%I #24 Dec 28 2021 14:13:17
%S 1,0,0,2,0,4,0,12,12,0,48,80,0,240,520,120,0,1440,3696,1680,0,10080,
%T 29232,19040,1680,0,80640,256896,211456,40320,0,725760,2493504,
%U 2429280,705600,30240,0,7257600,26547840,29430720,11285120,1108800,0,79833600,307992960,378595008,177580480,27720000,665280
%N Irregular triangle read by rows: T(n,k) = S1(n,k)*2^k, where S1(n,k) is the associated Stirling number of the first kind (cf. A008306) (n >= 0, 0 <= k <= floor(n/2)).
%C T(n,k) is the number of cycle-colored n-derangements possessing exactly k cycles; two colors are available.
%H Steven Finch, <a href="https://arxiv.org/abs/2111.14487">Rounds, Color, Parity, Squares</a>, arXiv:2111.14487 [math.CO], 2021.
%e Triangle begins:
%e [0] 1;
%e [1] 0;
%e [2] 0, 2;
%e [3] 0, 4;
%e [4] 0, 12, 12;
%e [5] 0, 48, 80;
%e [6] 0, 240, 520, 120;
%e [7] 0, 1440, 3696, 1680;
%e [8] 0, 10080, 29232, 19040, 1680;
%e [9] 0, 80640, 256896, 211456, 40320;
%e ...
%p b:= proc(n) option remember; expand(`if`(n=0, 1, add(
%p 2*x*b(n-j)*binomial(n-1, j-1)*(j-1)!, j=2..n)))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=0..floor(n/2)))(b(n)):
%p seq(T(n), n=0..14); # _Alois P. Heinz_, Nov 19 2021
%t S1[0, 0] = 1; S1[_, 0] = 0; S1[n_, k_] /; k > Quotient[n, 2] = 0;
%t S1[n_, k_] := S1[n, k] = (n-1)*(S1[n-1, k] + S1[n-2, k-1]);
%t T[n_, k_] := S1[n, k]*2^k;
%t Table[T[n, k], {n, 0, 14}, {k, 0, Quotient[n, 2]}] // Flatten (* _Jean-François Alcover_, Dec 28 2021 *)
%Y Row sums give A087981.
%Y Cf. A008275, A125553.
%K nonn,tabf
%O 0,4
%A _Steven Finch_, Nov 19 2021
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