On Graham partitions twisted by the Legendre symbol

Byungchan Kim; Ji Young Kim; Chong Gyu Lee; Sang June Lee; Poo-Sung Park; Yoon Kyung Park

doi:10.1515/math-2023-0134

Article Open Access

On Graham partitions twisted by the Legendre symbol

Byungchan Kim , Ji Young Kim , Chong Gyu Lee , Sang June Lee , Poo-Sung Park and Yoon Kyung Park

Published/Copyright: October 26, 2023

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$Open Mathematics$

From the journal Open Mathematics Volume 21 Issue 1

Abstract

We investigate when there is a partition of a positive integer n ,

n = f ( λ 1 ) + f ( λ 2 ) + ⋯ + f ( λ ℓ ) ,

satisfying that

1 = χ p ( λ 1 ) λ 1 + χ p ( λ 2 ) λ 2 + ⋯ + χ p ( λ ℓ ) λ ℓ ,

where χ p is the Legendre symbol modulo prime p and f ( k ) = k or the k th m -gonal number with m = 3 , 4, or 5.

Keywords: Graham partition; sum of reciprocals; quadratic twist; Legendre symbol; polygonal number

MSC 2010: 11P81; 05A17

1 Introduction

In a seminal article [1], Graham initiated a quest to search for a partition of a positive integer n ,

n = λ 1 + λ 2 + ⋯ + λ ℓ ,

satisfying that

1 = 1 λ 1 + 1 λ 2 + ⋯ + 1 λ ℓ .

In a recent article by Kim et al. [2] in honor of Graham, we call such a partition a Graham partition. For example, ( 12 , 6 , 4 , 2 ) is a Graham partition of 24 as

24 = 12 + 6 + 4 + 2 and 1 = 1 12 + 1 6 + 1 4 + 1 2 .

In 1963, Graham [1] proved that there is a Graham partition of n into distinct parts, provided that n ≥ 78 . Though there is no written proof of the ordinary partition version, by adopting Graham’s idea, one can easily prove that there is a Graham partition of a positive integer n for n ≥ 24 . The number of ordinary Graham partitions is listed at OEIS [3].

A common theme in the theory of q -series and partitions considers a weight by a quadratic character. For example, due to a connection with the Rogers-Ramanujan continued fraction, Berndt and Sohn [4] studied the asymptotic expansion of

∏ n = 1 ∞ ( 1 − x n ) χ 5 ( n ) ,

where χ 5 ( n ) is the Legendre symbol modulo 5. A quadratic twist is also common in the theory of elliptic curves and modular forms [5–7].

This article deals with Graham partitions twisted by the Legendre symbol modulo odd primes. Let χ p ( n ) be the Legendre symbol modulo p , and let G χ p ( n ) be the set of partitions λ = ( λ 1 , λ 2 , … , λ ℓ ) such that

n = ∑ i = 1 ℓ λ i and 1 = ∑ i = 1 ℓ χ p ( λ i ) λ i .

We call a partition λ ∈ G χ p ( n ) a Graham partition twisted by χ p . For example, ( 6 , 4 , 4 , 3 , 1 ) ∈ G χ 7 ( 18 ) as

18 = 6 + 4 + 4 + 3 + 1 and 1 = − 1 6 + 2 ⋅ 1 4 − 1 3 + 1 .

To show the nonemptiness of the set G χ p ( n ) for a fixed prime p , one can adopt Graham’s idea [1], which uses recurrence relations. For example, when p = 13 , we can observe that if λ ∈ G χ 13 ( n ) , then

3 λ ∪ ( 3 , 3 ) ∈ G χ 13 ( 3 n + 6 ) , 3 λ ∪ ( 3 , 3 , 2 , 2 , 1 ) ∈ G χ 13 ( 3 n + 11 ) ,

and

3 λ ∪ ( 6 , 6 , 1 ) ∈ G χ 13 ( 3 n + 13 ) ,

where 3 λ is the partition consisting of three times of parts in λ and the partition λ ∪ μ is one whose parts are a union of parts from λ and μ . While the idea of using recurrence relations is powerful for a fixed prime p , the nature of the proof is ad hoc on p , so we need a different idea to show the existence of Graham partitions twisted by χ p for varying primes p . Indeed, the set G χ p ( n ) is a nonempty set for odd primes p ≠ 5 and for a sufficiently large integer n . On the other hand, the set G χ 5 ( n ) is empty unless n is in a certain arithmetic progression.

Theorem 1

The set G χ 5 ( n ) is not empty if and only if n is congruent to 1 modulo 5. For an odd prime p ≠ 5 and an integer n ≥ n χ p , the set G χ p ( n ) is not empty, where

n χ p = 18.8 p 3 ⁄ 2 ( log p ) 2 .

Remark

Our proof is constructive so that we can derive a sharper n χ p for a fixed prime p . For example, we can obtain n χ 3 = 9 (resp. n χ 7 = 55 ) by taking α = 2 (resp. α = 3 ) in the proof of Theorem 1.

We remark that the lower bound n χ p is far from the tight bound for the nonemptiness of G χ p ( n ) . Let N χ p be the smallest integer such that G χ p ( n ) is nonempty for integers n ≥ N χ p . Table 1 provides some numerics on N χ p and n χ p .

Table 1

Lower bounds in Theorem 1

p	N χ p	n χ p
3	9	118
7	14	1,319
13	19	5,798

One can verify N χ p by checking G χ p ( n ) for n < n χ p . However, the computational cost for the exhaustive search could be expensive for large primes p as the number of partitions of n grows exponentially.

Motivated by Euler’s partition identity, i.e., the number of partitions of n into distinct parts is equal to the number of partitions of n into odd parts, and Kim et al. [2] investigated the existence of a Graham partition of n into odd parts. Here, we show that there is also a twisted Graham partition of n into odd parts.

Corollary 2

For an odd prime p ≥ 7 and an integer n ≥ p ( p + 6 ) 2 , there is a partition λ ∈ G χ p ( n ) such that all parts of λ are odd.

Graham [1] proposed a general conjecture, which claims that for a sufficiently large integer n , there is a partition of n , i.e.,

n = f ( k 1 ) + f ( k 2 ) + ⋯ + f ( k ℓ ) ,

satisfying that

1 = 1 k 1 + 1 k 2 + ⋯ + 1 k ℓ ,

where f ( k ) is an integer-valued positive definite polynomial.

In the case f ( k ) = k 2 , Alekseyev [8] proved that for integers n ≥ 8,543 , there are positive integers k 1 > k 2 > ⋯ > k ℓ satisfying that

n = k 1 2 + k 2 2 + ⋯ + k ℓ 2 and 1 = 1 k 1 + 1 k 2 + ⋯ + 1 k ℓ .

The main idea of the proof is that there is a recurrence relation that generates a Graham partition of a larger integer from a given Graham partition of an integer. However, this idea seems infeasible to apply to general polynomials other than f ( k ) = k 2 .

Here, we consider a twisted version of the original conjecture for the case that f ( k ) is the k th m -gonal number:

f ( k ) = P m ( k ) ≔ ( m − 2 ) k 2 − ( m − 4 ) k 2 .

To be more specific, we search for a partition of a positive integer n satisfying that

(1.1) n = P m ( k 1 ) + P m ( k 2 ) + ⋯ + P m ( k ℓ ) , 1 = χ p ( k 1 ) k 1 + χ p ( k 2 ) k 2 + ⋯ + χ p ( k ℓ ) k ℓ .

We define G m , χ p ( n ) to be the set of partitions satisfying equation (1.1). For example,

( 3 , 3 , 3 , 3 , 1 , 1 , 1 ) ∈ G 3 , χ 3 ( 15 )

15 = 3 + 3 + 3 + 3 + 1 + 1 + 1 and 1 = 4 ⋅ χ 3 ( 2 ) 2 + 3 ⋅ χ 3 ( 1 ) 1 .

By incorporating the character χ p , one can find more algorithms to generate a partition of G m , χ p ( n ) from such a partition of a smaller integer. Here, we prove that G m , χ p ( n ) is nonempty for a large enough integer n for m = 3 , 4, or 5.

Theorem 3

The following statements hold true:

For an odd prime p and n ≥ n 3 , χ p = p 6 ⁄ 4 , the set G 3 , χ p ( n ) is not empty.
For an odd prime p ≠ 3 , 7 and n ≥ n 4 , χ p , the set G 4 , χ p ( n ) is not empty, where
n 4 , χ p = 193 , if p = 5 , p 5 , otherwise.
For an odd prime p and n ≥ n 5 , χ p , the set G 5 , χ p ( n ) is not empty, where
n 5 , χ p = 341 , if p = 5 , 9 p 5 ⁄ 4 , otherwise.

Here, for simplicity, we roughly state n m , χ p in terms of p . A sharper bound n ˜ m , χ p will be given in Section 3 during the proof. As in Theorem 1, the bounds n m , χ p in Theorem 3 are far from the tight lower bounds. Let N m , χ p be the smallest integer such that G m , χ p ( n ) is not empty for all n ≥ N m , χ p . Then, Table 2 provides some values of N m , χ p and n m , χ p .

Table 2

Lower bounds in Theorem 3

m	p	N m , χ p	n m , χ p
3	3	24	183
4	5	96	193
5	3	78	547

We verify N 4 , χ 5 using the recurrence relations that if λ ∈ G 4 , χ 5 ( n ) , then

4 λ ∪ ( 4 , 1 , 1 ) ∈ G 4 , χ 5 ( 4 n + 6 ) , 4 λ ∪ ( 4 , 4 , 4 , 1 , 1 , 1 ) ∈ G 4 , χ 5 ( 4 n + 15 ) , 4 λ ∪ ( 4 , 4 , 4 , 4 , 4 , 1 , 1 , 1 , 1 ) ∈ G 4 , χ 5 ( 4 n + 24 ) , and 4 λ ∪ ( 16 , 16 , 1 ) ∈ G 4 , χ 5 ( 4 n + 33 ) .

Since there is no known recurrence relation for the case m ≠ 4 , computational cost to verify N m , χ p could be very expensive, even for small primes p .

The rest of this article is organized as follows. In Section 2, we prove Theorem 1 and Corollary 2; in Section 3, we prove Theorem 3 and give a sharper lower bound n ˜ m , χ p than n m , χ p . We conclude the article with some remarks in Section 4.

2 Proof of Theorem 1

We first provide a famous result of Sylvester [9,10], which we use to find the lower bound of an integer n having a twisted Graham partition. Let F ( a 1 , a 2 ) be the Frobenius number of positive integers a 1 and a 2 , that is, the largest integer b such that

a 1 x 1 + a 2 x 2 = b

does not hold for nonnegative integers x 1 and x 2 .

Lemma 4

(Sylvester [9, 10]) If ( a 1 , a 2 ) = 1 , then F ( a 1 , a 2 ) = ( a 1 − 1 ) ( a 2 − 1 ) − 1 .

The following two lemmas are useful to determine the lower bound n χ p of n with nonempty set G χ p ( n ) .

Lemma 5

(Treviño [11, Theorem 1.2]) Let p > 3 be an odd prime and k ≥ 2 be an integer such that k ∣ p − 1 . Let g ( p , k ) be the least kth power nonresidue modulo p . Then,

g ( p , k ) ≤ 1.1 p 1 ⁄ 4 log p , i f k = 2 a n d p ≡ 3 ( mod 4 ) , 0.9 p 1 ⁄ 4 log p , o t h e r w i s e .

Lemma 6

(Treviño [12, 13, Theorem A]) If χ is any nonprincipal Dirichlet character to the prime modulus p, which is constant on ( N , N + H ] , then

H < π 2 e 3 + o ( 1 ) p 1 ⁄ 4 log p ,

where the o ( 1 ) term depends only on p . Furthermore,

H ≤ 1.55 p 1 ⁄ 4 log p , f o r p ≥ 1 0 13 , 3.38 p 1 ⁄ 4 log p , f o r a l l o d d p .

Since the proof of Theorem 1 is lengthy, we first consider the case p = 5 as a proposition.

Proposition 7

The set G χ 5 ( n ) is not empty if and only if n is congruent to 1 modulo 5.

Proof

We first assume that n is a positive integer congruent to 1 modulo 5. Let k be a nonnegative integer such that n = 5 k + 1 . Then, λ = ( 5 , … , 5 , 1 ) is a partition of n with k + 1 parts, which is in G χ 5 ( n ) .

Next, assume that λ ∈ G χ 5 ( n ) . We may rewrite the parts λ i of λ as a i , b i , c i , d i , and e i such that

a i ≡ 1 , b i ≡ 2 , c i ≡ 3 , d i ≡ 4 , e i ≡ 0 ( mod 5 ) .

Then, we have

(2.1) n = ∑ i = 1 r 1 a i + ∑ i = 1 r 2 b i + ∑ i = 1 r 3 c i + ∑ i = 1 r 4 d i + ∑ i = 1 r 5 e i

and

(2.2) 1 = ∑ i = 1 r 1 1 a i − ∑ i = 1 r 2 1 b i − ∑ i = 1 r 3 1 c i + ∑ i = 1 r 4 1 d i .

Let

N = ∏ i = 1 r 1 a i ∏ i = 1 r 2 b i ∏ i = 1 r 3 c i ∏ i = 1 r 4 d i ,

and then, N ≡ ( − 1 ) r 3 + r 4 2 r 2 + r 3 ( mod 5 ) .

Multiplying both sides of equation (2.2) by N , we see that

( − 1 ) r 3 + r 4 2 r 2 + r 3 ≡ ( − 1 ) r 3 + r 4 2 r 2 + r 3 ( r 1 − r 4 ) − ( − 1 ) r 3 + r 4 2 r 2 + r 3 − 1 ( r 2 − r 3 ) ( mod 5 ) ,

which implies that ( r 1 − r 4 ) + 2 ( r 2 − r 3 ) ≡ 1 ( mod 5 ) . Because equation (2.1) gives

n ≡ ( r 1 − r 4 ) + 2 ( r 2 − r 3 ) ( mod 5 ) ,

we complete our proof.□

We now turn to the case that p is an odd prime other than 5.

Proof of Theorem 1

We first claim that, for an odd prime p ≠ 5 , there is an integer α ∈ { 2 , 3 , … , p − 1 } such that

(2.3) χ p ( α ) = − 1 and α 2 + 1 ≢ 0 ( mod p ) .

Indeed, for p = 3 , we can take α = 2 . When p ≥ 7 , there are ( p − 1 ) ⁄ 2 ≥ 3 quadratic nonresidues modulo p , while there are at most two solutions for x 2 + 1 ≡ 0 ( mod p ) .

From now on, we may assume α as the smallest integer satisfying equation (2.3). Lemma 4 gives that F ( α 2 + 1 , p ) = α 2 ( p − 1 ) − 1 . For n ≥ α 2 ( p − 1 ) + 1 , there are nonnegative integers j 1 and j 2 such that

n = ( α 2 + 1 ) j 1 + p j 2 + 1 .

The partition

λ = ( α , α , … , α ︸ ) α j 1 ∪ ( 1 , 1 , … , 1 ︸ ) j 1 + 1 ∪ ( p , p , … , p ︸ ) j 2

is a twisted Graham partition of n because

(2.4) ∑ i χ p ( λ i ) λ i = α j 1 ⋅ χ p ( α ) α + ( j 1 + 1 ) ⋅ χ p ( 1 ) 1 + j 2 ⋅ χ p ( p ) p = 1 .

Now we estimate α to obtain n χ p . First, suppose that p ≡ 3 ( mod 4 ) . Note that there is no solution for x 2 + 1 ≡ 0 ( mod p ) . By Lemma 5, the smallest quadratic nonresidue α satisfies that

α < 1.1 p 1 4 log p .

Next, suppose that p ≡ 1 ( mod 4 ) . Lemma 5 says that the smallest quadratic nonresidue q 1 satisfies the following:

q 1 < 0.9 p 1 4 log p .

If q 1 is not a solution of x 2 + 1 ≡ 0 ( mod p ) , then it is the desired α . Otherwise, we obtain the second-smallest quadratic nonresidue q 2 . If q 2 = q 1 + 1 , then

q 1 , q 2 < 0.9 p 1 4 log p + 1 < p − 1 2 .

If both q 1 and q 2 are solutions of x 2 + 1 ≡ 0 ( mod p ) , then

2 q 1 + 1 < 1.8 p 1 4 log p + 1 < p − 1 ,

which is a contradiction.

If q 2 > q 1 + 1 , then

χ p ( q 1 + 1 ) = ⋯ = χ p ( q 2 − 1 ) = − 1 ,

so that

q 2 − q 1 − 1 < 3.38 p 1 4 log p

by Lemma 6. Hence, we obtain

q 2 < 4.28 p 1 4 log p + 1 < p − 1 2 ,

so that not all of q 1 and q 2 are solutions of x 2 + 1 ≡ 0 ( mod p ) . So, we can take α = q 1 or q 2 .

At any case, we have α ≤ 4.28 p 1 4 log p + 1 . We arrive at

□ F ( α 2 + 1 , p ) + 2 ≤ 4.28 p 1 4 log p + 1 2 ( p − 1 ) + 1 ≤ 18.8 ( log p ) 2 p 3 ⁄ 2 ≕ n χ p .

Now we see if there is a partition λ ∈ G χ p ( n ) consisting of only odd parts. To this end, we use the following lemmas.

Lemma 8

(Somer [14, Lemma 2]) Let N 1 ( p ) be the number of odd quadratic residues modulo p. If p is congruent to 1 modulo 4, then N 1 ( p ) = ( p − 1 ) ⁄ 4 .

Lemma 9

(Knapp et al. [15, Theorem 2]) Let p be a prime ≥ 29 and ν p be the smallest odd quadratic nonresidue modulo p. Then,

ν p ≤ p − 12 ,

except for p = 59 , 109, or 131.

Now we are ready to prove Corollary 2.

Proof of Corollary 2

It is suffices to choose an odd integer α ∈ { 3 , … , p − 2 } satisfying equation (2.3). Note that all parts of λ are 1, α or p which are odd. Hence we construct a twisted Graham partition λ only with odd parts. After taking such an α , one can proceed exactly as in the proof of Theorem 1.

Suppose that p ≡ 1 ( mod 4 ) , and then p ≥ 13 as p ≠ 5 . Then, Lemma 8 gives that the number of odd quadratic nonresidues modulo p is

p − 1 2 − N 1 ( p ) = p − 1 4 ≥ 3 ,

where N 1 ( p ) is given in Lemma 8. Because there are at most two solutions for x 2 + 1 ≡ 0 ( mod p ) , there is an odd integer α satisfying equation (2.3).

Suppose that p ≡ 3 ( mod 4 ) . We may take α = p − 4 because p − 4 is odd,

χ p ( p − 4 ) = χ p ( − 1 ) = − 1 and ( p − 4 ) 2 + 1 ≡ 17 ≢ 0 ( mod p ) .

Let q be the smallest odd quadratic nonresidue modulo p . Then, by Lemma 9, we find that q ≤ p − 12 when p ≥ 137 . Since q 2 < p − 1 , it is clear that q 2 + 1 ≢ 0 ( mod p ) . By checking primes up to 131, we verify that we can choose α ≤ p − 1 + 6 . Therefore, we can conclude that

F ( α 2 + 1 , p ) + 2 ≤ ( p − 1 + 6 ) 2 ( p − 1 ) + 2 ≤ p ( p + 6 ) 2

is desired.□

3 Proof of Theorem 3

Before proving Theorem 3, we start with a lemma.

Lemma 10

Let p be an odd prime, and let m ≥ 3 be an integer. Assume that there is a positive integer α satisfying the following conditions:

α is a quadratic nonresidue modulo p.
α P m ( α ) + 1 is relatively prime to P m ( t p ) for a positive integer t.

Then, the set G m , χ p ( n ) is not empty for n ≥ α P m ( α ) ( P m ( t p ) − 1 ) + 1 .

Proof

Suppose that α is a positive integer satisfying (a) and (b) and also that

n ≥ α P m ( α ) ( P m ( t p ) − 1 ) + 1 .

By assumption (b), we have

α P m ( α ) ( P m ( t p ) − 1 ) + 1 = F ( α P m ( α ) + 1 , P m ( t p ) ) + 2 ,

and there are nonnegative integers j 1 and j 2 such that

n − 1 = ( α P m ( α ) + 1 ) j 1 + P m ( t p ) j 2 .

Define the partition λ of n as follows:

λ = ( P m ( α ) , P m ( α ) , … , P m ( α ) ︸ ) α j 1 ∪ ( P m ( 1 ) , P m ( 1 ) , … , P m ( 1 ) ︸ ) j 1 + 1 ∪ ( P m ( t p ) , P m ( t p ) , … , P m ( t p ) ︸ ) j 2 .

Then, λ ∈ G m , χ p ( n ) , because

□ ∑ i χ p ( k i ) k i = α j 1 ⋅ χ p ( α ) α + ( j 1 + 1 ) ⋅ χ p ( 1 ) 1 + j 2 ⋅ χ p ( t p ) p = 1 .

For each case m = 3 , 4, or 5, we present a separate proposition to provide more details on lower bounds.

Proposition 11

For an odd prime p and an integer n ≥ n ˜ 3 , χ p , there is a partition in G 3 , χ p ( n ) , where

n ˜ 3 , χ p = α P 3 ( α ) ( α P 3 ( α ) + 5 ) + 1 , if p ≡ 1 ( mod 8 ) , 6 P 3 ( p ) − 5 , if p ≡ 5 ( mod 8 ) a n d ( p + 1 , 7 ) = 1 , 6 P 3 ( 2 p ) − 5 , if p ≡ 5 ( mod 8 ) a n d ( p + 1 , 7 ) = 7 , ( p − 1 ) P 3 ( p − 1 ) ( P 3 ( p ) − 1 ) + 1 , if p ≡ 3 ( mod 4 ) ,

where α is the smallest positive integer satisfying (a) and (b) in Lemma 10 with m = 3 and t = 1 .

Proof

First, let p ≡ 1 ( mod 8 ) and α be the smallest quadratic nonresidue modulo p . Then, α is a prime number greater than 2 because χ p ( 2 ) = 1 . Note that α P 3 ( α ) + 1 = α 2 ( α + 1 ) ⁄ 2 + 1 ≡ 1 , 2 , or 4 ( mod 5 ) , and thus

( α P 3 ( α ) + 1 , α P 3 ( α ) + 6 ) = ( α P 3 ( α ) + 1 , 5 ) = 1 .

Therefore, for n ≥ F ( α P 3 ( α ) + 1 , α P 3 ( α ) + 6 ) + 2 , there exist nonnegative integers j 1 and j 2 such that

n = ( α P 3 ( α ) + 1 ) j 1 + ( α P 3 ( α ) + 6 ) j 2 + 1 = ( j 1 + 1 ) P 3 ( 1 ) + 2 j 2 P 3 ( 2 ) + ( α j 1 + α j 2 ) P 3 ( α ) .

We note that

λ = ( P 3 ( 1 ) , P 3 ( 1 ) , … , P 3 ( 1 ) ︸ ) j 1 + 1 ∪ ( P 3 ( 2 ) , P 3 ( 2 ) , … , P 3 ( 2 ) ︸ ) 2 j 2 ∪ ( P 3 ( α ) , P 3 ( α ) , … , P 3 ( α ) ︸ ) α j 1 + α j 2

is in G 3 , χ p ( n ) as it satisfies

∑ χ p ( k i ) k i = ( j 1 + 1 ) ⋅ χ p ( 1 ) 1 + 2 j 2 ⋅ χ p ( 2 ) 2 + ( α j 1 + α j 2 ) ⋅ χ p ( α ) α = 1 .

Now we assume that p ≡ 5 ( mod 8 ) and ( p + 1 , 7 ) = 1 . Then, we choose α = 2 in Lemma 10 with t = 1 because ( 2 P 3 ( 2 ) + 1 , P 3 ( p ) ) = ( 7 , p ( p + 1 ) ⁄ 2 ) = 1 and 2 is always a quadratic nonresidue modulo p . If p ≡ 5 ( mod 8 ) and ( p + 1 , 7 ) = 7 , then we choose α = 2 in Lemma 10 with m = 3 and t = 2 . Thus, we have a desirable partition for all n greater than

F ( 2 P 3 ( 2 ) + 1 , P 3 ( 2 p ) ) + 2 .

Finally, we focus on p ≡ 3 ( mod 4 ) . For α = p − 1 , let d = ( α P 3 ( α ) + 1 , P 3 ( p ) ) . Then, d is a divisor of p 2 because

p ( α P 3 ( α ) + 1 ) − ( p 2 − 3 p + 2 ) P 3 ( p ) = p 2 .

Since d is a divisor of α P 3 ( α ) + 1 , which is equivalent to 1 modulo p , we have that d = 1 . Moreover χ p ( p − 1 ) = − 1 , α = p − 1 , satisfies conditions (a) and (b) in Lemma 10.□

Now we investigate G 4 , χ p ( n ) . When p = 3 or p = 7 , we obtain a sufficient and necessary condition for the nonemptyness of G 4 , χ p ( n ) .

Proposition 12

The following statements hold true:

The set G 4 , χ 3 ( n ) is not empty if and only if n ≡ 1 ( mod 9 ) .
The set G 4 , χ 7 ( n ) is not empty if and only if n ≡ 1 ( mod 7 ) and n ≠ 15 , 22 .

Proof

We first prove (1). If n = 9 k + 1 for a nonnegative integer k , then the partition λ = ( P 4 ( 3 ) , … , P 4 ( 3 ) , 1 ) is in G 4 , χ 3 ( n ) .

Assume that there is a partition λ ∈ G 4 , χ 3 ( n ) . We rewrite the parts of λ as P 4 ( a j ( i ) ) such that

a j ( i ) ≡ i ( mod 9 ) , ( i = 0 , ⋯ , 8 ) .

Then,

(3.1) n = ∑ i = 0 8 ∑ j = 1 r i ( a j ( i ) ) 2

and

(3.2) 1 = ∑ j = 1 r 1 1 a j ( 1 ) − ∑ j = 1 r 2 1 a j ( 2 ) + ∑ j = 1 r 4 1 a j ( 4 ) − ∑ j = 1 r 5 1 a j ( 5 ) + ∑ j = 1 r 7 1 a j ( 7 ) − ∑ j = 1 r 8 1 a j ( 8 ) .

Let N be the positive integer defined as

N = ∏ j = 1 r 1 a j ( 1 ) ∏ j = 1 r 2 a j ( 2 ) ∏ j = 1 r 4 a j ( 4 ) ∏ j = 1 r 5 a j ( 5 ) ∏ j = 1 r 7 a j ( 7 ) ∏ j = 1 r 8 a j ( 8 ) .

Multiplying both sides of equation (3.2) by N , we have that

( − 1 ) r 5 + r 7 + r 8 2 r 2 + r 7 4 r 4 + r 5 ≡ ( − 1 ) r 5 + r 7 + r 8 r 1 2 r 2 + r 7 4 r 4 + r 5 + ( − 1 ) r 5 + r 7 + r 8 + 1 r 2 2 r 2 + r 7 − 1 4 r 4 + r 5 + ( − 1 ) r 5 + r 7 + r 8 r 4 2 r 2 + r 7 4 r 4 + r 5 − 1 + ( − 1 ) r 5 + r 7 + r 8 r 5 2 r 2 + r 7 4 r 4 + r 5 − 1 + ( − 1 ) r 5 + r 7 + r 8 + 1 r 7 2 r 2 + r 7 − 1 4 r 4 + r 5 + ( − 1 ) r 5 + r 7 + r 8 r 8 2 r 2 + r 7 4 r 4 + r 5 ( mod 9 )

and

1 ≡ r 1 + 4 r 2 + 7 r 4 + 7 r 5 + 4 r 7 + r 8 ( mod 9 ) .

In equation (3.1), we see that

n ≡ r 1 + r 8 + 4 ( r 2 + r 7 ) + 7 ( r 4 + r 5 ) ( mod 9 ) .

Therefore, n ≡ 1 ( mod 9 ) .

Next we prove (2). The proof for p = 7 is similar to the proof for (1). Suppose that n is a positive integer congruent to 1 modulo 7. In each residue class modulo 49, we can construct a partition λ in G 4 , χ 7 ( n ) as follows:

λ = ( P 4 ( 1 ) ) ∪ ( P 4 ( 7 ) , … , P 4 ( 7 ) ) , if n ≡ 1 ( mod 49 ) , ( P 4 ( 2 ) , P 4 ( 2 ) ) ∪ ( P 4 ( 7 ) , … , P 4 ( 7 ) ) , if n ≡ 8 ( mod 49 ) , ( P 4 ( 4 ) , P 4 ( 4 ) , P 4 ( 4 ) , P 4 ( 4 ) ) ∪ ( P 4 ( 7 ) , … , P 4 ( 7 ) ) , if n ≡ 15 ( mod 49 ) , ( P 4 ( 2 ) , P 4 ( 2 ) , P 4 ( 2 ) , P 4 ( 3 ) , P 4 ( 3 ) , P 4 ( 3 ) , P 4 ( 4 ) , P 4 ( 4 ) ) ∪ ( P 4 ( 7 ) , … , P 4 ( 7 ) ) , if n ≡ 22 ( mod 49 ) , ( P 4 ( 1 ) , P 4 ( 1 ) , P 4 ( 3 ) , P 4 ( 3 ) , P 4 ( 3 ) ) ∪ ( P 4 ( 7 ) , … , P 4 ( 7 ) ) , if n ≡ 29 ( mod 49 ) , ( P 4 ( 2 ) , P 4 ( 4 ) , P 4 ( 4 ) ) ∪ ( P 4 ( 7 ) , … , P 4 ( 7 ) ) , if n ≡ 36 ( mod 49 ) , ( P 4 ( 2 ) , P 4 ( 2 ) , P 4 ( 2 ) , P 4 ( 2 ) , P 4 ( 3 ) , P 4 ( 3 ) , P 4 ( 3 ) ) ∪ ( P 4 ( 7 ) , … , P 4 ( 7 ) ) , if n ≡ 43 ( mod 49 ) .

By an exhaustive search on partitions of 15 and 22, we can check that G 4 , χ 7 ( 15 ) and G 4 , χ 7 ( 22 ) are empty.

Now we assume that λ ∈ G 4 , χ 7 ( n ) . Then, we may rewrite the parts of λ as P 4 ( a j ( i ) ) such that

a j ( i ) ≡ i ( mod 7 ) , ( i = 0 , ⋯ , 6 )

and

n = ∑ i = 0 6 ∑ j = 1 r i P 4 ( a j ( i ) ) .

Multiplying

N = ∏ j = 1 r 1 a j ( 1 ) ∏ j = 1 r 2 a j ( 2 ) ∏ j = 1 r 3 a j ( 3 ) ∏ j = 1 r 4 a j ( 4 ) ∏ j = 1 r 5 a j ( 5 ) ∏ j = 1 r 6 a j ( 6 )

on both sides of

(3.3) 1 = ∑ j = 1 r 1 1 a j ( 1 ) + ∑ j = 1 r 2 1 a j ( 2 ) − ∑ j = 1 r 3 1 a j ( 3 ) + ∑ j = 1 r 4 1 a j ( 4 ) − ∑ j = 1 r 5 1 a j ( 5 ) − ∑ j = 1 r 6 1 a j ( 6 ) ,

we complete our proof.□

For an odd prime p other than 3 and 7, the set G 4 , χ p ( n ) is not empty for a sufficiently large integer n .

Proposition 13

For an odd prime p ≠ 3 and 7 and a positive integer n ≥ n ˜ 4 , χ p , there is a partition in G 4 , χ p ( n ) , where

n ˜ 4 , χ p = 193 , i f p = 5 , α P 4 ( α ) ( P 4 ( p ) − 1 ) + 1 , i f p ≥ 11 ,

where α is the smallest positive integer satisfying (a) and (b) in Lemma 10 with m = 4 and t = 1 .

Proof

When p = 5 , we let α = 2 , m = 4 , and t = 1 to satisfy two conditions in Lemma 10 and hence n ˜ 4 , χ 5 = α P m ( α ) ( P m ( t p ) − 1 ) + 1 = 193 .

When p ≥ 11 , there are ( p − 1 ) ⁄ 2 ≥ 5 quadratic nonresidues. However, the number of solutions to x 3 + 1 ≡ 0 ( mod p ) is at most three. Therefore, we can deduce that there is an integer α in Lemma 10 with m = 4 and t = 1 . Hence, we find that n ˜ 4 , χ p = α P 4 ( α ) ( P 4 ( p ) − 1 ) + 1 .□

Now we give a lower bound for the existence of twisted pentagonal Graham partitions.

Proposition 14

For an odd prime p and a positive integer n ≥ n ˜ 5 , χ p , there is a partition in G 5 , χ p ( n ) , where

n ˜ 5 , χ p = α P 5 ( α ) ( α P 5 ( α ) + 9 ) + 1 , if p ≡ 1 ( mod 8 ) , 10 P 5 ( p ) − 9 , if p ≡ 5 ( mod 8 ) a n d ( 11 , 6 p − 1 ) = 1 , 10 P 5 ( 2 p ) − 9 , if p ≡ 5 ( mod 8 ) a n d ( 11 , 6 p − 1 ) = 11 , ( p − 1 ) P 5 ( p − 1 ) ( P 5 ( p ) − 1 ) + 1 , if p ≡ 3 ( mod 4 ) ,

where α is the smallest positive integer satisfying (a) and (b) in Lemma 10 with m = 5 and t = 1 .

As the proof of Proposition 14 is similar to the proof of Propositions 11 and 13, we only give a brief sketch.

Proof

Let p ≡ 1 ( mod 8 ) and α be the smallest quadratic nonresidue modulo p . Then, α ≥ 3 is a prime number. Note that ( α P 5 ( α ) + 1 , α P 5 ( α ) + 10 ) = ( α P 5 ( α ) + 1 , 9 ) = 1 because α P 5 ( α ) + 1 ≢ 0 ( mod 3 ) . Therefore, we find that

n ˜ 5 , χ p = α P 5 ( α ) ( α P 5 ( α ) + 9 ) + 1

as there are nonnegative integers j 1 and j 2 such that

n = ( α P 5 ( α ) + 1 ) j 1 + ( α P 5 ( α ) + 10 ) j 2 + 1 = ( j 1 + 1 ) P 5 ( 1 ) + 2 j 2 P 5 ( 2 ) + α ( j 1 + j 2 ) P 5 ( α ) .

Therefore, we can construct a partition λ ∈ G 5 , χ p ( n ) as follows:

λ = ( P 5 ( α ) , … , P 5 ( α ) ︸ ) α ( j 1 + j 2 ) ∪ ( P 5 ( 1 ) , … , P 5 ( 1 ) ︸ ) j 1 + 1 ∪ ( P 5 ( 2 ) , … , P 5 ( 2 ) ︸ ) 2 j 2 .

If p ≡ 5 ( mod 8 ) with ( 11 , 6 p − 1 ) = 1 , then we can choose α = 2 with m = 5 and t = 1 in Lemma 10, which implies

n ˜ 5 , χ p = F ( 11 , P 5 ( p ) ) + 2 = 10 P 5 ( p ) − 9 .

If p ≡ 5 ( mod 8 ) with ( 11 , 6 p − 1 ) ≠ 1 , then we can choose α = 2 with m = 5 and t = 2 in Lemma 10, so we derive that

n ˜ 5 , χ p = F ( 11 , P 5 ( 2 p ) ) + 2 = 10 P 5 ( 2 p ) − 9 .

Finally, for p ≡ 3 ( mod 4 ) , one can check that ( p − 1 ) P 5 ( p − 1 ) + 1 is always relative prime to P 5 ( p ) . Since p − 1 is a quadratic nonresidue, we may choose α = p − 1 in Lemma 10 with t = 1 . Therefore, we can derive that

□ n ˜ 5 , χ p = ( p − 1 ) P 5 ( p − 1 ) ( P 5 ( p ) − 1 ) + 1 .

To complete the proof of Theorem 3, it suffices to check that the lower bound n m , χ p is not smaller than n ˜ m , χ p .

Proof of Theorem 3

Since checking the inequality n ˜ ≤ n is tedious, we only give the worst case for each m .

For the triangular case, we estimate n ˜ 3 , χ p when p ≡ 1 ( mod 8 ) . Since α ≤ p − 2 and p ≥ 17 , we see that

n ˜ 3 , χ p = α P 3 ( α ) ( α P 3 ( α ) + 5 ) + 1 ≤ ( α P 3 ( α ) + 3 ) 2 ≤ ( p − 2 ) 3 2 + 3 2 ≤ 1 4 p 6 .

For the square case, we see that

n ˜ 4 , χ p = α 3 ( p 2 − 1 ) + 1 < p 5 .

Finally, for the pentagonal case, we estimate n ˜ 5 , χ p when p ≡ 3 ( mod 4 ) . We find that

n ˜ 5 , χ p = ( p − 1 ) 2 3 p − 4 2 p ( 3 p − 1 ) 2 − 1 + 1 < 9 4 p 5

is desired.□

4 Concluding remarks

For completeness, we give the results on the existence of Graham partitions twisted by the Legendre symbol modulo 2 without proofs:

The set G χ 2 ( n ) is not empty if and only if n is odd.
The set G 3 , χ 2 ( n ) is not empty for n ≥ 16 .
The set G 4 , χ 2 ( n ) is not empty for odd n ≥ 25 .
The set G 5 , χ 2 ( n ) is not empty for n ≥ 70 .

Numerical experiments support that the set G m , χ p ( n ) is not empty for an odd prime p , an integer m ≥ 6 , and a sufficiently large integer n . It would be interesting if one could obtain an effective bound for the nonemptiness of G m , χ p ( n ) . It is also desirable to find an efficient algorithm to find a twisted m -gonal Graham partition of n . In generating functionology,

∑ n ≥ 0 ∣ G m , χ p ( n ) ∣ q n = [ z ] ∏ k ≥ 1 1 1 − z χ p ( k ) ⁄ k q P m ( k ) ,

where ∣ G m , χ p ( n ) ∣ is the number of elements of G m , χ p ( n ) and [ z ] g ( z , q ) ∈ Z [ [ z , q ] ] is the coefficient of z of the power series expansion of g ( z , q ) . In this sense, we have studied the positivity of the coefficients of z q n for the generating function. It would be nice if one could obtain other interesting arithmetic properties of twisted Graham partitions from the above generating function.

Acknowledgements

The authors are grateful for the valuable comments of the referees. This work was partially done while the authors were visiting the Korea Institute for Advanced Study.

Funding information: Byungchan Kim, Sang June Lee, Poo-Sung Park and Yoon Kyung Park were supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science and ICT (NRF-2022R1F1A1063530, NRF-2022R1F1A1074886, NRF-2021R1A2C1092930, NRF-2021R1F1A1055200). Ji Young Kim and Chong Gyu Lee were supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (NRF-2020R1I1A1A01053318, NRF-2021R1A6A1A10044154).
Conflict of interest: The authors state that there is no conflict of interest.

References

[1] R. L. Graham, A theorem on partitions, J. Aust. Math. Soc. 3 (1963), 435–441. 10.1017/S1446788700039045Search in Google Scholar

[2] B. Kim, J. Y. Kim, C. G. Lee, S. J. Lee, and P.-S. Park, On the existence of Graham partitions with congruence conditions, Bull. Korean Math. Soc. 59 (2022), no. 1, 15–25, DOI: https://doi.org/10.4134/BKMS.b200730. Search in Google Scholar

[3] N. Sloane, The on-line encyclopedia of integer sequences, DOI: http://oeis.org/A051908, Sequence A051908.Search in Google Scholar

[4] B. C. Berndt and J. Sohn, Asymptotic formulas for two continued fractions in Ramanujan’s lost notebook, J. Lond. Math. Soc. 65 (2002), no. 2, 271–284, DOI: https://doi.org/10.1112/S0024610701002952. 10.1112/S0024610701002952Search in Google Scholar

[5] H. Iwaniec and E. Kowalski, Analytic Number Theory, American Mathematical Society Colloquium Publications, Vol. 53, American Mathematical Society, Providence, 2004. 10.1090/coll/053Search in Google Scholar

[6] K. Ono, The Web of Modularity: Arithmetic of the Coefficients of Modular Forms and q-Series, CBMS Regional Conference Series in Mathematics, Vol. 102, American Mathematical Society, Providence, 2004. 10.1090/cbms/102Search in Google Scholar

[7] J. H. Silverman, The Arithmetic of Elliptic Curves, 2nd ed., Graduate Texts in Mathematics, Vol. 106, Springer, 2009. 10.1007/978-0-387-09494-6Search in Google Scholar

[8] M. A. Alekseyev, On partitions into squares of distinct integers whose reciprocals sum to 1, in: The Mathematics of Various Entertaining Subjects, Vol. 3, Princeton University Press, Princeton, 2019, pp. 213–221. 10.2307/j.ctvd58spj.18Search in Google Scholar

[9] J. J. Sylvester, On subinvariants, i.e., semi-invariants to binary quantics of an unlimited order, Amer. J. Math. 5 (1882), 79–136, DOI: https://doi.org/10.2307/2369536. 10.2307/2369536Search in Google Scholar

[10] J. J. Sylvester, Problem 7382, The Educational Times and Journal of the College of Preceptors, New Series 36 (1883), no. 266, 177.Search in Google Scholar

[11] E. Treviño, The least k-th power non-residue, J. Number Theory 149 (2015), 201–224, DOI: https://doi.org/10.1016/j.jnt.2014.10.019. 10.1016/j.jnt.2014.10.019Search in Google Scholar

[12] E. Treviño, On the maximum number of consecutive integers on which a character is constant, Mosc. J. Comb. Number Theory 2 (2012), no. 1, 56–72. Search in Google Scholar

[13] E. Treviño, Corrigendum to On the maximum number of consecutive integers on which a character is constant, Mosc. J. Comb. Number Theory 7 (2017), no. 3, 79–80. Search in Google Scholar

[14] L. Somer, The residues of nn modulo p, Fibonacci Quart. 19 (1981), no. 2, 110–117. Search in Google Scholar

[15] W. Knapp, M. Köcher, and P. Schmid, On the least odd quadratic non-residue, Beitr. Algebra Geom. 54 (2013), no. 2, 483–491, DOI: http://doi.org/10.1007/s13366-012-0086-6. 10.1007/s13366-012-0086-6Search in Google Scholar

Received: 2022-11-24

Revised: 2023-08-12

Accepted: 2023-09-24

Published Online: 2023-10-26

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2023-0134

Keywords for this article

Graham partition; sum of reciprocals; quadratic twist; Legendre symbol; polygonal number

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