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ARITHMETIC PROPERTIES SATISFIED BY A RECENT INTEGER PARTITION FUNCTION OF DOMBOS

Part of: Additive number theory; partitions Combinatorics

Published online by Cambridge University Press:  03 November 2026

ROBSON DA SILVA Open the ORCID record for ROBSON DA SILVA [Opens in a new window]  and
JAMES A. SELLERS Open the ORCID record for JAMES A. SELLERS [Opens in a new window]
ROBSON DA SILVA
Affiliation:
Universidade Federal de São Paulo , Av. Cesare M. G. Lattes, 1201, São José dos Campos, SP, 12247–014, Brazil e-mail: silva.robson@unifesp.br
JAMES A. SELLERS*
Affiliation:
Department of Mathematics and Statistics, University of Minnesota Duluth , Duluth, MN 55812, USA
*
e-mail: jsellers@d.umn.edu
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Abstract

The set of integer partitions of n, wherein the parts are either divisible by 4 or congruent to $\pm 1 \pmod {6}$, arose in a natural way in work of Dombos [‘A new grounded partition identity of type $D_4^{(3)}$’, SIGMA 22 (2026), Article no. 042]. Denote the function that counts the number of such partitions of n by $\mathit {dp}(n)$. Using elementary generating function manipulations and classical q-series results, we prove several congruences satisfied by $\mathit {dp}(n)$. As an example, we prove that, for all $\alpha \geq 1$ and all $n \geq 0$,

$$ \begin{align*} {\it dp} \bigg( 3^{2\alpha + 1}n + \frac{7 \cdot 9^\alpha + 1}{4} \bigg) \equiv 0 \pmod{3}. \end{align*} $$

Keywords

partitionsgrounded partitionscongruencesgenerating functionsq-seriesdissections

MSC classification

Primary: 11P83: Partitions; congruences and congruential restrictions
Secondary: 05A17: Partitions of integers

Information

Type
Research Article
Information
Bulletin of the Australian Mathematical Society , First View , pp. 1 - 9
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Copyright
© The Author(s), 2026. Published by Cambridge University Press on behalf of Australian Mathematical Publishing Association Inc.

1 Introduction

The arithmetic properties of partition functions have been a central theme in the theory of partitions since the pioneering work of Ramanujan [Reference Ramanujan6]. His celebrated congruences for the ordinary partition function (see below) initiated a vast literature devoted to the congruence properties of both the ordinary partition function and numerous families of restricted partition functions. More recently, many new partition functions arising from Rogers–Ramanujan-type identities and related q-series have also been shown to satisfy remarkable arithmetic properties.

We recall that a partition of a positive integer n is a nonincreasing sequence of positive integers $\lambda _1, \lambda _2, \ldots , \lambda _r$ lamda 1 comma lamda 2 comma ellipsis comma lamda Subscript r Baseline such that $n=\lambda _1+ \lambda _2+ \cdots + \lambda _r$ n equals lamda 1 plus lamda 2 plus midline horizontal ellipsis plus lamda Subscript r . Each $\lambda _i$ lamda Subscript i , $1\leq i\leq r$ 1 less than or equals i less than or equals r , is called a part of the partition. We denote the number of partitions of n by $p(n)$ p left parenthesis n right parenthesis , the ordinary partition function.

In the late 1910s, Ramanujan [Reference Ramanujan6] proved the following congruences satisfied by $p(n)$ p left parenthesis n right parenthesis for the moduli $5, 7$ 5 comma 7 and $11$ 11 using generating function manipulations: for all $n\geq 0$ n greater than or equals 0 ,

$$ \begin{align*} p(5n+4) \equiv 0 \pmod{5}, \quad p(7n+5) \equiv 0 \pmod{7} \quad\text{and}\quad p(11n+6) \equiv 0 \pmod{11}. \end{align*} $$ p left parenthesis 5 n plus 4 right parenthesis identical to 0 left parenthesis mod 5 right parenthesis comma p left parenthesis 7 n plus 5 right parenthesis identical to 0 left parenthesis mod 7 right parenthesis and p left parenthesis 11 n plus 6 right parenthesis identical to 0 left parenthesis mod 11 right parenthesis period

These congruences, and others that have since been proven for numerous restricted partition functions, serve as the motivation for this work.

Recently, Dombos [Reference Dombos4] introduced a new Rogers–Ramanujan-type identity involving grounded partitions. One family of partitions arising naturally from this identity consists of partitions whose parts are either divisible by 4 or congruent to $\pm 1 \pmod {6}$ plus or minus 1 left parenthesis mod 6 right parenthesis . We refer the reader to [Reference Dombos4, Theorem 1.6] for more details. Let $\mathit {dp}(n)$ italic dp left parenthesis n right parenthesis denote the number of such partitions of n. In light of this definition, the generating function for $\mathit {dp}(n)$ italic dp left parenthesis n right parenthesis is given by

(1.1) $$ \begin{align} G(q):=\sum_{n=0}^{\infty} \mathit{dp}(n)q^n = \frac{1}{(q;q^6)_\infty (q^5;q^6)_\infty (q^4;q^4)_\infty}, \end{align} $$ upper G left parenthesis q right parenthesis colon equals sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis n right parenthesis q Superscript n Baseline equals StartFraction 1 Over left parenthesis q semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline left parenthesis q Superscript 5 Baseline semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline left parenthesis q Superscript 4 Baseline semicolon q Superscript 4 Baseline right parenthesis Subscript infinity Baseline EndFraction comma

where $(a;q)_\infty := (1-a)(1-aq)(1-aq^{2})\cdots $ left parenthesis a semicolon q right parenthesis Subscript infinity Baseline colon equals left parenthesis 1 minus a right parenthesis left parenthesis 1 minus a q right parenthesis left parenthesis 1 minus a q squared right parenthesis midline horizontal ellipsis is the usual q-Pochhammer symbol.

Since the generating function of $\mathit {dp}(n)$ italic dp left parenthesis n right parenthesis has a simple infinite product representation, it is natural to investigate whether $\mathit {dp}(n)$ italic dp left parenthesis n right parenthesis satisfies Ramanujan-type congruences analogous to those enjoyed by many other restricted partition functions. To the best of our knowledge, no arithmetic properties of $\mathit {dp}(n)$ italic dp left parenthesis n right parenthesis have previously been established.

The purpose of this paper is to initiate the study of the arithmetic properties of the partition function $\mathit {dp}(n)$ italic dp left parenthesis n right parenthesis by establishing several infinite families of Ramanujan-type congruences. Our proofs rely entirely on elementary generating function manipulations together with classical 2- and 3-dissection identities, and several well-known q-series identities, avoiding the use of modular forms. In addition to congruences modulo powers of 2 and modulo 3, we obtain an infinite family of congruences modulo 4 indexed by primes in two residue classes. These results are summarised in the following theorems.

Theorem 1.1. For all $n \geq 0$ n greater than or equals 0 ,

(1.2) $$ \begin{align} \mathit{dp}(6n+4) & \equiv 0 \pmod{2}, \end{align} $$ StartLayout 1st Row 1st Column italic dp left parenthesis 6 n plus 4 right parenthesis 2nd Column identical to 0 left parenthesis mod 2 right parenthesis comma EndLayout
(1.3) $$ \begin{align} \mathit{dp}(18n+10) & \equiv 0 \pmod{4}, \end{align} $$ StartLayout 1st Row 1st Column italic dp left parenthesis 18 n plus 10 right parenthesis 2nd Column identical to 0 left parenthesis mod 4 right parenthesis comma EndLayout
(1.4) $$ \begin{align} \mathit{dp}(54n+52) & \equiv 0 \pmod{8}. \end{align} $$ StartLayout 1st Row 1st Column italic dp left parenthesis 54 n plus 52 right parenthesis 2nd Column identical to 0 left parenthesis mod 8 right parenthesis period EndLayout

Theorem 1.2. Let p be a prime such that $p \equiv 13, 17, 19 \text { or } 23 \pmod {24}$ p identical to 13 comma 17 comma 19 or 23 left parenthesis mod 24 right parenthesis . Then, for all $k,m \geq 0$ k comma m greater than or equals 0 with $p \nmid m$ p does not divide m ,

$$ \begin{align*} \mathit{dp}\bigg( 6p^{2k+1}m + \frac{15p^{2k+2}+1}{4} \bigg) \equiv 0 \pmod{4}. \end{align*} $$ italic dp left parenthesis 6 p Superscript 2 k plus 1 Baseline m plus StartFraction 15 p Superscript 2 k plus 2 Baseline plus 1 Over 4 EndFraction right parenthesis identical to 0 left parenthesis mod 4 right parenthesis period

Theorem 1.3. For all $n \geq 0$ n greater than or equals 0 ,

(1.5) $$ \begin{align} \mathit{dp}(27n+16) & \equiv 0 \pmod{3}. \end{align} $$ StartLayout 1st Row 1st Column italic dp left parenthesis 27 n plus 16 right parenthesis 2nd Column identical to 0 left parenthesis mod 3 right parenthesis period EndLayout

Theorem 1.4. For all $n \geq 0$ n greater than or equals 0 ,

$$ \begin{align*} \mathit{dp}(27n+7) \equiv \mathit{dp}(3n+1) \pmod{3}. \end{align*} $$ italic dp left parenthesis 27 n plus 7 right parenthesis identical to italic dp left parenthesis 3 n plus 1 right parenthesis left parenthesis mod 3 right parenthesis period

Theorem 1.5. For all $\alpha \geq 1$ alpha greater than or equals 1 and all $n \geq 0$ n greater than or equals 0 ,

$$ \begin{align*} \mathit{dp} \bigg( 3^{2\alpha + 1}n + \frac{7 \cdot 9^\alpha + 1}{4} \bigg) \equiv 0 \pmod{3}. \end{align*} $$ italic dp left parenthesis 3 Superscript 2 alpha plus 1 Baseline n plus StartFraction 7 dot 9 Superscript alpha Baseline plus 1 Over 4 EndFraction right parenthesis identical to 0 left parenthesis mod 3 right parenthesis period

2 Preliminaries

Throughout the remainder of this paper, we define

$$ \begin{align*}f_k := (q^k;q^k)_{\infty}\end{align*} $$ f Subscript k Baseline colon equals left parenthesis q Superscript k Baseline semicolon q Superscript k Baseline right parenthesis Subscript infinity

to shorten the notation. We begin by rewriting the generating function (1.1) in terms of the $f_k$ f Subscript k .

Lemma 2.1. We have

(2.1) $$ \begin{align} \sum_{n=0}^{\infty} \mathit{dp}(n)q^n = \frac{f_2f_3}{f_1f_4f_6}. \end{align} $$ sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis n right parenthesis q Superscript n Baseline equals StartFraction f 2 f 3 Over f 1 f 4 f 6 EndFraction period

Proof. Initially, we note that

$$ \begin{align*} \frac{1}{(q;q^6)_\infty (q^5;q^6)_\infty} & = \frac{(q^2;q^6)_\infty(q^3;q^6)_\infty(q^4;q^6)_\infty(q^6;q^6)_\infty}{(q;q^6)_\infty(q^2;q^6)_\infty(q^3;q^6)_\infty(q^4;q^6)_\infty(q^5;q^6)_\infty(q^6;q^6)_\infty} \\ & = \frac{(q^3;q^3)_\infty(q^2;q^6)_\infty(q^4;q^6)_\infty}{(q;q^3)_\infty(q^2;q^3)_\infty(q^3;q^3)_\infty} \\ & = \frac{(q^2;q^6)_\infty(q^4;q^6)_\infty}{(q;q^3)_\infty(q^2;q^3)_\infty} \\ & = \frac{(q^2;q^2)_\infty}{(q^6;q^6)_\infty}\cdot \frac{(q^3;q^3)_\infty}{(q;q)_\infty} = \frac{f_2f_3}{f_1f_6}. \end{align*} $$ StartLayout 1st Row 1st Column StartFraction 1 Over left parenthesis q semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline left parenthesis q Superscript 5 Baseline semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline EndFraction 2nd Column equals StartFraction left parenthesis q squared semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline left parenthesis q cubed semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline left parenthesis q Superscript 4 Baseline semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline left parenthesis q Superscript 6 Baseline semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline Over left parenthesis q semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline left parenthesis q squared semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline left parenthesis q cubed semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline left parenthesis q Superscript 4 Baseline semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline left parenthesis q Superscript 5 Baseline semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline left parenthesis q Superscript 6 Baseline semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline EndFraction 2nd Row 1st Column Blank 2nd Column equals StartFraction left parenthesis q cubed semicolon q cubed right parenthesis Subscript infinity Baseline left parenthesis q squared semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline left parenthesis q Superscript 4 Baseline semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline Over left parenthesis q semicolon q cubed right parenthesis Subscript infinity Baseline left parenthesis q squared semicolon q cubed right parenthesis Subscript infinity Baseline left parenthesis q cubed semicolon q cubed right parenthesis Subscript infinity Baseline EndFraction 3rd Row 1st Column Blank 2nd Column equals StartFraction left parenthesis q squared semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline left parenthesis q Superscript 4 Baseline semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline Over left parenthesis q semicolon q cubed right parenthesis Subscript infinity Baseline left parenthesis q squared semicolon q cubed right parenthesis Subscript infinity Baseline EndFraction 4th Row 1st Column Blank 2nd Column equals StartFraction left parenthesis q squared semicolon q squared right parenthesis Subscript infinity Baseline Over left parenthesis q Superscript 6 Baseline semicolon q Superscript 6 Baseline right parenthesis Subscript infinity Baseline EndFraction dot StartFraction left parenthesis q cubed semicolon q cubed right parenthesis Subscript infinity Baseline Over left parenthesis q semicolon q right parenthesis Subscript infinity Baseline EndFraction equals StartFraction f 2 f 3 Over f 1 f 6 EndFraction period EndLayout

Thus, it follows from (1.1) that

$$ \begin{align*} \sum_{n=0}^{\infty} \mathit{dp}(n)q^n & = \frac{f_2f_3}{f_1f_4f_6}, \end{align*} $$ StartLayout 1st Row 1st Column sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis n right parenthesis q Superscript n 2nd Column equals StartFraction f 2 f 3 Over f 1 f 4 f 6 EndFraction comma EndLayout

which is (2.1).

The proofs of the above-mentioned main results require a few well-known 2- and 3-dissections.

Lemma 2.2 [Reference da Silva and Sellers3, Lemma 1].

We have

(2.2) $$ \begin{align} {f_{1}^{2}} & = \frac{f_{2}f_8^5}{f_{4}^{2}f_{16}^{2}} -2q\frac{f_{2}f_{16}^{2}}{f_{8}}, \end{align} $$ StartLayout 1st Row 1st Column f 1 squared 2nd Column equals StartFraction f 2 f 8 Superscript 5 Baseline Over f 4 squared f 16 squared EndFraction minus 2 q StartFraction f 2 f 16 squared Over f 8 EndFraction comma EndLayout
(2.3) $$ \begin{align} \frac{1}{f_{1}^{2}} & = \frac{f_{8}^{5}}{f_{2}^{5}f_{16}^{2}} + 2q\frac{f_{4}^{2}f_{16}^{2}}{f_{2}^{5}f_{8}}. \end{align} $$ StartLayout 1st Row 1st Column StartFraction 1 Over f 1 squared EndFraction 2nd Column equals StartFraction f 8 Superscript 5 Baseline Over f 2 Superscript 5 Baseline f 16 squared EndFraction plus 2 q StartFraction f 4 squared f 16 squared Over f 2 Superscript 5 Baseline f 8 EndFraction period EndLayout

Lemma 2.3 [Reference Toh7, (2.1c)].

We have

(2.4) $$ \begin{align} \frac{f_{2}}{f_{1}f_{4}} = \displaystyle\frac{f_{18}^{9}}{f_{3}^2f_{9}^3f_{12}^2f_{36}^3} + q\displaystyle\frac{f_{6}^2f_{18}^{3}}{f_{3}^3f_{12}^3} +q^2 \displaystyle\frac{f_{6}^4f_{9}^3f_{36}^{3}}{f_{3}^4f_{12}^4f_{18}^3}. \end{align} $$ StartFraction f 2 Over f 1 f 4 EndFraction equals StartFraction f 18 Superscript 9 Baseline Over f 3 squared f 9 cubed f 12 squared f 36 cubed EndFraction plus q StartFraction f 6 squared f 18 cubed Over f 3 cubed f 12 cubed EndFraction plus q squared StartFraction f 6 Superscript 4 Baseline f 9 cubed f 36 cubed Over f 3 Superscript 4 Baseline f 12 Superscript 4 Baseline f 18 cubed EndFraction period

Lemma 2.4 [Reference Toh7, (2.1b)].

We have

(2.5) $$ \begin{align} \frac{f_{2}}{f_{1}^2} = \frac{f_{6}^4f_{9}^6}{f_{3}^8f_{18}^3} + 2q\displaystyle\frac{f_{6}^3f_{9}^3}{f_{3}^7} + 4q^2\displaystyle\frac{f_{6}^2f_{18}^3}{f_{3}^6}. \end{align} $$ StartFraction f 2 Over f 1 squared EndFraction equals StartFraction f 6 Superscript 4 Baseline f 9 Superscript 6 Baseline Over f 3 Superscript 8 Baseline f 18 cubed EndFraction plus 2 q StartFraction f 6 cubed f 9 cubed Over f 3 Superscript 7 Baseline EndFraction plus 4 q squared StartFraction f 6 squared f 18 cubed Over f 3 Superscript 6 Baseline EndFraction period

Lemma 2.5 [Reference Hirschhorn5, (14.3.2)].

We have

(2.6) $$ \begin{align} \frac{f_{1}^2}{f_{2}} = \displaystyle\frac{f_{9}^2}{f_{18}} - 2q\displaystyle\frac{f_{3}f_{18}^2}{f_{6}f_{9}}. \end{align} $$ StartFraction f 1 squared Over f 2 EndFraction equals StartFraction f 9 squared Over f 18 EndFraction minus 2 q StartFraction f 3 f 18 squared Over f 6 f 9 EndFraction period

Lemma 2.6 [Reference Toh7, (3.1)].

We have

(2.7) $$ \begin{align} \displaystyle\frac{f_{2}^3}{f_{1}^3} = \displaystyle\frac{f_{6}}{f_{3}} +3q\displaystyle\frac{f_{6}^4f_{9}^{5}}{f_{3}^8f_{18}} +6q^2\displaystyle\frac{f_{6}^3f_{9}^{2}f_{18}^{2}}{f_{3}^7} + 12q^3\displaystyle\frac{f_{6}^2f_{18}^5}{f_{3}^6f_{9}}. \end{align} $$ StartFraction f 2 cubed Over f 1 cubed EndFraction equals StartFraction f 6 Over f 3 EndFraction plus 3 q StartFraction f 6 Superscript 4 Baseline f 9 Superscript 5 Baseline Over f 3 Superscript 8 Baseline f 18 EndFraction plus 6 q squared StartFraction f 6 cubed f 9 squared f 18 squared Over f 3 Superscript 7 Baseline EndFraction plus 12 q cubed StartFraction f 6 squared f 18 Superscript 5 Baseline Over f 3 Superscript 6 Baseline f 9 EndFraction period

We recall Ramanujan’s theta functions

$$ \begin{align*} f(a,b) & :=\sum_{n=-\infty}^\infty a^{{n(n+1)}/{2}}b^{{n(n-1)}/{2}} \quad\text{for } |ab|<1, \nonumber \\ \phi(q) & := f(q,q) = \sum_{n=-\infty}^{\infty} q^{n^2} = \frac{f_2^5}{f_1^2f_4^2}, \\ \psi(q) & := f(q,q^3) = \sum_{n=0}^{\infty} q^{n(n+1)/2} = \frac{f_2^2}{f_1}. \end{align*} $$ StartLayout 1st Row 1st Column f left parenthesis a comma b right parenthesis 2nd Column colon equals sigma summation Underscript n equals negative infinity Overscript infinity Endscripts a Superscript n left parenthesis n plus 1 right parenthesis divided by 2 Baseline b Superscript n left parenthesis n minus 1 right parenthesis divided by 2 Baseline for StartAbsoluteValue a b EndAbsoluteValue less than 1 comma 2nd Row 1st Column phi left parenthesis q right parenthesis 2nd Column colon equals f left parenthesis q comma q right parenthesis equals sigma summation Underscript n equals negative infinity Overscript infinity Endscripts q Superscript n squared Baseline equals StartFraction f 2 Superscript 5 Baseline Over f 1 squared f 4 squared EndFraction comma 3rd Row 1st Column psi left parenthesis q right parenthesis 2nd Column colon equals f left parenthesis q comma q cubed right parenthesis equals sigma summation Underscript n equals 0 Overscript infinity Endscripts q Superscript n left parenthesis n plus 1 right parenthesis divided by 2 Baseline equals StartFraction f 2 squared Over f 1 EndFraction period EndLayout

With the above definitions, we can state the following 3-dissection of $1/\psi (q)$ 1 divided by psi left parenthesis q right parenthesis .

Lemma 2.7 ([Reference Chern and Hao2, Lemma 2.2]).

We have

(2.8) $$ \begin{align} \frac{1}{\psi(q)} = \frac{f_3^2f_9^3}{f_6^6} - q\frac{f_3^3f_{18}^3}{f_6^7} + q^2\frac{f_3^4f_{18}^6}{f_6^8f_9^3}. \end{align} $$ StartFraction 1 Over psi left parenthesis q right parenthesis EndFraction equals StartFraction f 3 squared f 9 cubed Over f 6 Superscript 6 Baseline EndFraction minus q StartFraction f 3 cubed f 18 cubed Over f 6 Superscript 7 Baseline EndFraction plus q squared StartFraction f 3 Superscript 4 Baseline f 18 Superscript 6 Baseline Over f 6 Superscript 8 Baseline f 9 cubed EndFraction period

Lastly, we require the classical identity of Jacobi (see [Reference Berndt1, Theorem 1.3.9]):

(2.9) $$ \begin{align} f_1^3 = \sum_{n=0}^{\infty} (-1)^n (2n+1) q^{n(n+1)/2}. \end{align} $$ f 1 cubed equals sigma summation Underscript n equals 0 Overscript infinity Endscripts left parenthesis negative 1 right parenthesis Superscript n Baseline left parenthesis 2 n plus 1 right parenthesis q Superscript n left parenthesis n plus 1 right parenthesis divided by 2 Baseline period

3 Proof of Theorem 1.1

Let

$$ \begin{align*}H(q) = \sum_{n=0}^{\infty}h(n)q^n = G(-q) = \sum_{n=0}^{\infty}\mathit{dp}(n)(-q)^n.\end{align*} $$ upper H left parenthesis q right parenthesis equals sigma summation Underscript n equals 0 Overscript infinity Endscripts h left parenthesis n right parenthesis q Superscript n Baseline equals upper G left parenthesis negative q right parenthesis equals sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis n right parenthesis left parenthesis negative q right parenthesis Superscript n Baseline period

Thus, $h(n) = (-1)^n\mathit {dp}(n)$ h left parenthesis n right parenthesis equals left parenthesis negative 1 right parenthesis Superscript n Baseline italic dp left parenthesis n right parenthesis for all n. We note that

$$ \begin{align*}(-q;-q)_\infty = \frac{f_2^3}{f_1f_4}.\end{align*} $$ left parenthesis negative q semicolon negative q right parenthesis Subscript infinity Baseline equals StartFraction f 2 cubed Over f 1 f 4 EndFraction period

Using this identity and (2.1),

$$ \begin{align*} H(q) = \frac{f_1f_6^2}{f_2^2f_3f_{12}} = \frac{f_6^2}{f_3f_{12}} \frac{1}{\psi(q)}. \end{align*} $$ upper H left parenthesis q right parenthesis equals StartFraction f 1 f 6 squared Over f 2 squared f 3 f 12 EndFraction equals StartFraction f 6 squared Over f 3 f 12 EndFraction StartFraction 1 Over psi left parenthesis q right parenthesis EndFraction period

Thanks to (2.8),

$$ \begin{align*} \sum_{n=0}^{\infty}h(3n+1)q^{3n+1} & = -q \frac{f_3^2f_{18}^3}{f_6^5f_{12}}, \end{align*} $$ StartLayout 1st Row 1st Column sigma summation Underscript n equals 0 Overscript infinity Endscripts h left parenthesis 3 n plus 1 right parenthesis q Superscript 3 n plus 1 2nd Column equals minus q StartFraction f 3 squared f 18 cubed Over f 6 Superscript 5 Baseline f 12 EndFraction comma EndLayout

which, after dividing by q and replacing $q^3$ q cubed by q, yields

$$ \begin{align*} \sum_{n=0}^{\infty}h(3n+1)q^{n} & = - \frac{f_1^2f_{6}^3}{f_2^5f_{4}}. \end{align*} $$ StartLayout 1st Row 1st Column sigma summation Underscript n equals 0 Overscript infinity Endscripts h left parenthesis 3 n plus 1 right parenthesis q Superscript n 2nd Column equals minus StartFraction f 1 squared f 6 cubed Over f 2 Superscript 5 Baseline f 4 EndFraction period EndLayout

Using (2.2), we can 2-dissect the expression above to obtain

$$ \begin{align*} \sum_{n=0}^{\infty}h(6n+4)q^{2n+1} & = 2q \frac{f_6^3f_{16}^2}{f_2^4f_{4}f_8}. \end{align*} $$ StartLayout 1st Row 1st Column sigma summation Underscript n equals 0 Overscript infinity Endscripts h left parenthesis 6 n plus 4 right parenthesis q Superscript 2 n plus 1 2nd Column equals 2 q StartFraction f 6 cubed f 16 squared Over f 2 Superscript 4 Baseline f 4 f 8 EndFraction period EndLayout

Thus,

$$ \begin{align*} \sum_{n=0}^{\infty}h(6n+4)q^{n} & = 2 \frac{f_3^3f_{8}^2}{f_1^4f_{2}f_4}, \end{align*} $$ StartLayout 1st Row 1st Column sigma summation Underscript n equals 0 Overscript infinity Endscripts h left parenthesis 6 n plus 4 right parenthesis q Superscript n 2nd Column equals 2 StartFraction f 3 cubed f 8 squared Over f 1 Superscript 4 Baseline f 2 f 4 EndFraction comma EndLayout

which proves (1.2) since $\mathit {dp}(6n+4) = h(6n+4) \equiv 0 \pmod {2}$ italic dp left parenthesis 6 n plus 4 right parenthesis equals h left parenthesis 6 n plus 4 right parenthesis identical to 0 left parenthesis mod 2 right parenthesis .

Working modulo 8,

(3.1) $$ \begin{align} \sum_{n=0}^{\infty}h(6n+4)q^{n} & = 2 \frac{f_3^3f_{8}^2}{f_1^4f_{2}f_4} \equiv 2 f_3^3\frac{f_4^3}{f_2^3} \pmod{8}. \end{align} $$ StartLayout 1st Row 1st Column sigma summation Underscript n equals 0 Overscript infinity Endscripts h left parenthesis 6 n plus 4 right parenthesis q Superscript n 2nd Column equals 2 StartFraction f 3 cubed f 8 squared Over f 1 Superscript 4 Baseline f 2 f 4 EndFraction identical to 2 f 3 cubed StartFraction f 4 cubed Over f 2 cubed EndFraction left parenthesis mod 8 right parenthesis period EndLayout

Thanks to (2.7),

$$ \begin{align*} \sum_{n=0}^{\infty}h(18n+10)q^{3n+1} & \equiv 4q^4 \frac{f_3^3f_{12}^3f_{18}^2f_{36}^2}{f_6^7} \pmod{8}, \end{align*} $$ StartLayout 1st Row 1st Column sigma summation Underscript n equals 0 Overscript infinity Endscripts h left parenthesis 18 n plus 10 right parenthesis q Superscript 3 n plus 1 2nd Column identical to 4 q Superscript 4 Baseline StartFraction f 3 cubed f 12 cubed f 18 squared f 36 squared Over f 6 Superscript 7 Baseline EndFraction left parenthesis mod 8 right parenthesis comma EndLayout

which simplifies to

$$ \begin{align*} \sum_{n=0}^{\infty}h(18n+10)q^{n} & \equiv 4q \frac{f_1^3f_{4}^3f_{6}^2f_{12}^2}{f_2^7} \pmod{8}. \end{align*} $$ StartLayout 1st Row 1st Column sigma summation Underscript n equals 0 Overscript infinity Endscripts h left parenthesis 18 n plus 10 right parenthesis q Superscript n 2nd Column identical to 4 q StartFraction f 1 cubed f 4 cubed f 6 squared f 12 squared Over f 2 Superscript 7 Baseline EndFraction left parenthesis mod 8 right parenthesis period EndLayout

It follows that $h(18n+10) \equiv 0 \pmod {4}$ h left parenthesis 18 n plus 10 right parenthesis identical to 0 left parenthesis mod 4 right parenthesis , which completes the proof of (1.3) since ${h(n) = (-1)^n \mathit {dp}(n)}$ h left parenthesis n right parenthesis equals left parenthesis negative 1 right parenthesis Superscript n Baseline italic dp left parenthesis n right parenthesis .

To prove (1.4), we use (2.7) to extract the terms of the form $q^{3n+2}$ q Superscript 3 n plus 2 from (3.1). The resulting congruence is

(3.2) $$ \begin{align} \sum_{n=0}^{\infty}h(18n+16)q^{n} & \equiv 6\frac{f_1^3f_{4}^4f_{6}^5}{f_2^8f_{12}} \pmod{8}\nonumber \\ & \equiv 6\frac{f_6^5}{f_{12}} f_1^3 \pmod{8} \nonumber \\ & \equiv 6\frac{f_6^5}{f_{12}} \sum_{n=0}^{\infty} (-1)^n (2n+1) q^{n(n+1)/2} \pmod{8} \end{align} $$ StartLayout 1st Row 1st Column sigma summation Underscript n equals 0 Overscript infinity Endscripts h left parenthesis 18 n plus 16 right parenthesis q Superscript n 2nd Column identical to 6 StartFraction f 1 cubed f 4 Superscript 4 Baseline f 6 Superscript 5 Baseline Over f 2 Superscript 8 Baseline f 12 EndFraction left parenthesis mod 8 right parenthesis 2nd Row 1st Column Blank 2nd Column identical to 6 StartFraction f 6 Superscript 5 Baseline Over f 12 EndFraction f 1 cubed left parenthesis mod 8 right parenthesis 3rd Row 1st Column Blank 2nd Column identical to 6 StartFraction f 6 Superscript 5 Baseline Over f 12 EndFraction sigma summation Underscript n equals 0 Overscript infinity Endscripts left parenthesis negative 1 right parenthesis Superscript n Baseline left parenthesis 2 n plus 1 right parenthesis q Superscript n left parenthesis n plus 1 right parenthesis divided by 2 Baseline left parenthesis mod 8 right parenthesis EndLayout

by (2.9). We note that $f_6^5/f_{12}$ f 6 Superscript 5 Baseline divided by f 12 is a function of $q^3$ q cubed and $n(n+1)/2$ n left parenthesis n plus 1 right parenthesis divided by 2 is never congruent to 2 modulo 3. Hence, once we expand (3.2) as a power series in q, there will be no powers of the form $q^{3n+2}$ q Superscript 3 n plus 2 . Therefore, for all $n \geq 0$ n greater than or equals 0 ,

$$ \begin{align*}h(18(3n+2)+16) = h(54n+52) \equiv 0 \pmod{8},\end{align*} $$ h left parenthesis 18 left parenthesis 3 n plus 2 right parenthesis plus 16 right parenthesis equals h left parenthesis 54 n plus 52 right parenthesis identical to 0 left parenthesis mod 8 right parenthesis comma

which completes the proof of (1.4).

4 Proof of Theorem 1.2

Initially, we use (2.4) to extract the terms of the form $q^{3n+1}$ q Superscript 3 n plus 1 from (2.1), which gives

$$ \begin{align*}\sum_{n=0}^{\infty} \mathit{dp}(3n+1)q^{3n+1} = q\frac{f_{6}f_{18}^3}{f_3^2f_{12}^3}.\end{align*} $$ sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis 3 n plus 1 right parenthesis q Superscript 3 n plus 1 Baseline equals q StartFraction f 6 f 18 cubed Over f 3 squared f 12 cubed EndFraction period

After dividing by q and replacing $q^3$ q cubed by q, we are left with

(4.1) $$ \begin{align} \sum_{n=0}^{\infty} \mathit{dp}(3n+1)q^{n} & = \frac{f_{2}f_{6}^3}{f_1^2f_{4}^3}. \end{align} $$ StartLayout 1st Row 1st Column sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis 3 n plus 1 right parenthesis q Superscript n 2nd Column equals StartFraction f 2 f 6 cubed Over f 1 squared f 4 cubed EndFraction period EndLayout

Now, we use (2.3) to extract the odd parts of (4.1), which gives

(4.2) $$ \begin{align} \sum_{n=0}^{\infty} \mathit{dp}(6n+4)q^{n} = 2\frac{f_3^3f_8^2}{f_1^4f_2f_4} \equiv 2f_2^3 f_3^3\pmod{4}. \end{align} $$ sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis 6 n plus 4 right parenthesis q Superscript n Baseline equals 2 StartFraction f 3 cubed f 8 squared Over f 1 Superscript 4 Baseline f 2 f 4 EndFraction identical to 2 f 2 cubed f 3 cubed left parenthesis mod 4 right parenthesis period

Using (2.9), it follows from (4.2) that

$$ \begin{align*} \sum_{n=0}^{\infty} \mathit{dp}(6n+4)q^{n} & \equiv 2f_2^3f_3^3 \pmod{4}\\ & = 2 \sum_{k,l=0}^{\infty} (-1)^{k+l}(2k+1)(2l+1)q^{k(k+1)+3l(l+1)/2} \\ & \equiv 2\sum_{k,l=0}^{\infty} q^{k(k+1)+3l(l+1)/2} \pmod{4}. \end{align*} $$ StartLayout 1st Row 1st Column sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis 6 n plus 4 right parenthesis q Superscript n 2nd Column identical to 2 f 2 cubed f 3 cubed left parenthesis mod 4 right parenthesis 2nd Row 1st Column Blank 2nd Column equals 2 sigma summation Underscript k comma l equals 0 Overscript infinity Endscripts left parenthesis negative 1 right parenthesis Superscript k plus l Baseline left parenthesis 2 k plus 1 right parenthesis left parenthesis 2 l plus 1 right parenthesis q Superscript k left parenthesis k plus 1 right parenthesis plus 3 l left parenthesis l plus 1 right parenthesis divided by 2 Baseline 3rd Row 1st Column Blank 2nd Column identical to 2 sigma summation Underscript k comma l equals 0 Overscript infinity Endscripts q Superscript k left parenthesis k plus 1 right parenthesis plus 3 l left parenthesis l plus 1 right parenthesis divided by 2 Baseline left parenthesis mod 4 right parenthesis period EndLayout

Thus,

$$ \begin{align*} \sum_{n=0}^{\infty} \mathit{dp}(6n+4)q^{24n+15} & \equiv 2\sum_{k,l=0}^{\infty} q^{6(2k+1)^2+(6l+3)^2} \pmod{4}. \end{align*} $$ StartLayout 1st Row 1st Column sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis 6 n plus 4 right parenthesis q Superscript 24 n plus 15 2nd Column identical to 2 sigma summation Underscript k comma l equals 0 Overscript infinity Endscripts q Superscript 6 left parenthesis 2 k plus 1 right parenthesis squared plus left parenthesis 6 l plus 3 right parenthesis squared Baseline left parenthesis mod 4 right parenthesis period EndLayout

It follows that $\mathit {dp}(6n+4) \equiv 0 \pmod {4}$ italic dp left parenthesis 6 n plus 4 right parenthesis identical to 0 left parenthesis mod 4 right parenthesis if $24n+15$ 24 n plus 15 is not of the form $6x^2+y^2$ 6 x squared plus y squared .

Since $p \equiv 13, 17, 19 \text { or } 23 \pmod {24}$ p identical to 13 comma 17 comma 19 or 23 left parenthesis mod 24 right parenthesis , it follows that $ ({-6}/{p}) = -1$ left parenthesis negative 6 divided by p right parenthesis equals negative 1 , which implies that $\nu _p(N)$ nu Subscript p Baseline left parenthesis upper N right parenthesis is even if N is of the form $6x^2+y^2$ 6 x squared plus y squared . Taking $n = p^{2k+1}m + 5({p^{2k+2}-1})/{8}$ n equals p Superscript 2 k plus 1 Baseline m plus 5 left parenthesis p Superscript 2 k plus 2 Baseline minus 1 right parenthesis divided by 8 , we have

$$ \begin{align*}24n+15 = 24p^{2k+1}m + 15p^{2k+2} = p^{2k+1}(24m+15p).\end{align*} $$ 24 n plus 15 equals 24 p Superscript 2 k plus 1 Baseline m plus 15 p Superscript 2 k plus 2 Baseline equals p Superscript 2 k plus 1 Baseline left parenthesis 24 m plus 15 p right parenthesis period

Therefore, $\nu _p(24n+15)$ nu Subscript p Baseline left parenthesis 24 n plus 15 right parenthesis is odd and the result follows by the fact that

$$ \begin{align*}6\bigg( p^{2k+1}m + \frac{5(p^{2k+2}-1)}{8} \bigg) + 4 = 6p^{2k+1}m + \frac{15p^{2k+2}+1}{4}.\end{align*} $$ 6 left parenthesis p Superscript 2 k plus 1 Baseline m plus StartFraction 5 left parenthesis p Superscript 2 k plus 2 Baseline minus 1 right parenthesis Over 8 EndFraction right parenthesis plus 4 equals 6 p Superscript 2 k plus 1 Baseline m plus StartFraction 15 p Superscript 2 k plus 2 Baseline plus 1 Over 4 EndFraction period

5 Proofs of Theorems 1.3–1.5

We now transition to proving Theorems 1.3–1.5 where the focus is on arithmetic properties modulo 3. Let us begin by proving (1.5). It follows from (4.1) that

$$ \begin{align*} \sum_{n=0}^{\infty} \mathit{dp}(3n+1)q^{n} & \equiv \frac{f_{2}f_{6}^3}{f_1^2f_{12}} \pmod{3}. \end{align*} $$ StartLayout 1st Row 1st Column sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis 3 n plus 1 right parenthesis q Superscript n 2nd Column identical to StartFraction f 2 f 6 cubed Over f 1 squared f 12 EndFraction left parenthesis mod 3 right parenthesis period EndLayout

Thanks to (2.5),

$$ \begin{align*} \sum_{n=0}^{\infty} \mathit{dp}(9n+7)q^{3n+2} \equiv 4q^2\frac{f_{6}^5f_{18}^3}{f_3^6f_{12}} \pmod{3}. \end{align*} $$ sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis 9 n plus 7 right parenthesis q Superscript 3 n plus 2 Baseline identical to 4 q squared StartFraction f 6 Superscript 5 Baseline f 18 cubed Over f 3 Superscript 6 Baseline f 12 EndFraction left parenthesis mod 3 right parenthesis period

Dividing by q and replacing $q^3$ q cubed by q yields

$$ \begin{align*} \sum_{n=0}^{\infty} \mathit{dp}(9n+7)q^{n} \equiv \frac{f_{2}^5f_{6}^3}{f_1^6f_{4}} \equiv \frac{f_6^4f_2^2}{f_3^2f_4} \pmod{3}. \end{align*} $$ sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis 9 n plus 7 right parenthesis q Superscript n Baseline identical to StartFraction f 2 Superscript 5 Baseline f 6 cubed Over f 1 Superscript 6 Baseline f 4 EndFraction identical to StartFraction f 6 Superscript 4 Baseline f 2 squared Over f 3 squared f 4 EndFraction left parenthesis mod 3 right parenthesis period

Finally, we use (2.6) to get

(5.1) $$ \begin{align} \sum_{n=0}^{\infty} \mathit{dp}(9n+7)q^{n} \equiv \frac{f_6^4}{f_3^2} \bigg( \frac{f_{18}^2}{f_{36}} - 2q^2\displaystyle\frac{f_{6}f_{36}^2}{f_{12}f_{18}} \bigg) \pmod{3}, \end{align} $$ sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis 9 n plus 7 right parenthesis q Superscript n Baseline identical to StartFraction f 6 Superscript 4 Baseline Over f 3 squared EndFraction left parenthesis StartFraction f 18 squared Over f 36 EndFraction minus 2 q squared StartFraction f 6 f 36 squared Over f 12 f 18 EndFraction right parenthesis left parenthesis mod 3 right parenthesis comma

from which (1.5) follows.

We next consider the ‘internal congruence’ given in Theorem 1.4. It follows from (5.1) that

$$ \begin{align*} \sum_{n=0}^{\infty} \mathit{dp}(27n+7)q^{3n} & \equiv \frac{f_6^4f_{18}^2}{f_3^2f_{36}} \pmod{3}. \end{align*} $$ StartLayout 1st Row 1st Column sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis 27 n plus 7 right parenthesis q Superscript 3 n 2nd Column identical to StartFraction f 6 Superscript 4 Baseline f 18 squared Over f 3 squared f 36 EndFraction left parenthesis mod 3 right parenthesis period EndLayout

Replacing $q^3$ q cubed by q, we are left with

$$ \begin{align*} \sum_{n=0}^{\infty} \mathit{dp}(27n+7)q^{n} \equiv \frac{f_2^4f_{6}^2}{f_1^2f_{12}} \equiv \frac{f_2f_{6}^3}{f_1^2f_{4}^3} \pmod{3}. \end{align*} $$ sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis 27 n plus 7 right parenthesis q Superscript n Baseline identical to StartFraction f 2 Superscript 4 Baseline f 6 squared Over f 1 squared f 12 EndFraction identical to StartFraction f 2 f 6 cubed Over f 1 squared f 4 cubed EndFraction left parenthesis mod 3 right parenthesis period

Thus, it follows by (4.1) that

$$ \begin{align*} \sum_{n=0}^{\infty} \mathit{dp}(27n+7)q^{n} \equiv \sum_{n=0}^{\infty} \mathit{dp}(3n+1)q^{n} \pmod{3}, \end{align*} $$ sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis 27 n plus 7 right parenthesis q Superscript n Baseline identical to sigma summation Underscript n equals 0 Overscript infinity Endscripts italic dp left parenthesis 3 n plus 1 right parenthesis q Superscript n Baseline left parenthesis mod 3 right parenthesis comma

which completes the proof.

Lastly, we use the previous two theorems to prove Theorem 1.5. We prove this theorem by induction on $\alpha $ alpha . Note that the case $\alpha =1$ alpha equals 1 has already been established earlier in Theorem 1.3. We now use it as our base case in a proof by induction. Thus, we assume that, for some $\alpha \geq 1$ alpha greater than or equals 1 and all $n\geq 0$ n greater than or equals 0 ,

$$ \begin{align*} \mathit{dp} \bigg( 3^{2\alpha + 1}n + \frac{7 \cdot 9^\alpha + 1}{4} \bigg) \equiv 0 \pmod{3}. \end{align*} $$ italic dp left parenthesis 3 Superscript 2 alpha plus 1 Baseline n plus StartFraction 7 dot 9 Superscript alpha Baseline plus 1 Over 4 EndFraction right parenthesis identical to 0 left parenthesis mod 3 right parenthesis period

We then want to prove that

$$ \begin{align*} \mathit{dp} \bigg( 3^{2\alpha + 3}n + \frac{7 \cdot 9^{\alpha+1} + 1}{4} \bigg) \equiv 0 \pmod{3}. \end{align*} $$ italic dp left parenthesis 3 Superscript 2 alpha plus 3 Baseline n plus StartFraction 7 dot 9 Superscript alpha plus 1 Baseline plus 1 Over 4 EndFraction right parenthesis identical to 0 left parenthesis mod 3 right parenthesis period

We note that

$$ \begin{align*} \mathit{dp} \bigg( 3 \bigg( 3^{2\alpha}n + \frac{3\cdot 7 \cdot 9^{\alpha-1} -1}{4} \bigg) +1 \bigg) & = \mathit{dp} \bigg( 3^{2\alpha+1}n + \frac{7 \cdot 9^{\alpha} -3}{4} +1 \bigg) \\ & = \mathit{dp} \bigg( 3^{2\alpha+1}n + \frac{7 \cdot 9^{\alpha} +1}{4} \bigg) \equiv 0 \pmod{3}, \end{align*} $$ StartLayout 1st Row 1st Column italic dp left parenthesis 3 left parenthesis 3 Superscript 2 alpha Baseline n plus StartFraction 3 dot 7 dot 9 Superscript alpha minus 1 Baseline minus 1 Over 4 EndFraction right parenthesis plus 1 right parenthesis 2nd Column equals italic dp left parenthesis 3 Superscript 2 alpha plus 1 Baseline n plus StartFraction 7 dot 9 Superscript alpha Baseline minus 3 Over 4 EndFraction plus 1 right parenthesis 2nd Row 1st Column Blank 2nd Column equals italic dp left parenthesis 3 Superscript 2 alpha plus 1 Baseline n plus StartFraction 7 dot 9 Superscript alpha Baseline plus 1 Over 4 EndFraction right parenthesis identical to 0 left parenthesis mod 3 right parenthesis comma EndLayout

by the induction hypothesis. Finally, by Theorem 1.4,

$$ \begin{align*} \mathit{dp} \bigg( 3^{2\alpha + 3}n + \frac{7 \cdot 9^{\alpha+1} + 1}{4} \bigg) & = \mathit{dp} \bigg( 3^{2\alpha + 3}n + \frac{7 \cdot 9^{\alpha+1} -27}{4} + 7 \bigg) \\ & = \mathit{dp} \bigg( 27 \bigg( 3^{2\alpha}n + \frac{3 \cdot 7 \cdot 9^{\alpha-1} - 1}{4} \bigg) + 7 \bigg) \\ & \equiv \mathit{dp} \bigg( 3 \bigg( 3^{2\alpha}n + \frac{3\cdot 7 \cdot 9^{\alpha-1} -1}{4} \bigg) +1 \bigg) \equiv 0 \pmod{3}, \end{align*} $$ StartLayout 1st Row 1st Column italic dp left parenthesis 3 Superscript 2 alpha plus 3 Baseline n plus StartFraction 7 dot 9 Superscript alpha plus 1 Baseline plus 1 Over 4 EndFraction right parenthesis 2nd Column equals italic dp left parenthesis 3 Superscript 2 alpha plus 3 Baseline n plus StartFraction 7 dot 9 Superscript alpha plus 1 Baseline minus 27 Over 4 EndFraction plus 7 right parenthesis 2nd Row 1st Column Blank 2nd Column equals italic dp left parenthesis 27 left parenthesis 3 Superscript 2 alpha Baseline n plus StartFraction 3 dot 7 dot 9 Superscript alpha minus 1 Baseline minus 1 Over 4 EndFraction right parenthesis plus 7 right parenthesis 3rd Row 1st Column Blank 2nd Column identical to italic dp left parenthesis 3 left parenthesis 3 Superscript 2 alpha Baseline n plus StartFraction 3 dot 7 dot 9 Superscript alpha minus 1 Baseline minus 1 Over 4 EndFraction right parenthesis plus 1 right parenthesis identical to 0 left parenthesis mod 3 right parenthesis comma EndLayout

which concludes the inductive proof.

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