中文

无穷乘积展开中消失系数的一些结果

数论 2023-01-30 v1

摘要

近来,M. D. Hirschhorn证明了,若 n=0anqn:=(q,q4;q5)(q,q9;q10)3\sum_{n=0}^\infty a_nq^n := (-q,-q^4;q^5)_\infty(q,q^9;q^{10})_\infty^3n=0bnqn:=(q2,q3;q5)(q3,q7;q10)3\sum_{n=0}^\infty b_nq^n:=(-q^2,-q^3;q^5)_\infty(q^3,q^7;q^{10})_\infty^3,则 a5n+2=a5n+4=0a_{5n+2}=a_{5n+4}=0b5n+1=b5n+4=0b_{5n+1}=b_{5n+4}=0。受Hirschhorn工作的启发,D. Tang证明了若干类似结果,包括如下:若 n=0cnqn:=(q,q4;q5)3(q3,q7;q10) \sum_{n=0}^\infty c_nq^n := (-q,-q^4;q^5)_\infty^3(q^3,q^7;q^{10})_\inftyn=0dnqn:=(q2,q3;q5)3(q,q9;q10)\sum_{n=0}^\infty d_nq^n := (-q^2,-q^3;q^5)_\infty^3(q,q^9;q^{10})_\infty,则 c5n+3=c5n+4=0c_{5n+3}=c_{5n+4}=0d5n+3=d5n+4=0d_{5n+3}=d_{5n+4}=0。本文中,我们证明 a5n=b5n+2a_{5n}=b_{5n+2}a5n+1=b5n+3a_{5n+1}=b_{5n+3}a5n+2=b5n+4a_{5n+2}=b_{5n+4}a5n1=b5n+1a_{5n-1}=b_{5n+1}c5n+3=d5n+3c_{5n+3}=d_{5n+3}c5n+4=d5n+4c_{5n+4}=d_{5n+4}c5n=d5nc_{5n}=d_{5n}c5n+2=d5n+2c_{5n+2}=d_{5n+2},以及 c5n+1>d5n+1c_{5n+1}>d_{5n+1}。我们还记录了Tang未列出的一些其他类似结果。

关键词

引用

@article{arxiv.1908.07737,
  title  = {Some results on vanishing coefficients in infinite product expansions},
  author = {Nayandeep Deka Baruah and Mandeep Kaur},
  journal= {arXiv preprint arXiv:1908.07737},
  year   = {2023}
}

备注

15 pages