A217093 Number of partitions of n objects of 3 colors.
1, 3, 12, 38, 117, 330, 906, 2367, 6027, 14873, 35892, 84657, 196018, 445746, 997962, 2201438, 4792005, 10300950, 21889368, 46012119, 95746284, 197344937, 403121547, 816501180, 1640549317, 3271188702, 6475456896, 12730032791, 24861111315, 48246729411, 93065426256
Offset: 0
Examples
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. 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.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
- S. Benvenuti, B. Feng, A. Hanany and Y. H. He, Counting BPS operators in gauge theories: Quivers, syzygies and plethystics, arXiv:hep-th/0608050, Aug 2006, p.42.
- Carlos A. A. Florentino, Plethystic exponential calculus and permutation polynomials, arXiv:2105.13049 [math.CO], 2021. Mentions this sequence.
- Vaclav Kotesovec, Graph - The asymptotic ratio
Crossrefs
Programs
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Maple
with(numtheory): a:= proc(n) option remember; `if`(n=0, 1, add(add( d*binomial(d+2, 2), d=divisors(j))*a(n-j), j=1..n)/n) end: seq(a(n), n=0..30); # Alois P. Heinz, Sep 26 2012 with(numtheory): series(exp(add(((1/2)*sigma[3](k) + (3/2)*sigma[2](k) + sigma[1](k))*x^k/k, k = 1..30)), x, 31): seq(coeftayl(%, x = 0, n), n = 0..30); # Peter Bala, Jan 16 2025 -
Mathematica
nn=30; p=Product[1/(1- x^i)^Binomial[i+2,2],{i,1,nn}]; CoefficientList[Series[p,{x,0,nn}],x] -
Python
from functools import lru_cache from sympy import divisors @lru_cache(maxsize=None) def A217093_aux(n): return sum(d*(d+1)*(d+2)>>1 for d in divisors(n,generator=True)) @lru_cache(maxsize=None) def A217093(n): return 1 if n == 0 else (A217093_aux(n)+sum(A217093_aux(k)*A217093(n-k) for k in range(1,n)))//n # Chai Wah Wu, Mar 19 2025
Formula
G.f.: Product_{i>=1} 1/(1-x^i)^binomial(i+2,2).
EULER transform of 3, 6, 10, 15, ... .
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).
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
G.f.: exp(Sum_{k >= 1} ((1/2)*sigma_3(k) + (3/2)*sigma_2(k) + sigma_1(k))*x^k/k) = 1 + 3*x + 12*x^2 + 38*x^3 + 117*x^4 + .... - Peter Bala, Jan 16 2025
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