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A(n, k) = Stirling2(n + k, k)*A053657(n)*k!/(n + k)!, array read by ascending antidiagonals for n >= 0 and k >= 0.
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%I #16 Jan 05 2024 13:30:06

%S 1,0,1,0,1,1,0,4,2,1,0,2,14,3,1,0,48,12,30,4,1,0,16,496,36,52,5,1,0,

%T 576,288,2064,80,80,6,1,0,144,18288,1656,5832,150,114,7,1,0,3840,8160,

%U 145200,5920,13240,252,154,8,1

%N A(n, k) = Stirling2(n + k, k)*A053657(n)*k!/(n + k)!, array read by ascending antidiagonals for n >= 0 and k >= 0.

%e [0] 1, 1, 1, 1, 1, 1, 1, 1, ... A000012

%e [1] 0, 1, 2, 3, 4, 5, 6, 7, ... A001477

%e [2] 0, 4, 14, 30, 52, 80, 114, 154, ... A049451

%e [3] 0, 2, 12, 36, 80, 150, 252, 392, ... A011379

%e [4] 0, 48, 496, 2064, 5832, 13240, 26088, 46536, ...

%e [5] 0, 16, 288, 1656, 5920, 16200, 37296, 76048, ...

%e [6] 0, 576, 18288, 145200, 654816, 2153280, 5775936, 13429248, ...

%e A163176

%p A := (n, k) -> Stirling2(n + k, k)*A053657(n)*k!/(n + k)!:

%p seq(seq(A(n - k, k), k=0..n), n=0..10);

%t a053657[n_] := Product[p^Sum[Floor[(n-1) / ((p-1) p^k)], {k, 0, n}], {p, Prime[Range[n]]}];

%t A[n_, k_] := StirlingS2[n+k, k] a053657[n+1] k! / (n+k)!;

%t Table[A[n-k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 21 2019 *)

%Y Rows include A001477, A049451, A011379. Columns include A163176.

%Y Cf. A053657.

%K nonn,tabl

%O 0,8

%A _Peter Luschny_, May 22 2019