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A326480 T(n, k) = 2^n * n! * [x^k] [z^n] (4*exp(x*z))/(exp(z) + 1)^2, triangle read by rows, for 0 <= k <= n. Coefficients of Euler polynomials of order 2.

%I A326480 #21 Jul 15 2019 02:44:33
%S A326480 1,-2,2,2,-8,4,4,12,-24,8,-16,32,48,-64,16,-32,-160,160,160,-160,32,
%T A326480 272,-384,-960,640,480,-384,64,544,3808,-2688,-4480,2240,1344,-896,
%U A326480 128,-7936,8704,30464,-14336,-17920,7168,3584,-2048,256
%N A326480 T(n, k) = 2^n * n! * [x^k] [z^n] (4*exp(x*z))/(exp(z) + 1)^2, triangle read by rows, for 0 <= k <= n. Coefficients of Euler polynomials of order 2.
%C A326480 T(m, n, k) = 2^n * n! * [x^k] [z^n] (2^m*exp(x*z))/(exp(z) + 1)^m are the coefficients of the generalized Euler polynomials (or Euler polynomials of higher order).
%C A326480 The classical case (m=1) is in A004174, this sequence is case m=2. A different normalization for m=1 is given in A058940 and for m=2 in A326485.
%C A326480 Generalized Euler numbers are 2^n*Sum_{k=0..n} T(m, n, k)*(1/2)^k. The classical Euler numbers are in A122045 and for m=2 in A326483.
%H A326480 NIST Digital Library of Mathematical Functions, §24.16(i), <a href="https://dlmf.nist.gov/24.16#i">Higher-Order Analogs (of Bernoulli and Euler Polynomials)</a>, Release 1.0.23 of 2019-06-15.
%e A326480 Triangle starts:
%e A326480 [0] [     1]
%e A326480 [1] [    -2,       2]
%e A326480 [2] [     2,      -8,     4]
%e A326480 [3] [     4,      12,   -24,      8]
%e A326480 [4] [   -16,      32,    48,    -64,     16]
%e A326480 [5] [   -32,    -160,   160,    160,   -160,     32]
%e A326480 [6] [   272,    -384,  -960,    640,    480,   -384,    64]
%e A326480 [7] [   544,    3808, -2688,  -4480,   2240,   1344,  -896,   128]
%e A326480 [8] [ -7936,    8704, 30464, -14336, -17920,   7168,  3584, -2048,   256]
%e A326480 [9] [-15872, -142848, 78336, 182784, -64512, -64512, 21504,  9216, -4608, 512]
%p A326480 E2 := proc(n) (4*exp(x*z))/(exp(z) + 1)^2;
%p A326480 series(%, z, 48); 2^n*n!*coeff(%, z, n) end:
%p A326480 ListTools:-Flatten([seq(PolynomialTools:-CoefficientList(E2(n), x), n=0..9)]);
%t A326480 T[n_, k_] := 2^n n! SeriesCoefficient[4 Exp[x z]/(Exp[z]+1)^2, {z, 0, n}, {x, 0, k}];
%t A326480 Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Jul 15 2019 *)
%Y A326480 Let E2_{n}(x) = Sum_{k=0..n} T(n,k) x^k. Then E2_{n}(1) = A155585(n+1),
%Y A326480 E2_{n}(0) = A326481(n), E2_{n}(-1) = A326482(n), 2^n*E2_{n}(1/2) = A326483(n),
%Y A326480 2^n*E2_{n}(-1/2) = A326484(n), [x^n] E2_{n}(x) = A000079(n).
%Y A326480 Cf. A004174, A058940, A326485, A122045, A081733.
%K A326480 sign,tabl
%O A326480 0,2
%A A326480 _Peter Luschny_, Jul 11 2019