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T(n, k) = 2^n * n! * [x^k] [z^n] (exp(z) + 1)^2/(4*exp(x*z)), triangle read by rows, for 0 <= k <= n.
0

%I #10 Jul 21 2019 08:56:40

%S 1,2,-2,6,-8,4,20,-36,24,-8,72,-160,144,-64,16,272,-720,800,-480,160,

%T -32,1056,-3264,4320,-3200,1440,-384,64,4160,-14784,22848,-20160,

%U 11200,-4032,896,-128,16512,-66560,118272,-121856,80640,-35840,10752,-2048,256

%N T(n, k) = 2^n * n! * [x^k] [z^n] (exp(z) + 1)^2/(4*exp(x*z)), triangle read by rows, for 0 <= k <= n.

%F Generated by 1/A326480(z), where A326480(z) denotes the generating function of A326480 which generates the Euler polynomials of order 2.

%e [0] [ 1]

%e [1] [ 2, -2]

%e [2] [ 6, -8, 4]

%e [3] [ 20, -36, 24, -8]

%e [4] [ 72, -160, 144, -64, 16]

%e [5] [ 272, -720, 800, -480, 160, -32]

%e [6] [ 1056, -3264, 4320, -3200, 1440, -384, 64]

%e [7] [ 4160, -14784, 22848, -20160, 11200, -4032, 896, -128]

%e [8] [16512, -66560, 118272, -121856, 80640, -35840, 10752, -2048, 256]

%e [9] [65792, -297216, 599040, -709632, 548352, -290304, 107520, -27648, 4608, -512]

%p IE2 := proc(n) (exp(z) + 1)^2/(4*exp(x*z));

%p series(%, z, 48); 2^n*n!*coeff(%, z, n) end:

%p for n from 0 to 9 do PolynomialTools:-CoefficientList(IE2(n), x) od;

%t T[n_, k_] := 2^n n! SeriesCoefficient[(E^z + 1)^2/(4 E^(x z)), {x, 0, k}, {z, 0, n}]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 21 2019 *)

%Y Cf. A326480, A063376.

%K sign,tabl

%O 0,2

%A _Peter Luschny_, Jul 12 2019