A291973 a(n) = (3*n)! * [z^(3*n)] exp(exp(z)/3 + 2*exp(-z/2)*cos(z*sqrt(3)/2)/3 - 1).
1, 1, 11, 365, 25323, 3068521, 583027547, 161601254725, 62042488237755, 31728742163212641, 20963751508027371691, 17461136553331587079965, 17967906090023681913528523, 22459900935806853610377326041, 33617974358392980795259947648187, 59515082206147526028817472280664565
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Keywords
Examples
For n=2, there are a(2)=11 partitions of {1,2,...,6} with every block size a multiple of 3: 123456, 123|456, 124|356, 125|346, 126|345, 134|256, 135|246, 136|245, 145|236, 146|235 and 156|234. - _Per W. Alexandersson_, Jun 20 2024
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..100
Crossrefs
Cf. A291451.
Programs
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Maple
A291973 := proc(n) exp(exp(z)/3+2*exp(-z/2)*cos(z*sqrt(3)/2)/3-1): (3*n)!*coeff(series(%, z, 3*(n+1)), z, 3*n) end: seq(A291973(n), n=0..15);
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Mathematica
P[m_, n_] := P[m, n] = If[n == 0, 1, Sum[Binomial[m*n, m*k]*P[m, n - k]*x, {k, 1, n}]]; a[n_] := Module[{cl = CoefficientList[P[3, n], x]}, Sum[cl[[k + 1]]/k!, {k, 0, n}]]; Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Jul 23 2019, after Peter Luschny in A291451 *)
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PARI
seq(n)={my(a=vector(n+1)); a[1]=1; for(n=1, n, a[1+n]=sum(k=1, n, binomial(3*n-1, 3*k-1) * a[1+n-k])); a} \\ Andrew Howroyd, Jan 21 2020
Formula
a(0) = 1; a(n) = Sum_{k=1..n} binomial(3*n-1,3*k-1) * a(n-k). - Ilya Gutkovskiy, Jan 21 2020
Comments