^{i-1}. \end{aligned}
(6.1)
If we apply the operator U to (6.1) and use (5.11), we find
\begin{aligned}&\sum _{n\ge 0}p_D\left( 5^{2\alpha -1}(5n)+\frac{5^{2\alpha }-1}{24}\right) q^n\\&\quad =\sum _{i=1}^{(5^{2\alpha }-1)/24}x_{2\alpha -1,i}U(\gamma \zeta ^{i-1})\\&\quad =\sum _{i=1}^{(5^{2\alpha }-1)/24}x_{2\alpha -1,i}\delta \sum _{j=1}^{5i}\alpha _{i,j}\zeta ^{j-1}\\&\quad =\delta \sum _{j=1}^{(5^{2\alpha +1}-5)/24}\left( \sum _{i=1}^{(5^{2\alpha }-1)/24}x_{2\alpha -1,i}\alpha _{i,j}\right) \zeta ^{j-1}\\&\quad =\delta \sum _{j=1}^{(5^{2\alpha +1}-5)/24}x_{2\alpha ,j}\zeta ^{j-1}, \end{aligned}
or,
\begin{aligned} \sum _{n\ge 0}p_D\left( 5^{2\alpha }n+\frac{5^{2\alpha }-1}{24}\right) q^n=\delta \sum _{j=1}^{(5^{2\alpha +1}-5)/24}x_{2\alpha ,j}\zeta ^{j-1}, \end{aligned}
which is (1.4).
Now suppose (1.4) is true for some $$\alpha \ge 1$$. Then
\begin{aligned} \sum _{n\ge 0}p_D\left( 5^{2\alpha }n+\frac{5^{2\alpha }-1}{24}\right) q^{n-1}=q^{-1}\delta \sum _{i=1}^{(5^{2\alpha +1}-5)/24}x_{2\alpha ,i}\zeta ^{i-1}. \end{aligned}
(6.2)
If we apply the operator U to (6.2) and use (5.17), we find
\begin{aligned}&\sum _{n\ge 0}p_D\left( 5^{2\alpha }(5n+1)+\frac{5^{2\alpha }-1}{24}\right) q^n\\&\quad =\sum _{i=1}^{(5^{2\alpha +1}-5)/24}x_{2\alpha ,i}U(q^{-1}\delta \zeta ^{i-1})\\&\quad =\sum _{i=1}^{(5^{2\alpha +1}-5)/24}x_{2\alpha ,i}\gamma \sum _{j=1}^{5i+1}\beta _{i,j}\zeta ^{j-1}\\&\quad =\gamma \sum _{j=1}^{(5^{2\alpha +2}-1)/24}\left( \sum _{i=1}^{(5^{2\alpha +1}-5)/24}x_{2\alpha ,i}\beta _{i,j}\right) \zeta ^{j-1}\\&\quad =\gamma \sum _{j=1}^{(5^{2\alpha +2}-1)/24}x_{2\alpha +1,j}\zeta ^{j-1}, \end{aligned}
or,
\begin{aligned} \sum _{n\ge 0}p_D\left( 5^{2\alpha +1}n+\frac{5^{2\alpha +2}-1}{24}\right) q^n=\gamma \sum _{j=1}^{(5^{2\alpha +2}-1)/24}x_{2\alpha +1,j}\zeta ^{j-1}, \end{aligned}
which is (1.3) with $$\alpha +1$$ for $$\alpha$$. $$\square$$

## 7 Proof of the Second Part of Theorem 1.1

Let $$\nu (n)$$ denote the (highest) power of 5 that divides n.

We prove the following theorem.

### Theorem 7.1

\begin{aligned} \nu (\alpha _{i,j})&\ge \left\lfloor \frac{5j-i-1}{6}} \right\rfloor , \end{aligned
(7.1)
\begin{aligned} \nu (\beta _{i,j})&\ge \left\lfloor \frac{5j-i-1}{6}}\right\rfloor . \end{aligned
(7.2)

### Proof

Let $$\lambda _{i,j}=\nu (\alpha _{i,j})$$, $$\rho _{i,j}=\left\lfloor \frac{5j-i-1}{6}}\right\rfloor$$.

Observe that from the recurrence (5.1), for $$i,j\ge 6$$,
\begin{aligned} \lambda _{i,j}&\ge \min ( \lambda _{i-1,j-1}+1,\lambda _{i-1,j-2}+2,\lambda _{i-1,j-3}+3,\lambda _{i-1,j-4}+4,\nonumber \\&\quad \lambda _{i-1,j-5}+4,\lambda _{i-2,j-1}+1,\lambda _{i-2,j-2}+2,\lambda _{i-2,j-3}+4,\nonumber \\&\quad \lambda _{i-2,j-4}+3,\lambda _{i-2,j-5}+4,\lambda _{i-3,j-1}+1,\lambda _{i-3,j-2}+3,\nonumber \\&\quad \lambda _{i-3,j-3}+2,\lambda _{i-3,j-4}+3,\lambda _{i-3,j-5}+4,\lambda _{i-4,j-1}+1,\nonumber \\&\quad \lambda _{i-4,j-2}+1,\lambda _{i-4,j-3}+2,\lambda _{i-4,j-4}+3,\lambda _{i-4,j-5}+4,\nonumber \\&\quad \lambda _{i-5,j-1}+0,\lambda _{i-5,j-2}+1,\lambda _{i-5,j-3}+2,\lambda _{i-5,j-4}+3,\nonumber \\&\quad \lambda _{i-5,j-5}+4). \end{aligned}
(7.3)
On the other hand,
\begin{aligned} \rho _{i,j}&=\min (\rho _{i-1,j-1}+1,\rho _{i-1,j-2}+2,\rho _{i-1,j-3}+3,\rho _{i-1,j-4}+4,\nonumber \\&\quad \rho _{i-1,j-5}+4,\rho _{i-2,j-1}+1,\rho _{i-2,j-2}+2,\rho _{i-2,j-3}+4,\nonumber \\&\quad \rho _{i-2,j-4}+3,\rho _{i-2,j-5}+4,\rho _{i-3,j-1}+1,\rho _{i-3,j-2}+3,\nonumber \\&\quad \rho _{i-3,j-3}+2,\rho _{i-3,j-4}+3,\rho _{i-3,j-5}+4,\rho _{i-4,j-1}+1,\nonumber \\&\quad \rho _{i-4,j-2}+1,\rho _{i-4,j-3}+2,\rho _{i-4,j-4}+3,\rho _{i-4,j-5}+4,\nonumber \\&\quad \rho _{i-5,j-1}+0,\rho _{i-5,j-2}+1,\rho _{i-5,j-3}+2,\rho _{i-5,j-4}+3,\nonumber \\&\quad \rho _{i-5,j-5}+4). \end{aligned}
(7.4)
The right side of (7.4)
\begin{aligned}&=\min \left( \left\lfloor \frac{5j-i+1}{6}}\right\rfloor ,\left\lfloor \frac{5j-i+2}{6}}\right\rfloor ,\left\lfloor \frac{5u-i+3}{6}}\right\rfloor ,\left\lfloor \frac{5j-i+4}{6}}\right\rfloor ,\left\lfloor \frac{5j-i-1}{6}}\right\rfloor ,\right. \\&\left. \quad \left\lfloor \frac{5j-i+2}{6}}\right\rfloor ,\left\lfloor \frac{5j-i+3}{6}}\right\rfloor ,\left\lfloor \frac{5j-i+10}{6}}\right\rfloor ,\left\lfloor \frac{5j-i-1}{6}}\right\rfloor ,\left\lfloor \frac{5j-i}{6}}\right\rfloor ,\right. \\&\quad \left. \left\lfloor \frac{5j-i+3}{6}}\right\rfloor ,\left\lfloor \frac{5j-i+10}{6}}\right\rfloor ,\left\lfloor \frac{5j-i-1}{6}}\right\rfloor ,\left\lfloor \frac{5j-i}{6}}\right\rfloor ,\left\lfloor \frac{5j-i+1}{6}}\right\rfloor ,\right. \\&\quad \left. \left\lfloor \frac{5j-i+4}{6}}\right\rfloor ,\left\lfloor \frac{5j-i-1}{6}}\right\rfloor ,\left\lfloor \frac{5j-i}{6}}\right\rfloor ,\left\lfloor \frac{5j-i+1}{6}}\right\rfloor ,\left\lfloor \frac{5j-i+3}{6}}\right\rfloor ,\right. \\&\quad \left. \left\lfloor \frac{5j-i-1}{6}}\right\rfloor ,\left\lfloor \frac{5j-i}{6}}\right\rfloor ,\left\lfloor \frac{5j-i+1}{6}}\right\rfloor ,\left\lfloor \frac{5j-i+2}{6}}\right\rfloor ,\left\lfloor \frac{5j-i+3}{6}}\right\rfloor \right) \\&=\left\lfloor \frac{5j-i-1}{6}}\right\rfloor =\rho _{i,j}. \end{aligned
The values of $$\lambda _{i,j}-\rho _{i,j}$$ for $$1\le i\le 5$$ and for $$1\le j\le 5$$ are given in the following tables. Note that they are all non-negative. (We use $$\bullet$$ for $$\infty$$.)We see that (7.1) follows from (7.3)–(7.6) by induction.

### Theorem 7.2

For$$\alpha \ge 0$$,
\begin{aligned} \nu (x_{2\alpha +1,1}&)\ge \alpha , \ \ \nu (x_{2\alpha +1,i})\ge \alpha +\left\lfloor \frac{5i-8}{6}}\right\rfloor \ \text {for}\ i\ge 2,\\ \nu (x_{2\alpha +2,i})&\ge \alpha +\left\lfloor \frac{5i-2}{6}}\right\rfloor . \end{aligned

### Proof

If we replace $$\nu (A)$$ by
\begin{aligned} \left( \left\lfloor \frac{5j-i-1}{6}}\right\rfloor \right) _{i,j\ge 1} \end{aligned
and $$\nu (B)$$ by
\begin{aligned} \left( \left\lfloor \frac{5j-i-1}{6}}\right\rfloor \right) _{i,j\ge 1} \end{aligned
with the exception $$\nu (b_{1,1})=1$$, and we start with $$\nu ({{\mathbf {x}}}_1)=(0,\,\infty ,\ \ldots )$$, the results follow by induction. $$\square$$

This completes the proof of Theorem 1.1. $$\square$$

## 8 Calculations

We find that
\begin{aligned} {{\mathbf {x}}}_1&=(1,\,0,\,\ldots \ ),\\ {{\mathbf {x}}}_2&=(1,\,160,\,2800,\,16000,\,32000,\,0,\ \ldots \ ),\\ {{\mathbf {x}}}_3&=(5*33,\,2^2*5*1039573,\,2^4*5^2*84358511,\,2^6*5^3*1519417629,\\&\ \ 2^8*5^3*57468885219,\,2^{10}*5^4*239126250621,\,2^{20}*5^6*493702983,\,\\&\ \ 2^{16}*5^7*57851635449,\,2^{17}*5^8*155363323153,\ 2^{22}*5^8*99443868167,\\&\ \ 2^{20}*5^9*1277863945093,\,2^{23}*5^{11}*82117001559,\,2^{24}*5^{12}*85675198911,\\&\ \ 2^{29}*5^{14}*916288433,\,2^{29}*5^{13}*32357578059,\,2^{33}*5^{14}*2366343709,\\&\ \ 2^{36}*5^{16}*57370733,\ 2^{37}*5^{17}*22998577,\,2^{36}*5^{18}*30309607,\\&\ \ 2^{38}*5^{18}*20313321,\,2^{40}*5^{19}*2181069,\,2^{43}*5^{21}*18319,\,\\&\ \ 2^{48}*5^{23}*29,\,2^{46}*5^{22}*521,\,2^{49}*5^{22}*37,\,2^{50}*5^{23},\,0,\,\ \ldots \ ), \end{aligned}
in agreement with Baruah and Begum and
\begin{aligned}&\nu ({{\mathbf {x}}}_1)=(0,\,\infty ,\ \ldots \ ),\\&\nu ({{\mathbf {x}}}_2)=(0,\,1,\,2,\,3,\,3,\,\infty ,\ \ldots \ ),\\&\nu ({{\mathbf {x}}}_3)=(1,\,1,\,2,\,3,\,3,\,4,\,6,\,7,\,8,\,8,\,9,\,11,\,12,\,14,\,13,\,14,\,16,\,17,\,18,\,18,\,19,\,21,\\&\qquad \qquad 23,\,22,\,22,\,23,\,\infty ,\ \ldots \ ). \end{aligned}

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