Advertisement

Annals of Combinatorics

, Volume 23, Issue 3–4, pp 659–682 | Cite as

Partitions into Distinct Parts Modulo Powers of 5

  • Shane Chern
  • Michael D. HirschhornEmail author
Article
  • 62 Downloads

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 [1] 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 [6] and independently to Gordon and Hughes [3]. See also Lovejoy [5].

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 [1].

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

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

Proof

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

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)
$$\begin{aligned}&=\min \left( \left\lfloor {\displaystyle \frac{5j-i+1}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i+2}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5u-i+3}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i+4}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i-1}{6}}\right\rfloor ,\right. \\&\left. \quad \left\lfloor {\displaystyle \frac{5j-i+2}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i+3}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i+10}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i-1}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i}{6}}\right\rfloor ,\right. \\&\quad \left. \left\lfloor {\displaystyle \frac{5j-i+3}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i+10}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i-1}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i+1}{6}}\right\rfloor ,\right. \\&\quad \left. \left\lfloor {\displaystyle \frac{5j-i+4}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i-1}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i+1}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i+3}{6}}\right\rfloor ,\right. \\&\quad \left. \left\lfloor {\displaystyle \frac{5j-i-1}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i+1}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i+2}{6}}\right\rfloor ,\left\lfloor {\displaystyle \frac{5j-i+3}{6}}\right\rfloor \right) \\&=\left\lfloor {\displaystyle \frac{5j-i-1}{6}}\right\rfloor =\rho _{i,j}. \end{aligned}$$
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.
The proof of (7.2) is essentially the same as that of (7.1). The boundary values are given by the following tables.\(\square \)

Theorem 7.2

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

Proof

If we replace \(\nu (A)\) by
$$\begin{aligned} \left( \left\lfloor {\displaystyle \frac{5j-i-1}{6}}\right\rfloor \right) _{i,j\ge 1} \end{aligned}$$
and \(\nu (B)\) by
$$\begin{aligned} \left( \left\lfloor {\displaystyle \frac{5j-i-1}{6}}\right\rfloor \right) _{i,j\ge 1} \end{aligned}$$
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
$$\begin{aligned}&\nu ({{\mathbf {x}}}_1)=(0,\,\infty ,\ \ldots \ ),\\&\nu ({{\mathbf {x}}}_2)=(0,\,1,\,2,\,3,\,3,\,\infty ,\ \ldots \ ),\\&\nu ({{\mathbf {x}}}_3)=(1,\,1,\,2,\,3,\,3,\,4,\,6,\,7,\,8,\,8,\,9,\,11,\,12,\,14,\,13,\,14,\,16,\,17,\,18,\,18,\,19,\,21,\\&\qquad \qquad 23,\,22,\,22,\,23,\,\infty ,\ \ldots \ ). \end{aligned}$$

Notes

References

  1. 1.
    Baruah, N.D., Begum, N.M.: Exact generating functions for the number of partitions into distinct parts. Int. J. Number Theory 14(7), 1995–2011 (2018)MathSciNetCrossRefGoogle Scholar
  2. 2.
    Chern, S., Tang, D.: Representations involving the Rogers–Ramanujan continued fraction and their applications (submitted)Google Scholar
  3. 3.
    Gordon, B., Hughes, K.: Ramanujan congruences for \(q(n)\). In: Analytic Number Theory (Philadelphia, Pa., 1980), Lecture Notes in Mathematics, Vol. 899, pp. 333–359. Springer, Berlin-New York (1981)Google Scholar
  4. 4.
    Hirschhorn, M.D.: The Power of \(q\). Developments in Mathematics. Vol. 49. Springer, Cham (2017)Google Scholar
  5. 5.
    Lovejoy, J.: The number of partitions into distinct parts modulo powers of \(5\). Bull. London Math. Soc. 35(1), 41–46 (2003)MathSciNetCrossRefGoogle Scholar
  6. 6.
    Rødseth, Ø.: Congruence properties of the partition functions \(q(n)\) and \(q_0(n)\). Arbok Univ. Bergen Mat.-Natur. Ser. 1969(13), 3–27 (1970)zbMATHGoogle Scholar

Copyright information

© Springer Nature Switzerland AG 2019

Authors and Affiliations

  1. 1.Department of MathematicsThe Pennsylvania State UniversityState CollegeUSA
  2. 2.School of Mathematics and StatisticsUniversity of New South WalesSydneyAustralia

Personalised recommendations