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arXiv:1908.07737 [math.NT]AbstractReferencesReviewsResources

Some results on vanishing coefficients in infinite product expansions

Nayandeep Deka Baruah, Mandeep Kaur

Published 2019-08-21Version 1

Recently, M. D. Hirschhorn proved that, if $\sum_{n=0}^\infty a_nq^n := (-q,-q^4;q^5)_\infty(q,q^9;q^{10})_\infty^3$ and $\sum_{n=0}^\infty b_nq^n:=(-q^2,-q^3;q^5)_\infty(q^3,q^7;q^{10})_\infty^3$, then $a_{5n+2}=a_{5n+4}=0$ and $b_{5n+1}=b_{5n+4}=0$. Motivated by the work of Hirschhorn, D. Tang proved some comparable results including the following: If $ \sum_{n=0}^\infty c_nq^n := (-q,-q^4;q^5)_\infty^3(q^3,q^7;q^{10})_\infty$ and $\sum_{n=0}^\infty d_nq^n := (-q^2,-q^3;q^5)_\infty^3(q,q^9;q^{10})_\infty$, then $c_{5n+3}=c_{5n+4}=0$ and $d_{5n+3}=d_{5n+4}=0$. In this paper, we prove that $a_{5n}=b_{5n+2}$, $a_{5n+1}=b_{5n+3}$, $a_{5n+2}=b_{5n+4}$, $a_{5n-1}=b_{5n+1}$, $c_{5n+3}=d_{5n+3}$, $c_{5n+4}=d_{5n+4}$, $c_{5n}=d_{5n}$, $c_{5n+2}=d_{5n+2}$, and $c_{5n+1}>d_{5n+1}$. We also record some other comparable results not listed by Tang.

Comments: 15 pages
Journal: The Ramanujan Journal, 2019
Categories: math.NT
Subjects: 33D15, 11F33
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