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A349479 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)).

%I A349479 #24 Dec 28 2021 14:13:17
%S A349479 1,0,0,2,0,4,0,12,12,0,48,80,0,240,520,120,0,1440,3696,1680,0,10080,
%T A349479 29232,19040,1680,0,80640,256896,211456,40320,0,725760,2493504,
%U A349479 2429280,705600,30240,0,7257600,26547840,29430720,11285120,1108800,0,79833600,307992960,378595008,177580480,27720000,665280
%N A349479 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 A349479 T(n,k) is the number of cycle-colored n-derangements possessing exactly k cycles; two colors are available.
%H A349479 Steven Finch, <a href="https://arxiv.org/abs/2111.14487">Rounds, Color, Parity, Squares</a>, arXiv:2111.14487 [math.CO], 2021.
%e A349479 Triangle begins:
%e A349479 [0] 1;
%e A349479 [1] 0;
%e A349479 [2] 0,     2;
%e A349479 [3] 0,     4;
%e A349479 [4] 0,    12,     12;
%e A349479 [5] 0,    48,     80;
%e A349479 [6] 0,   240,    520,    120;
%e A349479 [7] 0,  1440,   3696,   1680;
%e A349479 [8] 0, 10080,  29232,  19040,  1680;
%e A349479 [9] 0, 80640, 256896, 211456, 40320;
%e A349479 ...
%p A349479 b:= proc(n) option remember; expand(`if`(n=0, 1, add(
%p A349479       2*x*b(n-j)*binomial(n-1, j-1)*(j-1)!, j=2..n)))
%p A349479     end:
%p A349479 T:= n-> (p-> seq(coeff(p, x, i), i=0..floor(n/2)))(b(n)):
%p A349479 seq(T(n), n=0..14);  # _Alois P. Heinz_, Nov 19 2021
%t A349479 S1[0, 0] = 1; S1[_, 0] = 0; S1[n_, k_] /; k > Quotient[n, 2] = 0;
%t A349479 S1[n_, k_] := S1[n, k] = (n-1)*(S1[n-1, k] + S1[n-2, k-1]);
%t A349479 T[n_, k_] := S1[n, k]*2^k;
%t A349479 Table[T[n, k], {n, 0, 14}, {k, 0, Quotient[n, 2]}] // Flatten (* _Jean-François Alcover_, Dec 28 2021 *)
%Y A349479 Row sums give A087981.
%Y A349479 Cf. A008275, A125553.
%K A349479 nonn,tabf
%O A349479 0,4
%A A349479 _Steven Finch_, Nov 19 2021