|
%I #15 Jun 08 2019 08:12:44
%S 1,5,115,7955,1179715,304888655,121350927565,68751844662605,
%T 52528700295424915,52031089992310711055,64835758857480094584265,
%U 99249388572274155967996505,183075972804988649078529524365,400493686169423616676960341062705,1025151296160300228944197705742007715
%N Expansion of exp( Sum_{n >= 1} A002438(n)*x^n/n ).
%C A002438(n+1) =(-1)^n*6^(2*n)*E(2*n,1/6), where E(n,x) denotes 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 A255881 (k = 2), A255882(k = 3) and A255883 (k = 4).
%H G. C. Greubel, <a href="/A255884/b255884.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( 5*x + 205*x^2/2 + 22265*x^3/3 + 4544185 *x^4/4 + ... ) = 1 + 5*x + 115*x^2 + 7955*x^3 + 1179715*x^4 + ....
%F a(0) = 1 and for n >= 1, n*a(n) = Sum_{k = 1..n} (-1)^k*6^(2*k)*E(2*k,1/6)*a(n-k).
%F a(n) ~ 2^(4*n + 2) * 3^(2*n) * n^(2*n - 1/2) / (exp(2*n) * Pi^(2*n + 1/2)). - _Vaclav Kotesovec_, Jun 08 2019
%p #A255884
%p k := 6:
%p exp(add(k^(2*n)*euler(2*n, 1/k)*(-x)^n/n, n = 1 .. 14)): seq(coeftayl(%, x = 0, n), n = 0 .. 14);
%t A000243[n_]:= (1 + 9^(n - 1))*Abs[EulerE[2*(n - 1)]]/2; a:= With[{nmax = 75}, CoefficientList[Series[Exp[Sum[A000243[k + 1]*x^(k)/(k), {k, 1, 85}]], {x, 0, nmax}], x]]; Table[a[[n]], {n, 1, 50}] (* _G. C. Greubel_, Aug 26 2018 *)
%Y Cf. A002438, A188514, A255881, A255882, A255883.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Mar 09 2015
|
