Ramanujan's theta functions and linear combinations of three triangular   numbers

Zhi-Hong Sun

arXiv:1812.03820·math.NT·December 12, 2018

Ramanujan's theta functions and linear combinations of three triangular numbers

Zhi-Hong Sun

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TL;DR

This paper explores the relationships between Ramanujan's theta functions and the counts of representations of integers as sums of quadratic and triangular forms, providing new formulas and conjectures for specific parameter cases.

Contribution

It establishes explicit relations between representations by quadratic and triangular forms using Ramanujan's theta functions and derives formulas for various parameter configurations.

Findings

01

Relations between $t(2,3,3;n)$ and $N(1,3,3;n+1)$

02

Formulas for $t(a,3a,4b;n)$, $t(a,7a,4b;n)$, $t(3a,5a,4b;n)$, and $t(a,15a,4b;n)$ under certain conditions

03

Numerous conjectures on $t(a,b,c;n)$ for special $(a,b,c)$ values

Abstract

Let $Z$ be the set of integers. For positive integers $a, b, c$ and $n$ let $N (a, b, c; n)$ be the number of representations of $n$ by $a x^{2} + b y^{2} + c z^{2}$ , and let $t (a, b, c; n)$ be the number of representations of $n$ by $a x (x + 1) /2 + b y (y + 1) /2 + cz (z + 1) /2$ $(x, y, z \in Z)$ . In this paper, by using Ramanujan's theta functions $φ (q)$ and $ψ (q)$ we reveal the relation between $t (2, 3, 3; n)$ and $N (1, 3, 3; n + 1)$ , and the relation between $t (1, 1, 6; n)$ and $N (1, 1, 3; n + 1)$ . We also obtain formulas for $t (a, 3 a, 4 b; n),$ $t (a, 7 a, 4 b; n), t (3 a, 5 a, 4 b; n)$ and $t (a, 15 a, 4 b; n)$ under certain congruence conditions, where $a$ and $b$ are positive odd integers. In addition, we pose many conjectures on $t (a, b, c; n)$ for some special values of $(a, b, c)$ .

Equations512

\displaystyle N(a,b,c;n)=\big{|}\{(x,y,z)\in\mathbb{Z}^{3}\ |\ n=ax^{2}+by^{2}+cz^{2}\}\big{|},

\displaystyle N(a,b,c;n)=\big{|}\{(x,y,z)\in\mathbb{Z}^{3}\ |\ n=ax^{2}+by^{2}+cz^{2}\}\big{|},

\displaystyle t(a,b,c;n)=\Big{|}\Big{\{}(x,y,z)\in\mathbb{Z}^{3}\ \big{|}\ n\ =a\frac{x(x+1)}{2}+b\frac{y(y+1)}{2}+c\frac{z(z+1)}{2}\Big{\}}\Big{|},

\displaystyle T(a,b,c;n)=\Big{|}\Big{\{}(x,y,z)\in\mathbb{N}^{3}\ \big{|}\ n\ =a\frac{x(x+1)}{2}+b\frac{y(y+1)}{2}+c\frac{z(z+1)}{2}\Big{\}}\Big{|}.

t (a, b, c; n) = 8 T (a, b, c; n) ..1

t (a, b, c; n) = 8 T (a, b, c; n) ..1

C (a, b, c) = \frac{i _{1} ( i _{1} - 1 ) ( i _{1} - 2 ) ( i _{1} - 3 )}{4} + \frac{i _{1} ( i _{1} - 1 ) i _{2}}{2} + i_{1} i_{3}, .2

C (a, b, c) = \frac{i _{1} ( i _{1} - 1 ) ( i _{1} - 2 ) ( i _{1} - 3 )}{4} + \frac{i _{1} ( i _{1} - 1 ) i _{2}}{2} + i_{1} i_{3}, .2

t (a, b, c; n) = \frac{2}{2 + C ( a , b , c )} N (a, b, c; 8 n + a + b + c) for a + b + c \leq 7 ..3

t (a, b, c; n) = \frac{2}{2 + C ( a , b , c )} N (a, b, c; 8 n + a + b + c) for a + b + c \leq 7 ..3

t (a, b, c; n) = \frac{2}{2 + C ( a , b , c )} (N (a, b, c; 8 n + 8) - N (a, b, c; 2 n + 2)) for a + b + c = 8 . .4

t (a, b, c; n) = \frac{2}{2 + C ( a , b , c )} (N (a, b, c; 8 n + 8) - N (a, b, c; 2 n + 2)) for a + b + c = 8 . .4

φ (q) = n = - \infty \sum \infty q^{n^{2}} = 1 + 2 n = 1 \sum \infty q^{n^{2}} and ψ (q) = n = 0 \sum \infty q^{n (n + 1) /2} (∣ q ∣ < 1) ..5

φ (q) = n = - \infty \sum \infty q^{n^{2}} = 1 + 2 n = 1 \sum \infty q^{n^{2}} and ψ (q) = n = 0 \sum \infty q^{n (n + 1) /2} (∣ q ∣ < 1) ..5

n = 0 \sum \infty N (a_{1}, \dots, a_{k}; n) q^{n} = φ (q^{a_{1}}) \dots φ (q^{a_{k}}), .6

n = 0 \sum \infty N (a_{1}, \dots, a_{k}; n) q^{n} = φ (q^{a_{1}}) \dots φ (q^{a_{k}}), .6

n = 0 \sum \infty t (a_{1}, \dots, a_{k}; n) q^{n} = 2^{k} ψ (q^{a_{1}}) \dots ψ (q^{a_{k}}) ..7

ψ (q)^{2} = φ (q) ψ (q^{2}), .8

ψ (q)^{2} = φ (q) ψ (q^{2}), .8

φ (q) = φ (q^{4}) + 2 q ψ (q^{8}) = φ (q^{16}) + 2 q^{4} ψ (q^{32}) + 2 q ψ (q^{8}), .9

φ (q)^{2} = φ (q^{2})^{2} + 4 q ψ (q^{4})^{2} = φ (q^{4})^{2} + 4 q^{2} ψ (q^{8})^{2} + 4 q ψ (q^{4})^{2}, .10

ψ (q) ψ (q^{3}) = φ (q^{6}) ψ (q^{4}) + q φ (q^{2}) ψ (q^{12}) ..11

φ (q)^{2} = φ (q^{8})^{2} + 4 q^{4} ψ (q^{16})^{2} + 4 q^{2} ψ (q^{8})^{2} + 4 q φ (q^{16}) ψ (q^{8}) + 8 q^{5} ψ (q^{8}) ψ (q^{32}) ..12

φ (q)^{2} = φ (q^{8})^{2} + 4 q^{4} ψ (q^{16})^{2} + 4 q^{2} ψ (q^{8})^{2} + 4 q φ (q^{16}) ψ (q^{8}) + 8 q^{5} ψ (q^{8}) ψ (q^{32}) ..12

φ (q) φ (q^{3}) = φ (q^{16}) φ (q^{48}) + 4 q^{16} ψ (q^{32}) ψ (q^{96}) + 2 q φ (q^{48}) ψ (q^{8}) + 2 q^{3} φ (q^{16}) ψ (q^{24}) + 6 q^{4} ψ (q^{8}) ψ (q^{24}) + 4 q^{13} ψ (q^{8}) ψ (q^{96}) + 4 q^{7} ψ (q^{24}) ψ (q^{32}) . .13

φ (q) φ (q^{3}) = φ (q^{16}) φ (q^{48}) + 4 q^{16} ψ (q^{32}) ψ (q^{96}) + 2 q φ (q^{48}) ψ (q^{8}) + 2 q^{3} φ (q^{16}) ψ (q^{24}) + 6 q^{4} ψ (q^{8}) ψ (q^{24}) + 4 q^{13} ψ (q^{8}) ψ (q^{96}) + 4 q^{7} ψ (q^{24}) ψ (q^{32}) . .13

t (2, 3, 3; n) = N (2, 3, 3; 8 n + 8) - N (2, 3, 3; 2 n + 2) .

t (2, 3, 3; n) = N (2, 3, 3; 8 n + 8) - N (2, 3, 3; 2 n + 2) .

t (a, b, c; n) = N (a, b, c; 8 n + a + b + c) - N (a, b, c; 2 n + (a + b + c) /4) .

t (a, b, c; n) = N (a, b, c; 8 n + a + b + c) - N (a, b, c; 2 n + (a + b + c) /4) .

N (2, 3, 3; 2 m) = N (1, 3, 3; m),

N (2, 3, 3; 2 m) = N (1, 3, 3; m),

t (2, 3, 3; 8 m + 6) = 2 t (1, 3, 3; m), t (2, 3, 3; 32 m + 27) = 4 t (1, 3, 3; m) .

t(2,3,3;n)=\left\{\,\vbox{\openup 3.0pt\halign{$\displaystyle{#}$\hfil&\quad$\displaystyle{{}#}$\hfil\cr\mathstrut\crcr\kern-12.0pt\cr 4N(1,3,3;n+1)&\text{if $n\equiv 0,1\pmod{4}$,}\\2N(1,3,3;n+1)&\text{if $n\equiv 2\pmod{8}$,}\\N(1,3,3;n+1)&\text{if $n\equiv 6\pmod{8}$ or $n\equiv 27\pmod{32}$,}\\\frac{8}{5}N(1,3,3;n+1)&\text{if $n\equiv 3,7\pmod{16}$,}\\\frac{4}{3}N(1,3,3;n+1)&\text{if $n\equiv 11\pmod{32}$.}\crcr\mathstrut\crcr\kern-12.0pt\cr}}\,\right.

t(2,3,3;n)=\left\{\,\vbox{\openup 3.0pt\halign{$\displaystyle{#}$\hfil&\quad$\displaystyle{{}#}$\hfil\cr\mathstrut\crcr\kern-12.0pt\cr 4N(1,3,3;n+1)&\text{if $n\equiv 0,1\pmod{4}$,}\\2N(1,3,3;n+1)&\text{if $n\equiv 2\pmod{8}$,}\\N(1,3,3;n+1)&\text{if $n\equiv 6\pmod{8}$ or $n\equiv 27\pmod{32}$,}\\\frac{8}{5}N(1,3,3;n+1)&\text{if $n\equiv 3,7\pmod{16}$,}\\\frac{4}{3}N(1,3,3;n+1)&\text{if $n\equiv 11\pmod{32}$.}\crcr\mathstrut\crcr\kern-12.0pt\cr}}\,\right.

n = 0 \sum \infty N (2, 3, 3; n) q^{n} = φ (q^{2}) φ (q^{3})^{2} = φ (q^{2}) (φ (q^{6})^{2} + 4 q^{3} ψ (q^{12})^{2}) .

n = 0 \sum \infty N (2, 3, 3; n) q^{n} = φ (q^{2}) φ (q^{3})^{2} = φ (q^{2}) (φ (q^{6})^{2} + 4 q^{3} ψ (q^{12})^{2}) .

m = 0 \sum \infty N (2, 3, 3; 2 m) q^{2 m} = φ (q^{2}) φ (q^{6})^{2}

m = 0 \sum \infty N (2, 3, 3; 2 m) q^{2 m} = φ (q^{2}) φ (q^{6})^{2}

m = 0 \sum \infty N (2, 3, 3; 2 m) q^{m} = φ (q) φ (q^{3})^{2} = m = 0 \sum \infty N (1, 3, 3; m) q^{m} .

m = 0 \sum \infty N (2, 3, 3; 2 m) q^{m} = φ (q) φ (q^{3})^{2} = m = 0 \sum \infty N (1, 3, 3; m) q^{m} .

φ (q^{3})^{2} = φ (q^{24})^{2} + 4 q^{12} ψ (q^{48})^{2} + 4 q^{3} φ (q^{48}) ψ (q^{24}) + 8 q^{15} ψ (q^{24}) ψ (q^{96}) + 4 q^{6} ψ (q^{24})^{2} . .1

φ (q^{3})^{2} = φ (q^{24})^{2} + 4 q^{12} ψ (q^{48})^{2} + 4 q^{3} φ (q^{48}) ψ (q^{24}) + 8 q^{15} ψ (q^{24}) ψ (q^{96}) + 4 q^{6} ψ (q^{24})^{2} . .1

\begin{aligned} &\sum_{m=0}^{\infty}N(1,3,3;m)q^{m}\\ &=\varphi(q)\varphi(q^{3})^{2}=(\varphi(q^{4})+2q\psi(q^{8}))(\varphi(q^{12})^{2}+4q^{6}\psi(q^{24})^{2}+4q^{3}\psi(q^{12})^{2})\\ &=(\varphi(q^{16})+2q\psi(q^{8})+2q^{4}\psi(q^{32}))\big{(}\varphi(q^{24})^{2}+4q^{12}\psi(q^{48})^{2}\\ &\qquad\qquad\qquad\qquad+4q^{3}\varphi(q^{48})\psi(q^{24})+8q^{15}\psi(q^{24})\psi(q^{96})+4q^{6}\psi(q^{24})^{2}\big{)}.\end{aligned}.2

\begin{aligned} &\sum_{m=0}^{\infty}N(1,3,3;m)q^{m}\\ &=\varphi(q)\varphi(q^{3})^{2}=(\varphi(q^{4})+2q\psi(q^{8}))(\varphi(q^{12})^{2}+4q^{6}\psi(q^{24})^{2}+4q^{3}\psi(q^{12})^{2})\\ &=(\varphi(q^{16})+2q\psi(q^{8})+2q^{4}\psi(q^{32}))\big{(}\varphi(q^{24})^{2}+4q^{12}\psi(q^{48})^{2}\\ &\qquad\qquad\qquad\qquad+4q^{3}\varphi(q^{48})\psi(q^{24})+8q^{15}\psi(q^{24})\psi(q^{96})+4q^{6}\psi(q^{24})^{2}\big{)}.\end{aligned}.2

m = 0 \sum \infty N (1, 3, 3; 4 m + 1) q^{4 m + 1} = 2 q ψ (q^{8}) φ (q^{12})^{2},

m = 0 \sum \infty N (1, 3, 3; 4 m + 1) q^{4 m + 1} = 2 q ψ (q^{8}) φ (q^{12})^{2},

m = 0 \sum \infty N (1, 3, 3; 4 m + 2) q^{4 m + 2} = 4 q^{6} ψ (q^{24})^{2} φ (q^{4})

m = 0 \sum \infty N (1, 3, 3; 4 m + 1) q^{m} = 2 φ (q^{3})^{2} ψ (q^{2}), .3

m = 0 \sum \infty N (1, 3, 3; 4 m + 1) q^{m} = 2 φ (q^{3})^{2} ψ (q^{2}), .3

m = 0 \sum \infty N (1, 3, 3; 4 m + 2) q^{m} = 4 q φ (q) ψ (q^{6})^{2} ..4

m = 0 \sum \infty N (1, 3, 3; 8 m + 3) q^{8 m + 3} = φ (q^{16}) \cdot 4 q^{3} φ (q^{48}) ψ (q^{24}) + 2 q^{4} ψ (q^{32}) \cdot 8 q^{15} ψ (q^{24}) ψ (q^{96}),

m = 0 \sum \infty N (1, 3, 3; 8 m + 3) q^{8 m + 3} = φ (q^{16}) \cdot 4 q^{3} φ (q^{48}) ψ (q^{24}) + 2 q^{4} ψ (q^{32}) \cdot 8 q^{15} ψ (q^{24}) ψ (q^{96}),

m = 0 \sum \infty N (1, 3, 3; 8 m + 7) q^{8 m + 7}

= φ (q^{16}) \cdot 8 q^{15} ψ (q^{24}) ψ (q^{96}) + 2 q ψ (q^{8}) \cdot 4 q^{6} ψ (q^{24})^{2} + 2 q^{4} ψ (q^{32}) \cdot 4 q^{3} φ (q^{48}) ψ (q^{24})

m = 0 \sum \infty N (1, 3, 3; 8 m + 3) q^{m} = 4 φ (q^{2}) φ (q^{6}) ψ (q^{3}) + 16 q^{2} ψ (q^{4}) \cdot ψ (q^{3}) ψ (q^{12}), .5

m = 0 \sum \infty N (1, 3, 3; 8 m + 3) q^{m} = 4 φ (q^{2}) φ (q^{6}) ψ (q^{3}) + 16 q^{2} ψ (q^{4}) \cdot ψ (q^{3}) ψ (q^{12}), .5

m = 0 \sum \infty N (1, 3, 3; 8 m + 7) q^{m} .6

= 8 q φ (q^{2}) ψ (q^{3}) ψ (q^{12}) + 8 ψ (q) ψ (q^{3})^{2} + 8 ψ (q^{4}) φ (q^{6}) ψ (q^{3}) = 16 ψ (q) ψ (q^{3})^{2} .

n = 0 \sum \infty t (2, 3, 3; n) q^{n} = 8 ψ (q^{2}) ψ (q^{3})^{2} = 8 ψ (q^{2}) φ (q^{3}) ψ (q^{6}) = 8 (φ (q^{12}) + 2 q^{3} ψ (q^{24})) (φ (q^{12}) ψ (q^{8}) + q^{2} φ (q^{4}) ψ (q^{24})) . .7

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TopicsAdvanced Mathematical Identities · Analytic Number Theory Research · Advanced Combinatorial Mathematics

Full text

Ramanujan’s theta functions and linear combinations of three triangular numbers

[TABLE]

Zhi-Hong Sun

School of Mathematical Sciences

Huaiyin Normal University

Huaian, Jiangsu 223300, P.R. China

Email: [email protected]

Homepage: http://www.hytc.edu.cn/xsjl/szh

Abstract

Let $\mathbb{Z}$ be the set of integers. For positive integers $a,b,c$ and $n$ let $N(a,b,c;n)$ be the number of representations of $n$ by $ax^{2}+by^{2}+cz^{2}$ , and let $t(a,b,c;n)$ be the number of representations of $n$ by $ax(x+1)/2+by(y+1)/2+cz(z+1)/2$ $(x,y,z\in\mathbb{Z})$ . In this paper, by using Ramanujan’s theta functions $\varphi(q)$ and $\psi(q)$ we reveal the relation between $t(2,3,3;n)$ and $N(1,3,3;n+1)$ , and the relation between $t(1,1,6;n)$ and $N(1,1,3;n+1)$ . We also obtain formulas for $t(a,3a,4b;n),$ $t(a,7a,4b;n),t(3a,5a,4b;n)$ and $t(a,15a,4b;n)$ under certain congruence conditions, where $a$ and $b$ are positive odd integers. In addition, we pose many conjectures on $t(a,b,c;n)$ for some special values of $(a,b,c)$ .

Keywords: theta function; triangular number; ternary form

Mathematics Subject Classification 2010: 11D85, 11E25, 30B10, 33E20

1. Introduction

Let $\mathbb{Z}$ , $\mathbb{Z}^{+}$ and $\mathbb{N}$ be the set of integers, the set of positive integers and the set of nonnegative integers, respectively, and let $\mathbb{Z}^{3}=\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$ and $\mathbb{N}^{3}=\mathbb{N}\times\mathbb{N}\times\mathbb{N}$ . For $a,b,c\in\mathbb{Z}^{+}$ and $n\in\mathbb{N}$ set

[TABLE]

Since $\frac{x(x+1)}{2}=\frac{(-1-x)(-x)}{2}$ we see that

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For $a,b,c\in\mathbb{Z}^{+}$ let

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where $i_{j}$ denotes the number of elements in $\{a,b,c\}$ which are equal to $j$ . In 2005 Adiga, Cooper and Han [ACH] showed that

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In 2008 Baruah, Cooper and Hirschhorn [BCH] proved that

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Ramanujan’s theta functions $\varphi(q)$ and $\psi(q)$ are defined by

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It is evident that for positive integers $a_{1},\ldots,a_{k}$ and $|q|<1$ ,

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There are many identities involving $\varphi(q)$ and $\psi(q)$ . From [BCH, Lemma 4.1] or [Be] we know that for $|q|<1$ ,

[TABLE]

By [S1, Lemma 2.4],

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By [S1, Lemma 2.3], for $|q|<1$ we have

[TABLE]

Using theta function identities we may establish some relations between $t(a,b,c;n)$ and $N(a,b,c;8n+a+b+c)$ for some certain values of $(a,b,c)$ . See [S2,S3].

In this paper we reveal the relation between $t(2,3,3;n)$ and $N(1,3,3;n+1)$ , and the relation between $t(1,1,6;n)$ and $N(1,1,3;n+1)$ . We also obtain formulas for $t(a,3a,4b;n),$ $t(a,7a,4b;n),t(3a,5a,4b;n)$ and $t(a,15a,4b;n)$ under certain congruence conditions, where $a$ and $b$ are positive odd integers. In addition, we pose many conjectures on $t(a,b,c;n)$ for some special values of $(a,b,c)$ .

2. The relation between $t(2,3,3;n)$ and $N(1,3,3;n+1)$

By (1.4), for $n\in\mathbb{Z}^{+}$ we have

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By [S2, Theorem 6.1], if $a,b,c,n\in\mathbb{Z}^{+}$ , $2\nmid ab$ , $4\mid a-b$ and $4\mid c-2$ , then

[TABLE]

Now we present the following formula for $t(2,3,3;n)$ .

*Theorem 2.1. *** For $m\in\mathbb{N}$ we have

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For $n\in\mathbb{Z}^{+}$ with $n\not\equiv 15\pmod{16}$ we have

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Proof. By (1.6) and (1.9),

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Thus,

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and so

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This yields $N(2,3,3;2m)=N(1,3,3;m)$ . By (1.12),

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Thus, using (1.9), (1.10) and the above we see that

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This yields

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and so

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Also, from (2.2) we deduce that

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and so

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On the other hand,

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Hence

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and so

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Combining (2.3),(2.4), (2.7) and (2.8) yields

[TABLE]

Hence $t(2,3,3;n)=4N(1,3,3;n+1)$ for $n\equiv 0,1\pmod{4}$ .

Now we consider the case $n\equiv 2,6\pmod{8}$ . By (2.10),

[TABLE]

Thus,

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and so

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Thus,

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Combining (2.5),(2.6),(2.12) and (2.13) yields

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That is,

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By (2.11),

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Hence

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Therefore,

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From (2.2) we see that

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and so

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Therefore,

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and so

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Combining (2.17)-(2.18) with (2.14) and (2.15) yields

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That is,

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From (2.16) we see that

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Thus,

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Hence

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Thus, $t(2,3,3;32m+27)=4t(1,3,3;m).$ From (2.19) we see that

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Thus,

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Therefore,

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This together with (2.20) and (2.21) gives

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This completes the proof.

3. The relation between $t(1,1,6;n)$ and $N(1,1,3;n+1)$

By (1.4), for $n\in\mathbb{Z}^{+}$ we have

[TABLE]

*Theorem 3.1. *** For $n\in\mathbb{Z}^{+}$ we have

[TABLE]

For $n\in\mathbb{Z}^{+}$ with $n\not\equiv 15\pmod{16}$ we have

[TABLE]

Proof. By (1.7)-(1.11),

[TABLE]

Thus,

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This yields

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Since $\varphi(q)=\varphi(q^{4})+2q\psi(q^{8})$ , from (3.1) we see that

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This yields

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and

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Therefore,

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and

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By (1.7) and (3.6), $t(1,1,6;8n+4)=2t(1,1,3;n)$ . By (3.4),

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Thus

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Replacing $q$ with $q^{1/4}$ yields

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By (3.7),

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Thus,

[TABLE]

and

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Therefore,

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From (3.11) and (1.7) we find that $t(1,1,6;32n+19)=4t(1,1,3;n)$ . By (1.6) and (1.10),

[TABLE]

Thus,

[TABLE]

and so $N(1,1,6;2n)=N(1,1,3;n)$ . On the other hand,

[TABLE]

Thus,

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Substituting $q$ with $q^{1/4}$ yields

[TABLE]

Combining (3.2) with (3.14) gives $t(1,1,6;4n+1)=4N(1,1,3;4n+2)$ , and combining (3.3) with (3.15) yields $t(1,1,6;4n+2)=4N(1,1,3;4n+3)$ . By (3.13),

[TABLE]

Thus,

[TABLE]

and

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Therefore,

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and

[TABLE]

Combining (3.5) with (3.16) gives $t(1,1,6;8n)=2N(1,1,3;8n+1)$ , and combining (3.6) with (3.17) yields $t(1,1,6;8n+4)=N(1,1,3;8n+5)$ . By (3.12),

[TABLE]

Thus,

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Therefore,

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Combining (3.8) with (3.19) yields $t(1,1,6;16n+7)=\frac{8}{5}N(1,1,3;16n+8)$ , and combining (3.9) with (3.20) gives $t(1,1,6;16n+11)=\frac{8}{5}N(1,1,3;16n+8)$ .

By (1.8)-(1.11),

[TABLE]

Now from (3.18) and the above we deduce that

[TABLE]

Therefore,

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Replacing $q$ with $q^{1/2}$ yields

[TABLE]

Combining (3.10) and (3.21) gives $t(1,1,6;32n+3)=\frac{4}{3}N(1,1,3;32n+4)$ , and combining (3.11) with (3.22) yields $t(1,1,6;32n+19)=N(1,1,3;32n+20)$ .

Summarizing the above proves the theorem.

4. Formulas for $t(1,3,4;n),\ t(1,3,12;n),\ t(1,3,16;n),$

$t(1,3,36;n),t(1,3,48;n)$ and $t(3,4,9;n)$

*Lemma 4.1. *** Let $a,b\in\mathbb{Z}^{+}$ with $2\nmid a$ . For $n\in\mathbb{Z}^{+}$ we have

[TABLE]

Proof. By (1.9) and (1.11),

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Thus,

[TABLE]

Substituting $q$ with $q^{1/4}$ and then applying (1.7) yields the result.

*Lemma 4.2. *** Let $a,b\in\mathbb{Z}^{+}$ with $2\nmid a$ . If $2\mid b$ , then

[TABLE]

If $2\nmid b$ , then

[TABLE]

Proof. By (1.13),

[TABLE]

Thus,

[TABLE]

Substituting $q$ with $q^{1/8}$ in the last two formulas yields the result in the case $2\mid b$ . Now suppose $2\nmid b$ . Substituting $q$ with $q^{1/4}$ in the first formula gives

[TABLE]

Thus,

[TABLE]

Replacing $q$ with $q^{1/2}$ yields

[TABLE]

By (1.11), $\psi(q^{a})\psi(q^{3a})=\varphi(q^{6a})\psi(q^{4a})+q^{a}\varphi(q^{2a})\psi(q^{12a})$ . Thus the result in the case $2\nmid b$ follows from the above. The proof is now complete.

*Theorem 4.1. *** For $n\in\mathbb{Z}^{+}$ we have

[TABLE]

Proof. By Lemma 4.1 and (1.8),

[TABLE]

Thus, $t(1,3,16;4n+1)=t(2,2,3;n)$ . By Lemma 4.2,

[TABLE]

Thus, $t(1,3,16;4n+1)=t(2,2,3;n)=2N(1,3,16;8n+7)$ . By Lemma 4.1,

[TABLE]

Thus, $t(1,3,48;4n)=t(1,6,6;n)$ . On the other hand, using Lemma 4.2 we see that

[TABLE]

Hence

[TABLE]

Therefore, $t(1,3,48;4n)=t(1,6,6;n)=2N(1,3,48;8n+13)$ .

By Lemma 4.1, $t(1,3,4;8n+3)=2t(1,3,4;2n)$ . By (1.11),

[TABLE]

Thus,

[TABLE]

and so

[TABLE]

On the other hand, using Lemma 4.2 we see that

[TABLE]

Thus,

[TABLE]

and so

[TABLE]

which yields $t(1,3,4;8n+3)=\frac{4}{3}N(1,3,4;16n+8)$ . By Lemma 4.1 and (1.11),

[TABLE]

Thus,

[TABLE]

On the other hand, from Lemma 4.2 we know that

[TABLE]

Thus,

[TABLE]

and so $t(1,3,4;8n+5)=\frac{4}{3}N(1,3,4;16n+12)$ .

By the above, the formulas for $t(1,3,16;n),t(1,3,48;n)$ and $t(1,3,4;n)$ are true. Using Lemmas 4.1-4.2 and similar arguments we may deduce the remaining results for $t(1,3,12;n),$ $t(1,3,36;n)$ and $t(3,4,9;n)$ .

5. Formulas for $t(1,4,7;n),\ t(1,7,12;n),\ t(1,7,28;n),$

$t(3,4,21;n)$ and $t(3,21,28;n)$

We begin with an identity involving $\psi(q)\psi(q^{7})$ .

*Lemma 5.1. *** For $q|<1$ we have

[TABLE]

Proof. By [Be, p.315],

[TABLE]

Thus,

[TABLE]

Combining (5.1) with (5.2) gives the result.

*Lemma 5.2. *** Suppose $a,b\in\mathbb{Z}^{+}$ with $2\nmid a$ . Then

[TABLE]

Proof. Note that $\sum_{n=0}^{\infty}t(a,7a,4b;n)q^{n}=8\psi(q^{a})\psi(q^{7a})\psi(q^{4b}).$ Using Lemma 5.1 we see that

[TABLE]

Replacing $q$ with $q^{1/4}$ yields the result.

*Theorem 5.1. *** For $n\in\mathbb{Z}^{+}$ we have

[TABLE]

Proof. We only prove the formula for $t(1,4,7;n)$ . The other formulas can be proved similarly. By (1.9),

[TABLE]

Thus,

[TABLE]

Replacing $q$ with $q^{1/8}$ gives

[TABLE]

By Lemma 5.2,

[TABLE]

Thus $t(1,4,7;4n+1)=2N(1,4,7;2(4n+1)+3)$ . This completes the proof.

6. Formulas for $t(3,5,4b;n)$ and $t(1,15,4b;n)$

In order to deduce formulas for $t(3,5,4b;n)$ and $t(1,15,4b;n)$ , we begin with two useful identities.

*Lemma 6.1. *** For $|q|<1$ we have

[TABLE]

We note that (6.1) can be found in [Be,p.377], and (6.2) was deduced by Xia and Zhang in [XZ].

*Lemma 6.2. *** For $|q|<1$ we have

[TABLE]

Proof. By (6.2),

[TABLE]

This together with (6.1) yields the first identity. By (6.1),

[TABLE]

Combining this with (6.2) gives the second identity.

*Lemma 6.3. *** Let $b\in\mathbb{Z}^{+}$ . Then $t(3,5,4b;4n+3)=t(3,5,b;n)$ for $n\in\mathbb{Z}^{+}$ , and

[TABLE]

Proof. Note that $\sum_{n=0}^{\infty}t(3,5,4b;n)q^{n}=8\psi(q^{3})\psi(q^{5})\psi(q^{4b})$ . In view of Lemma 6.2 we see that

[TABLE]

Replacing $q$ with $q^{1/4}$ yields (6.3). Similarly,

[TABLE]

and so

[TABLE]

which yields $t(3,5,4b;4n+3)=t(3,5,b;n)$ . So the lemma is proved.

*Lemma 6.4. *** Suppose $b\in\mathbb{Z}^{+}$ with $2\nmid b$ . Then

[TABLE]

Proof. By (1.9),

[TABLE]

Thus,

[TABLE]

Now replacing $q$ with $q^{1/8}$ yields the result.

*Theorem 6.1. *** For $n\in\mathbb{Z}^{+}$ we have

[TABLE]

Proof. By Lemma 6.4,

[TABLE]

This together with Lemma 6.3 (with $b=1$ ) yields

[TABLE]

By Lemma 6.3,

[TABLE]

By Lemma 6.4 and the above,

[TABLE]

Hence $t(3,5,20;4n+5)=2N(3,5,20;8n+17)$ . The remaining results for $t(3,5,36;n)$ and $t(3,5,12;n)$ can be proved similarly.

*Theorem 6.2. *** For $n\in\mathbb{Z}^{+}$ we have

[TABLE]

Proof. Suppose $a,b\in\mathbb{Z}^{+}$ with $2\nmid ab$ . Then

[TABLE]

Applying Lemma 6.2 we see that

[TABLE]

Substituting $q$ with $q^{1/4}$ yields

[TABLE]

On the other hand,

[TABLE]

Thus,

[TABLE]

Replacing $q$ with $q^{1/8}$ yields

[TABLE]

Taking $a=b=1$ in (6.5) and (6.7) gives

[TABLE]

which implies $t(1,4,15;4n)=2N(1,4,15;8n+5)$ . Taking $a=1$ and $b=3$ in (6.6) yields

[TABLE]

Taking $a=1$ and $b=3$ in (6.7) yields

[TABLE]

Thus, $t(1,12,15;4n+2)=2N(1,12,15;8n+11)$ . The remaining parts for $t(1,15,20;n)$ and $t(3,4,45;n)$ can be proved similarly.

7. Some conjectures on $t(a,b,c;n)$

Let $a,b,c,n\in\mathbb{Z}^{+}$ . Based on calculations on Maple, in this section we pose many conjectures on $t(a,b,c;n)$ .

*Conjecture 7.1. *** Let $n\in\mathbb{Z}^{+}$ with $n\not\equiv 47,63\pmod{64}$ . Then

[TABLE]

*Conjecture 7.2. *** Let $n\in\mathbb{Z}^{+}$ with $n\not\equiv 14\pmod{16}$ . Then

[TABLE]

*Conjecture 7.3. *** Let $n\in\mathbb{Z}^{+}$ with $n\not\equiv 14\pmod{16}$ . Then

[TABLE]

*Conjecture 7.4. *** Let $n\in\mathbb{Z}^{+}$ and

[TABLE]

$(\text{\rm i})$ * If $n\equiv 1\pmod{3}$ , then $r(n)=0$ .*

$(\text{\rm ii})$ * If $n\equiv 0\pmod{3}$ , then*

[TABLE]

$(\text{\rm iii})$ * If $n\equiv-1\pmod{3}$ , then*

[TABLE]

*Conjecture 7.5. *** Let $n\in\mathbb{Z}^{+}$ with $n\equiv 1\pmod{8}$ . Then

[TABLE]

*Conjecture 7.6. *** Let $n\in\mathbb{Z}^{+}$ with $n\equiv 5\pmod{8}$ . Then

[TABLE]

*Conjecture 7.7. *** Let $n\in\mathbb{Z}^{+}$ with $n\equiv 3\pmod{8}$ . Then

[TABLE]

*Conjecture 7.8. *** Let $n\in\mathbb{Z}^{+}$ with $n\equiv 7\pmod{8}$ . Then

[TABLE]

*Conjecture 7.9. *** Let $n\in\mathbb{Z}^{+}$ with $n\equiv 0\pmod{8}$ . Then

[TABLE]

*Conjecture 7.10. *** Let $n\in\mathbb{Z}^{+}$ with $n\equiv 4\pmod{8}$ . Then

[TABLE]

*Conjecture 7.11. *** Let $n\in\mathbb{Z}^{+}$ with $n\equiv 2\pmod{8}$ . Then

[TABLE]

*Conjecture 7.12. *** Let $n\in\mathbb{Z}^{+}$ with $n\equiv 6\pmod{8}$ . Then

[TABLE]

Bibliography7

The reference list from the paper itself. Each links out to its DOI / PubMed record.

1[ACH] C. Adiga, S. Cooper and J. H. Han, A general relation between sums of squares and sums of triangular numbers , Int. J. Number Theory 1 (2005), 175-182.
2[BCH] N. D. Baruah, S. Cooper and M. Hirschhorn, Sums of squares and sums of triangular numbers induced by partitions of 8 8 8 , Int. J. Number Theory 4 (2008), 525-538.
3[Be] B.C. Berndt, Ramanujan’s Notebooks , Part III, Springer, New York, 1991.
4[S 1] Z.H. Sun, Some relations between t ( a , b , c , d ; n ) 𝑡 𝑎 𝑏 𝑐 𝑑 𝑛 t(a,b,c,d;n) and N ( a , b , c , d ; n ) 𝑁 𝑎 𝑏 𝑐 𝑑 𝑛 N(a,b,c,d;n) , Acta Arith. 175 (2016), 269-289.
5[S 2] Z.H. Sun, Ramanujan’s theta functions and sums of triangular numbers , Int. J. Number Theory, accepted.
6[S 3] Z.H. Sun, On the number of representations of n 𝑛 n as a linear combination of triangular numbers , Int. J. Number Theory, accepted.
7[XZ] E.X.W. Xia and Y. Zhang, Proofs of some conjectures of Sun on the relations between sums of squares and sums of triangular numbers , Int. J. Number Theory, accepted.

TL;DR

Contribution

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Abstract

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Abstract

1. Introduction

2. The relation between t(2,3,3;n)t(2,3,3;n)t(2,3,3;n) and N(1,3,3;n+1)N(1,3,3;n+1)N(1,3,3;n+1)

3. The relation between t(1,1,6;n)t(1,1,6;n)t(1,1,6;n) and N(1,1,3;n+1)N(1,1,3;n+1)N(1,1,3;n+1)

4. Formulas for t(1,3,4;n), t(1,3,12;n), t(1,3,16;n),t(1,3,4;n),\ t(1,3,12;n),\ t(1,3,16;n),t(1,3,4;n), t(1,3,12;n), t(1,3,16;n),

5. Formulas for t(1,4,7;n), t(1,7,12;n), t(1,7,28;n),t(1,4,7;n),\ t(1,7,12;n),\ t(1,7,28;n),t(1,4,7;n), t(1,7,12;n), t(1,7,28;n),

6. Formulas for t(3,5,4b;n)t(3,5,4b;n)t(3,5,4b;n) and t(1,15,4b;n)t(1,15,4b;n)t(1,15,4b;n)

7. Some conjectures on t(a,b,c;n)t(a,b,c;n)t(a,b,c;n)

2. The relation between $t(2,3,3;n)$ and $N(1,3,3;n+1)$

3. The relation between $t(1,1,6;n)$ and $N(1,1,3;n+1)$

4. Formulas for $t(1,3,4;n),\ t(1,3,12;n),\ t(1,3,16;n),$

5. Formulas for $t(1,4,7;n),\ t(1,7,12;n),\ t(1,7,28;n),$

6. Formulas for $t(3,5,4b;n)$ and $t(1,15,4b;n)$

7. Some conjectures on $t(a,b,c;n)$