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URL: https://oeis.org/A227937

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A227937
Partitions of n labeled elements into subsets of two or three elements.
3
1, 0, 1, 1, 3, 10, 25, 105, 385, 1540, 7245, 32725, 164395, 870870, 4689685, 27152125, 161786625, 997196200, 6443061625, 42702885225, 292938721075, 2078239413250, 15119319039825, 113390111659825, 873538909468225, 6894294734827500, 55855506234653125, 463151808682688125, 3927996120260086875, 34081631999814148750, 301951521812713898125, 2731127272307562253125, 25208456056107710010625, 237164027532948085570000
OFFSET
0,5
COMMENTS
Periodic modulo two and modulo three, but appears to eventually be divisible by other prime powers.
For n coprime to both 2 and 3 and for k >= 0, we conjecture that n divides a(n+k). - Peter Bala, Nov 21 2025
FORMULA
a(n) = (n-1)*a(n-2) + (n-1)*(n-2)*a(n-3)/2.
E.g.f.: exp( x^2/2 + x^3/6 ). [Joerg Arndt, Oct 07 2013]
a(n) ~ n^(2*n/3) * 2^(-n/3) * exp(2/9 - 2*n/3 - (2*n)^(1/3)/3 + (2*n)^(2/3)/2)/sqrt(3) * (1 + 34/(162*(2*n)^(1/3)) - 34802/(131220*(2*n)^(2/3))). - Vaclav Kotesovec, Oct 09 2013
EXAMPLE
The five elements a, b, c, d, e have ten partitions into sets of size two or three: ab/cde, ac/bde, ad/bce, ae/bcd, bc/ade, bd/ace, be/acd, cd/abe, ce/abd, and de/abc.
MATHEMATICA
Flatten[{1, RecurrenceTable[{2*a[n] - 2*(n-1)*a[n-2]-(n-2)*(n-1)*a[n-3] == 0, a[1]==0, a[2]==1, a[3]==1}, a, {n, 1, 20}]}] (* Vaclav Kotesovec, Oct 09 2013 *)
PROG
(PARI) x='x+O('x^66); Vec( serlaplace( exp( x^2/2 + x^3/6 ) ) ) \\ Joerg Arndt, Oct 07 2013
CROSSREFS
Sequence in context: A110713 A098702 A100001 * A299754 A190529 A343584
KEYWORD
nonn,easy
AUTHOR
David Eppstein, Oct 06 2013
STATUS
approved