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A344563 T(n, k) = binomial(n - 1, k - 1) * binomial(n, k) * 2^k, T(0, 0) = 1. Triangle read by rows, T(n, k) for 0 <= k <= n.

%I A344563 #14 May 30 2021 14:25:18
%S A344563 1,0,2,0,4,4,0,6,24,8,0,8,72,96,16,0,10,160,480,320,32,0,12,300,1600,
%T A344563 2400,960,64,0,14,504,4200,11200,10080,2688,128,0,16,784,9408,39200,
%U A344563 62720,37632,7168,256,0,18,1152,18816,112896,282240,301056,129024,18432,512
%N A344563 T(n, k) = binomial(n - 1, k - 1) * binomial(n, k) * 2^k, T(0, 0) = 1. Triangle read by rows, T(n, k) for 0 <= k <= n.
%H A344563 T. Amdeberhan, <a href="https://mathoverflow.net/q/393532">Power of 2 dividing a specialized Mittag-Leffler polynomial</a>, MathOverflow.
%e A344563 [0] 1;
%e A344563 [1] 0,  2;
%e A344563 [2] 0,  4,    4;
%e A344563 [3] 0,  6,   24,     8;
%e A344563 [4] 0,  8,   72,    96,     16;
%e A344563 [5] 0, 10,  160,   480,    320,     32;
%e A344563 [6] 0, 12,  300,  1600,   2400,    960,     64;
%e A344563 [7] 0, 14,  504,  4200,  11200,  10080,   2688,    128;
%e A344563 [8] 0, 16,  784,  9408,  39200,  62720,  37632,   7168,   256;
%e A344563 [9] 0, 18, 1152, 18816, 112896, 282240, 301056, 129024, 18432, 512.
%p A344563 aRow := n -> seq(binomial(n-1, k-1)*binomial(n,k)*2^k, k=0..n):
%p A344563 seq(print(aRow(n)), n=0..9);
%t A344563 T[n_, k_] := Binomial[n-1, k-1] * Binomial[n, k] * 2^k;
%t A344563 Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten
%o A344563 (Python)
%o A344563 from math import comb
%o A344563 def T(n, k):
%o A344563     return comb(n-1, k-1)*comb(n, k)*2**k if k > 0 else k**n
%o A344563 print([T(n, k) for n in range(10) for k in range(n+1)]) # _Michael S. Branicky_, May 30 2021
%Y A344563 Row sums are A002003 with a(0) = 1, cf. also A047781.
%Y A344563 The coefficients of the associated polynomials are in A103371.
%K A344563 nonn,tabl
%O A344563 0,3
%A A344563 _Peter Luschny_, May 30 2021