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#19 by Bruno Berselli at Wed Mar 28 03:58:43 EDT 2018
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#18 by Vaclav Kotesovec at Wed Mar 28 03:17:37 EDT 2018
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#17 by Ilya Gutkovskiy at Tue Mar 27 13:51:19 EDT 2018
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#16 by Ilya Gutkovskiy at Tue Mar 27 13:25:30 EDT 2018
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| FORMULA
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G.f.: exp(5*Sum_{k>=1} (-1)^k*x^k/(k*(1 - x^k)^2)). - Ilya Gutkovskiy, Mar 27 2018
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| STATUS
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approved
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#15 by Vaclav Kotesovec at Thu Apr 13 17:58:01 EDT 2017
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#14 by Vaclav Kotesovec at Thu Apr 13 17:56:04 EDT 2017
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| COMMENTS
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In general, if m >= 1 and g.f. = Product_{k>=1} 1/(1 + x^k)^(m*k), then a(n, m) ~ (-1)^n * exp(-m/12 + 3 * 2^(-5/3) * m^(1/3) * Zeta(3)^(1/3) * n^(2/3)) * 2^(m/18 - 5/6) * A^m * m^(1/6 - m/36) * Zeta(3)^(1/6 - m/36) * n^(m/36 - 2/3) / sqrt(3*Pi), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Apr 13 2017
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#13 by Vaclav Kotesovec at Thu Apr 13 17:47:25 EDT 2017
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a(n) ~ (-1)^n * exp(-5/12 + 3 * 2^(-5/3) * (5*Zeta(3))^(1/3) * n^(2/3)) * A^5 * (5*Zeta(3))^(1/36) / (2^(5/9) * sqrt(3*Pi) * n^(19/36)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Apr 13 2017
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approved
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#12 by Joerg Arndt at Thu Apr 13 08:28:26 EDT 2017
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#11 by Seiichi Manyama at Thu Apr 13 08:19:54 EDT 2017
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#10 by Seiichi Manyama at Thu Apr 13 08:19:48 EDT 2017
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| LINKS
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Seiichi Manyama, <a href="/A279932/b279932.txt">Table of n, a(n) for n = 0..10000</a>
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approved
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