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)).
1, 0, 0, 2, 0, 4, 0, 12, 12, 0, 48, 80, 0, 240, 520, 120, 0, 1440, 3696, 1680, 0, 10080, 29232, 19040, 1680, 0, 80640, 256896, 211456, 40320, 0, 725760, 2493504, 2429280, 705600, 30240, 0, 7257600, 26547840, 29430720, 11285120, 1108800, 0, 79833600, 307992960, 378595008, 177580480, 27720000, 665280
Offset: 0
Examples
Triangle begins: [0] 1; [1] 0; [2] 0, 2; [3] 0, 4; [4] 0, 12, 12; [5] 0, 48, 80; [6] 0, 240, 520, 120; [7] 0, 1440, 3696, 1680; [8] 0, 10080, 29232, 19040, 1680; [9] 0, 80640, 256896, 211456, 40320; ...
Links
- Steven Finch, Rounds, Color, Parity, Squares, arXiv:2111.14487 [math.CO], 2021.
Programs
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Maple
b:= proc(n) option remember; expand(`if`(n=0, 1, add( 2*x*b(n-j)*binomial(n-1, j-1)*(j-1)!, j=2..n))) end: T:= n-> (p-> seq(coeff(p, x, i), i=0..floor(n/2)))(b(n)): seq(T(n), n=0..14); # Alois P. Heinz, Nov 19 2021
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Mathematica
S1[0, 0] = 1; S1[, 0] = 0; S1[n, k_] /; k > Quotient[n, 2] = 0; S1[n_, k_] := S1[n, k] = (n-1)*(S1[n-1, k] + S1[n-2, k-1]); T[n_, k_] := S1[n, k]*2^k; Table[T[n, k], {n, 0, 14}, {k, 0, Quotient[n, 2]}] // Flatten (* Jean-François Alcover, Dec 28 2021 *)
Comments