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A190865 E.g.f. exp(x+x^3/6).

%I A190865 #45 Jun 04 2021 15:57:52
%S A190865 1,1,1,2,5,11,31,106,337,1205,5021,20186,86461,417847,1992355,9860306,
%T A190865 53734241,292816841,1633818457,9855157330,59926837141,370352343971,
%U A190865 2439935383271,16283034762842,109982177787505,783404343570301,5668314772422901,41412522553362026
%N A190865 E.g.f. exp(x+x^3/6).
%C A190865 a(n) is the number of set partitions of {1,2,...,n} such that the size of each block divides 3. - _Geoffrey Critzer_, Sep 23 2011
%H A190865 Alois P. Heinz, <a href="/A190865/b190865.txt">Table of n, a(n) for n = 0..666</a>
%F A190865 a(n) = n!*sum(k=0..n/3, 1/((k)!*(n-3*k)!*6^(k))), n>0, a(0)=1.
%F A190865 E.g.f.: E(0)/2, where E(k)= 1 + 1/(1 - x*(6+x^2)/(x*(6+x^2)+ 6*(k+1)/E(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Jun 01 2013
%F A190865 Recurrence: 2*a(n) = 2*a(n-1) + (n-2)*(n-1)*a(n-3). - _Vaclav Kotesovec_, Jun 27 2013
%F A190865 a(n) ~ n^(2*n/3) * exp(-2*n/3+(2*n)^(1/3)) / (sqrt(3)*2^(n/3)) * (1 - 2^(2/3)/(6*n^(1/3)) + 13*2^(1/3)/(36*n^(2/3))). - _Vaclav Kotesovec_, Jun 27 2013
%F A190865 a(n) = hypergeom([-n/3, (1 - n)/3, (2 - n)/3], [], -9/2). - _Peter Luschny_, Jun 04 2021
%e A190865 a(0) = 1 because (vacuously) all sizes of the blocks in the unique set partition of {} divide 3.
%e A190865 a(4) = 5 because there are 5 such set partitions of {1,2,3,4}: ({1},{2,3,4}) ({2},{1,3,4}) ({3},{1,2,4}) ({4},{1,2,3}) ({1},{2},{3},{4}).
%p A190865 a:= proc(n) option remember; `if`(n=0, 1, add(
%p A190865       `if`(j>n, 0, a(n-j)*binomial(n-1, j-1)), j=[1, 3]))
%p A190865     end:
%p A190865 seq(a(n), n=0..30);  # _Alois P. Heinz_, Jul 27 2016
%t A190865 Range[0, 25]! CoefficientList[Series[Exp[x + x^3/6] , {x, 0, 25}], x]
%o A190865 (Maxima)
%o A190865 a(n):=n!*sum(1/((k)!*(n-3*k)!*6^(k)),k,0,n/3);
%Y A190865 Cf. A001470.
%Y A190865 Column k=3 of A275422.
%K A190865 nonn
%O A190865 0,4
%A A190865 _Vladimir Kruchinin_, May 22 2011