Congruences modulo primes of the Romik sequence related to the Taylor expansion of the Jacobi theta constant $$\theta _3$$

Abstract

Recently, Romik determined in [8] the Taylor expansion of the Jacobi theta constant θ 3 , around the point x = 1. He discovered a new integer sequence, (d(n)) ∞ n=0 = 1, 1, −1, 51, 849, −26199, . . . , from which the Taylor coefficients are built, and conjectured that the numbers d(n) satisfy certain congruences. In this paper, we prove some of these conjectures, for example that d(n) ≡ (−1) n+1 (mod 5) for all n ≥ 1, and that for any prime p ≡ 3 (mod 4), d(n) vanishes modulo p for all large enough n.

$p$-adic Properties for Taylor Coefficients of Half-integral Weight Modular Forms on $Γ_1(4)$

“…We note that θ 2 (z) = 2e πiz 4 ψ(e 2πiz ) and θ 3 (z) = ϕ(e πiz ), where ψ and ϕ are defined in [1, pp. 323 From (10) and ∂Θ(i/2) = 0, we have that…”

“…around τ 0 = i (we refer to [8], [10], [2], [15], and [16]). Romik defined (d(n)) ∞ n=0 to be the sequence such that…”

Romik's conjecture for the Jacobi theta function

“…This triply nested definition makes it rather unwieldy to work with the Taylor coefficients directly, though Romik conjectured several nice properties of the d(n) coefficients modulo any prime. This paper is dedicated towards refining the second half of Romik's conjecture, which was proven by the combined efforts of Scherer [8] and Guerzhoy-Mertens-Rolen [3]. Guerzhoy, Mertens, and Rolen in fact prove a stronger statement in the context of an arbitrary half integer weight modular form.…”

“…This encodes more arithmetic information, and on iterating this p − 1 times we see that d(n) is periodic with period (p−1) 2 2 , which is not necessarily minimal. The proof structure is tripartite: (1) show that u(n), v(n) ≡ 0 (mod p) for n ≥ p+1 2 , a result conjectured in Scherer's paper [8]; (2) use this to simplify the expression for s(n, k) (mod p);…”

“…The last key ingredient we require is a higher order analog of [8,Lemma 14]: for n ≥ 3, n k=1 k even 2 n−k s(n, k) ≡ n k=1 k odd 2 n−k s(n, k) ≡ 0 (mod 5).…”