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%I #11 Jul 19 2024 14:26:11
%S 1,16,4,81,36,15,256,144,80,32,625,400,255,140,65,1296,900,624,396,
%T 240,108,2401,1764,1295,896,609,364,175,4096,3136,2400,1760,1280,864,
%U 544,256,6561,5184,4095,3132,2385,1728,1215,756,369,10000,8100,6560,5180,4080,3100,2320,1620,1040,500
%N Triangle T read by rows: T(n,k) = (n - k)*n*(4*n^2 - 4*n*k + 2*k^2 - 1 + (-1)^k)/4, with 0 <= k < n.
%C T(n, k) is the k-th super- and subdiagonal sum of the Hankel matrix M(n) whose permanent is A374668(n).
%F O.g.f.: x*(1 - 4*x^8*y^5 + x*(11 + 2*y) - x^7*y^4*(7 + 16*y) - x^2*(-11 + 6*y - 6*y^2) - x^5*y^2*(2 - 46*y - 3*y^2) - x^6*y^3*(-2 - 27*y + 4*y^2) - x^3*(-1 + 18*y + 38*y^2 - 2*y^3) - x^4*y*(2 + 14*y + 2*y^2 - y^3))/((1 - x)^5*(1 - x*y)^4*(1 + x*y)^2).
%F T(n,2) = A123865(n-1) for n > 1.
%e n\k| 0 1 2 3 4 5
%e ---+------------------------------
%e 1 | 1
%e 2 | 16 4
%e 3 | 81 36 15
%e 4 | 256 144 80 32
%e 5 | 625 400 255 140 65
%e 6 | 1296 900 624 396 240 108
%e ...
%e For n = 3 the matrix M is
%e [ 1, 4, 15]
%e [ 4, 15, 32]
%e [15, 32, 65]
%e and therefore T(3, 0) = 1 + 15 + 65 = 81, T(3, 1) = 4 + 32 = 36, and T(3, 2) = 15.
%t T[n_,k_]:=(n-k)*n*(4*n^2 - 4*n*k+2*k^2-1+(-1)^k)/4; Table[T[n,k],{n,10},{k,0,n-1}]//Flatten
%Y Cf. A317614 (diagonal), A374668.
%Y Cf. A333119, A330613.
%Y Cf. A000583 (k=0), A035287 (k=1), A123865, A374709 (row sums).
%K nonn,easy,tabl
%O 1,2
%A _Stefano Spezia_, Jul 17 2024
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