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%I #11 May 03 2017 12:38:35
%S 0,0,2,13,83,582,4554,39672,382248,4044240,46663920,583554240,
%T 7865622720,113711230080,1755484617600,28828769356800,501858148377600,
%U 9232213174732800,178968924600883200,3646603415927808000,77916767838981120000,1742147265551616000000
%N Sum of the second entries in all cycles of all permutations of [n].
%C Each cycle is written with the smallest element first and cycles are arranged in increasing order of their first elements.
%H Alois P. Heinz, <a href="/A285795/b285795.txt">Table of n, a(n) for n = 0..449</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Permutation">Permutation</a>
%e a(3) = 13 because the sum of the second entries in all cycles of all permutations of [3] ((123), (132), (12)(3), (13)(2), (1)(23), (1)(2)(3)) is 2+3+2+3+3+0 = 13.
%p a:= proc(n) option remember; `if`(n<3, n*(n-1),
%p ((2*n^2+3*n-1)*a(n-1)-(n-1)*n*(n+2)*a(n-2))/(n+1))
%p end:
%p seq(a(n), n=0..25);
%t Flatten[{0, Table[n! * (2*(n+1)*HarmonicNumber[n] - n - 3)/4, {n, 1, 25}]}] (* _Vaclav Kotesovec_, Apr 29 2017 *)
%Y Column k=2 of A285793.
%Y Cf. A001008, A002805.
%K nonn
%O 0,3
%A _Alois P. Heinz_, Apr 26 2017
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