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A237512 Number of solutions to Sum_{k=1..n} k*c(k) = n! , c(k) > 0. 6

%I #27 Sep 02 2018 02:07:22

%S 0,1,0,1,47,55496,2080571733,4441900888487987,

%T 849835826032526606030103,20540228659655619974131131927286681,

%U 82853643094578125257400348993596774353069331199,70898139566455107685443806945119782661588205935442233026505921

%N Number of solutions to Sum_{k=1..n} k*c(k) = n! , c(k) > 0.

%C a(n) is the number of partitions of n! - n*(n+1)/2 into parts that are at most n. - _Alois P. Heinz_, Feb 08 2014

%H Alois P. Heinz, <a href="/A237512/b237512.txt">Table of n, a(n) for n = 0..31</a>

%H A. V. Sills and D. Zeilberger, <a href="http://arxiv.org/abs/1108.4391">Formulae for the number of partitions of n into at most m parts (using the quasi-polynomial ansatz)</a> (arXiv:1108.4391 [math.CO])

%H StackExchange, <a href="http://math.stackexchange.com/questions/655541">Combinations sum_{k=1..m} k*n_k = m!</a>, Jan 29 2014

%F a(n) = [x^(n!)] Product_{k=1..n} x^k/(1-x^k).

%F a(n) = [x^(n!-n*(n+1)/2)] Product_{k=1..n} 1/(1-x^k). - _Alois P. Heinz_, Feb 08 2014

%F a(n) ~ n * (n!)^(n-3) ~ n^(n^2-5*n/2-1/2) * (2*Pi)^((n-3)/2) / exp(n*(n-3)-1/12). - _Vaclav Kotesovec_, Jun 05 2015

%t Table[Coefficient[Series[Product[x^k/(1-x^k),{k,n}],{x,0,n!}],x^(n!) ] ,{n,7}]

%Y Cf. A236810.

%K nonn

%O 0,5

%A _Wouter Meeussen_, Feb 08 2014

%E a(8)-a(11) from _Alois P. Heinz_, Feb 08 2014

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Last modified August 28 14:20 EDT 2024. Contains 375507 sequences. (Running on oeis4.)