用MacMahon型级数表示的某些无穷乘积
数论
2024-07-09 v1
摘要
最近，Ono和第三作者发现，theta级数( q ; q ) ∞ 3 (q;q)_\infty^3 ( q ; q ) ∞ 3 ( q 2 ; q 2 ) ∞ ( q ; q 2 ) ∞ 2 (q^2;q^2)_\infty(q;q^2)_\infty^2 ( q 2 ; q 2 ) ∞ ( q ; q 2 ) ∞ 2 A k ( q ) A_k(q) A k ( q ) C k ( q ) C_k(q) C k ( q ) Θ 6 ( q ) : = 1 2 ∑ n ∈ Z χ 6 ( n ) n q n 2 − 1 24 \Theta_{6}(q):=\frac{1}{2}\sum_{n\in\mathbb{Z}} \chi_6(n) n q^{\frac{n^2-1}{24}} Θ 6 ( q ) := 2 1 ∑ n ∈ Z χ 6 ( n ) n q 24 n 2 − 1 n n n 1 Θ 6 ( q ) = q − 3 n 2 + n 2 ∑ k = r 1 k ≡ n ( m o d 2 ) r 2 ( − 1 ) n − k 2 A k ( q ) C 3 n − k 2 ( q ) + O ( q n + 1 ) , \frac{1}{\Theta_{6}(q)}= q^{-\frac{3n^2+n}{2}}\sum_{\substack{k=r_1\ k\equiv n\hspace{-0.2cm}\pmod{2}}}^{r_2}(-1)^{\frac{n-k}{2}}A_{k}(q)C_{\frac{3n-k}{2}}(q)+O(q^{n+1}), Θ 6 ( q ) 1 = q − 2 3 n 2 + n k = r 1 k ≡ n ( mod 2 ) ∑ r 2 ( − 1 ) 2 n − k A k ( q ) C 2 3 n − k ( q ) + O ( q n + 1 ) , r 1 : = ⌊ 3 n − 1 − 12 n + 13 3 ⌋ + 1 r_1:=\lfloor\frac{3n-1-\sqrt{12n+13}}{3}\rfloor+1 r 1 := ⌊ 3 3 n − 1 − 12 n + 13 ⌋ + 1 r 2 : = ⌈ 3 n − 1 + 12 n + 13 3 ⌉ − 1 r_2:=\lceil\frac{3n-1+\sqrt{12n+13}}{3}\rceil-1 r 2 := ⌈ 3 3 n − 1 + 12 n + 13 ⌉ − 1
引用
@article{arxiv.2407.04798,
title = {Certain infinite products in terms of MacMahon type series},
author = {Seokho Jin and Badri Vishal Pandey and Ajit Singh},
journal= {arXiv preprint arXiv:2407.04798},
year = {2024}
}
备注
16 pages