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%I #10 Jun 23 2024 16:05:57
%S 0,2,4,-10,-56,362,2764,-24610,-250736,2873042,36581524,-512343610,
%T -7828053416,129570724922,2309644635484,-44110959165010,
%U -898621108880096,19450718635716002,445777636063460644,-10784052561125704810,-274613643571568682776,7342627959965776406282
%N a(n) = 2^n*n!*([z^n] exp(x*z)*tanh(z))(1/2).
%F a(n) = 1 - 4^n*Euler(n, 1/4).
%F Let p(n, x) = -x^n + Sum_{k=0..n} binomial(n,k)*Euler(k)*(x+1)^(n-k) (the polynomials defined in A162660), then a(n) = 2^n*p(n, 1/2).
%p seq(1 - 4^n*euler(n, 1/4), n=0..21);
%t p := CoefficientList[Series[Exp[x z] Tanh[z], {z, 0, 21}], z];
%t norm := Table[2^n n!, {n, 0, 21}]; norm (p /. x -> 1/2)
%Y Cf. A162660, A009832, A155585, A212435.
%K sign
%O 0,2
%A _Peter Luschny_, Jun 28 2019