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%I #53 May 28 2021 06:04:37
%S 1,3,12,38,117,330,906,2367,6027,14873,35892,84657,196018,445746,
%T 997962,2201438,4792005,10300950,21889368,46012119,95746284,197344937,
%U 403121547,816501180,1640549317,3271188702,6475456896,12730032791,24861111315,48246729411,93065426256
%N Number of partitions of n objects of 3 colors.
%C a(n) is also the number of unlabeled simple graphs with n nodes of 3 colors whose components are complete graphs.
%C Number of (integer) partitions of n into 3 sorts of part 1, 6 sorts of part 2, 10 sorts of part 3, ..., (k+2)*(k+1)/2 sorts of part k. - _Joerg Arndt_, Dec 07 2014
%C In general the g.f. 1 / prod(n>=1, (1-x^k)^m(k) ) gives the number of (integer) partitions where there are m(k) sorts of part k. - _Joerg Arndt_, Mar 10 2015
%H Alois P. Heinz, <a href="/A217093/b217093.txt">Table of n, a(n) for n = 0..1000</a>
%H S. Benvenuti, B. Feng, A. Hanany and Y. H. He, <a href="https://arxiv.org/abs/hep-th/0608050">Counting BPS operators in gauge theories: Quivers, syzygies and plethystics</a>, arXiv:hep-th/0608050, Aug 2006, p.42.
%H Carlos A. A. Florentino, <a href="https://arxiv.org/abs/2105.13049">Plethystic exponential calculus and permutation polynomials</a>, arXiv:2105.13049 [math.CO], 2021. Mentions this sequence.
%H Vaclav Kotesovec, <a href="/A217093/a217093.jpg">Graph - The asymptotic ratio</a>
%F G.f.: Product_{i>=1} 1/(1-x^i)^binomial(i+2,2).
%F EULER transform of 3, 6, 10, 15, ... .
%F Generally for the number of partitions of k colors the generating function is Product_{i>=1} 1/(1-x^i)^binomial(i+k-1,k-1).
%F a(n) ~ Pi^(1/8) * exp(1/8 + 3^4 * 5^2 * Zeta(3)^3 / (2*Pi^8) - 31*Zeta(3) / (8*Pi^2) + 5^(1/4) * Pi * n^(1/4) / 6^(3/4) - 3^(13/4) * 5^(5/4) * Zeta(3)^2 * n^(1/4) / (2^(7/4) * Pi^5) + 3^(3/2) * 5^(1/2) * Zeta(3) * n^(1/2) / (2^(1/2) * Pi^2) + 2^(7/4) * Pi * n^(3/4) / (3^(5/4) * 5^(1/4))) / (A^(3/2) * 2^(73/32) * 15^(9/32) * n^(25/32)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant and Zeta(3) = A002117 = 1.202056903... . - _Vaclav Kotesovec_, Mar 08 2015
%e We represent each summand, k, in a partition of n as k identical objects. Then we color each object. We have no regard for the order of the colored objects.
%e a(2) = 12 because we have: ww; wg; wb; gg; gb; bb; w + w; w + g; w + b; g + g; g + b; b + b, where the 3 colors are white w, gray g, and black b.
%p with(numtheory):
%p a:= proc(n) option remember; `if`(n=0, 1, add(add(
%p d*binomial(d+2, 2), d=divisors(j))*a(n-j), j=1..n)/n)
%p end:
%p seq(a(n), n=0..30); # _Alois P. Heinz_, Sep 26 2012
%t nn=30; p=Product[1/(1- x^i)^Binomial[i+2,2],{i,1,nn}]; CoefficientList[Series[p,{x,0,nn}],x]
%Y Cf. A005380, A255050, A255052.
%Y Column k=3 of A075196.
%K nonn
%O 0,2
%A _Geoffrey Critzer_, Sep 26 2012
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