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Square array T(n, k) = Product_{j=1..n} A129862(k+1, j) with T(n, 0) = n!, read by antidiagonals.
5

%I #11 Jun 25 2021 03:39:27

%S 1,1,1,1,1,2,1,1,0,6,1,1,-1,0,24,1,1,-2,-1,0,120,1,1,-3,-8,-1,0,720,1,

%T 1,-4,-27,32,2,0,5040,1,1,-5,-64,567,128,2,0,40320,1,1,-6,-125,3584,

%U 30618,-512,2,0,362880,1,1,-7,-216,14375,745472,-4317138,-2048,-4,0,3628800

%N Square array T(n, k) = Product_{j=1..n} A129862(k+1, j) with T(n, 0) = n!, read by antidiagonals.

%C Cartan_Dn refers to a Cartan matrix of type D_n. - _N. J. A. Sloane_, Jun 25 2021

%H G. C. Greubel, <a href="/A156612/b156612.txt">Antidiagonal rows n = 0..50, flattened</a>

%F T(n, k) = Product_{j=1..n} A129862(k+1, j) with T(n, 0) = n!.

%e Square array begins:

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

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

%e 2, 0, -1, -2, -3, -4 ...;

%e 6, 0, -1, -8, -27, -64 ...;

%e 24, 0, -1, 32, 567, 3584 ...;

%e 120, 0, 2, 128, 30618, 745472 ...;

%e Triangle begins as:

%e 1;

%e 1, 1;

%e 1, 1, 2;

%e 1, 1, 0, 6;

%e 1, 1, -1, 0, 24;

%e 1, 1, -2, -1, 0, 120;

%e 1, 1, -3, -8, -1, 0, 720;

%e 1, 1, -4, -27, 32, 2, 0, 5040;

%e 1, 1, -5, -64, 567, 128, 2, 0, 40320;

%e 1, 1, -6, -125, 3584, 30618, -512, 2, 0, 362880;

%e 1, 1, -7, -216, 14375, 745472, -4317138, -2048, -4, 0, 3628800;

%t (* First program *)

%t b[n_, k_, d_]:= If[n==k, 2, If[(k==d && n==d-2) || (n==d && k==d-2), -1, If[(k==n- 1 || k==n+1) && n<=d-1 && k<=d-1, -1, 0]]];

%t M[d_]:= Table[b[n, k, d], {n, d}, {k, d}];

%t p[x_, n_]:= If[n==0, 1, CharacteristicPolynomial[M[n], x]];

%t f = Table[p[x, n], {n, 0, 30}];

%t T[n_, k_]:= If[k==0, n!, Product[f[[j+1]], {j, n-1}]]/.x -> k+1;

%t Table[T[k, n - k], {n,0,15}, {k,0,n}]//Flatten (* modified by _G. C. Greubel_, Jun 25 2021 *)

%t (* Second program *)

%t f[n_, x_]:= f[n, x]= If[n<2, (2-x)^n, (2-x)*LucasL[2*(n-1), Sqrt[-x]]];

%t t[n_, k_]:= t[n, k]= If[k==0, n!, Product[f[j, x], {j, n-1}]]/.x -> (k+1);

%t Table[t[k, n-k], {n,0,15}, {k,0,n}]//Flatten (* _G. C. Greubel_, Jun 25 2021 *)

%o (Sage)

%o @CachedFunction

%o def f(n,x): return (2-x)^n if (n<2) else 2*(2-x)*sum( ((n-1)/(2*n-j-2))*binomial(2*n-j-2, j)*(-x)^(n-j-1) for j in (0..n-1) )

%o def T(n,k): return factorial(n) if (k==0) else product( f(j, k+1) for j in (1..n-1) )

%o flatten([[T(k,n-k) for k in (0..n)] for n in (0..15)]) # _G. C. Greubel_, Jun 25 2021

%Y Cf. A129862, A156608, A156609, A156610.

%K sign,tabl

%O 0,6

%A _Roger L. Bagula_, Feb 11 2009

%E Edited by _G. C. Greubel_, Jun 25 2021