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%I #13 Jun 08 2019 07:57:50
%S 1,1,3,23,371,10515,461869,28969177,2454072147,269732425859,
%T 37312477130105,6342352991066661,1299300852841580893,
%U 315702973949640373933,89765549161833322593411,29526682496433138896248775,11124674379405792463701519059
%N Expansion of exp( Sum_{n >= 1} A000364(n)*x^n/n ).
%C A000364(n) = (-1)^n*2^(2*n)*Euler(2*n,1/2), where E(n,x) is the n-th Euler polynomial. In general it appears that when k is a nonzero integer, the expansion of exp( Sum_{n >= 1} k^(2*n)*E(2*n,1/k)*(-x)^n/n ) has (positive) integer coefficients. See A255882 (k = 3), A255883(k = 4) and A255884 (k = 6).
%H G. C. Greubel, <a href="/A255881/b255881.txt">Table of n, a(n) for n = 0..200</a>
%H E. W. Weisstein, <a href="http://mathworld.wolfram.com/EulerPolynomial.html">Euler Polynomial</a>
%F O.g.f.: exp( x + 5*x^2/2 + 61*x^3/3 + 1385*x^4/4 + ... ) = 1 + x + 3*x^2 + 23*x^3 + 371*x^4 + ....
%F a(0) = 1 and for n >= 1, n*a(n) = Sum_{k = 1..n} (-1)^k*2^(2*k)*E(2*k,1/2)*a(n-k).
%F a(n) ~ 2^(4*n + 3) * n^(2*n - 1/2) / (Pi^(2*n + 1/2) * exp(2*n)). - _Vaclav Kotesovec_, Jun 08 2019
%p #A255881
%p k := 2:
%p exp(add(k^(2*n)*euler(2*n, 1/k)*(-x)^n/n, n = 1 .. 16)): seq(coeftayl(%, x = 0, n), n = 0 .. 16);
%t A000364:= Table[Abs[EulerE[2 n]], {n, 0, 80}]; a:= With[{nmax = 70}, CoefficientList[Series[Exp[Sum[A000364[[k + 1]]*x^(k)/(k), {k, 1, 75}]], {x, 0, nmax}], x]]; Table[a[[n]], {n, 1, 50}] (* _G. C. Greubel_, Aug 26 2018 *)
%Y Cf. A000364, A188514, A255882, A255883, A255884.
%K nonn,easy
%O 0,3
%A _Peter Bala_, Mar 09 2015
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