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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 *)
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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 >= 2, 1 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
2,1
0,4
[20] 21;
[31] 40;
[42] 12, 120, 2;
[53] 48, 800, 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;
T:= n-> (p-> seq(coeff(p, x, i), i=10..degreefloor(pn/2)))(b(n)):
seq(T(n), n=20..14); # Alois P. Heinz, Nov 19 2021
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[2] 2;
[3] 4;
[4] 12, 12;
[5] 48, 80;
[6] 240, 520, 120;
[7] 1440, 3696, 1680;
[8] 10080, 29232, 19040, 1680;
...
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