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Square array T(n, k) = Product_{j=1..n} ( Sum_{i=0..j-1} (i+1)*(k+1)^i ) with T(n, 0) = n!, read by antidiagonals.
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%I #9 Sep 08 2022 08:45:41

%S 1,1,1,1,1,2,1,1,5,6,1,1,7,85,24,1,1,9,238,4165,120,1,1,11,513,33796,

%T 537285,720,1,1,13,946,160569,18486412,172468485,5040,1,1,15,1573,

%U 554356,255786417,37065256060,132628264965,40320,1,1,17,2430,1549405,2057215116,1979019508329,263459840074480,237802479082245,362880

%N Square array T(n, k) = Product_{j=1..n} ( Sum_{i=0..j-1} (i+1)*(k+1)^i ) with T(n, 0) = n!, read by antidiagonals.

%H G. C. Greubel, <a href="/A156576/b156576.txt">Antidiagonal rows n = 0..50, flattened</a>

%F T(n, k) = Product_{j=1..n} ( Sum_{i=0..j-1} (i+1)*(k+1)^i ) with T(n, 0) = n! (square array).

%F T(n, k) = (1/k^(2*n))*Product_{j=1..n} (1 -(j+1)*(k+1)^j +j*(k+1)^(j+1)) with T(n, 0) = n! (square array). - _G. C. Greubel_, Jun 28 2021

%e Square array begins as:

%e 1, 1, 1, 1, 1, 1 ...;

%e 1, 1, 1, 1, 1, 1 ...;

%e 2, 5, 7, 9, 11, 13 ...;

%e 6, 85, 238, 513, 946, 1573 ...;

%e 24, 4165, 33796, 160569, 554356, 1549405 ...;

%e 120, 537285, 18486412, 255786417, 2057215116, 11566308325 ...;

%e Antidiagonal triangle begins as:

%e 1;

%e 1, 1;

%e 1, 1, 2;

%e 1, 1, 5, 6;

%e 1, 1, 7, 85, 24;

%e 1, 1, 9, 238, 4165, 120;

%e 1, 1, 11, 513, 33796, 537285, 720;

%e 1, 1, 13, 946, 160569, 18486412, 172468485, 5040;

%e 1, 1, 15, 1573, 554356, 255786417, 37065256060, 132628264965, 40320;

%t (* First program *)

%t T[n_, k_]:= T[n,k]= If[k==0, n!, Product[Sum[(i+1)*(k+1)^i, {i,0,j-1}] {j, n}]];

%t Table[T[k, n-k], {n,0,12}, {k,0,n}]//Flatten (* modified by _G. C. Greubel_, Jun 28 2021 *)

%t (* Second program *)

%t T[n_, k_]:= If[k==0, n!, Product[1 -(j+1)*(k+1)^j +j*(k+1)^(j+1), {j,n}]/k^(2*n)];

%t Table[T[k, n-k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Jun 28 2021 *)

%o (Magma)

%o A156576:= func< n,k | n eq 0 select 1 else k eq 0 select Factorial(n) else (1/k^(2*n))*(&*[1 -(j+1)*(k+1)^j +j*(k+1)^(j+1): j in [1..n]]) >;

%o [A156576(k,n-k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Jun 28 2021

%o (Sage)

%o def A156576(n,k): return factorial(n) if (k==0) else (1/k^(2*n))*product( 1 -(j+1)*(k+1)^j +j*(k+1)^(j+1) for j in [1..n])

%o flatten([[A156576(k,n-k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Jun 28 2021

%K nonn,tabl

%O 0,6

%A _Roger L. Bagula_, Feb 10 2009

%E Edited by _G. C. Greubel_, Jun 28 2021