A255900 - OEIS

%I #7 Aug 27 2018 03:46:37

%S 1,1,6,126,6291,581499,86010084,18599726484,5532984567639,

%T 2166830287921959,1080602568966548022,668603866168566179982,

%U 502601850887023486736757,451167540495577093495790397,476690705981608679350490956032,585587125411920596898761442409728

%N Expansion of exp( Sum_{n >= 1} A000464(n-1)*x^n/n ).

%C For similar results see A188514 and A255881.

%C A000464(n-1) = (-1)^n*4^(2*n-1)*E(2*n-1,1/4), where E(n,x) denotes the n-th Euler polynomial.

%C More generally, calculation suggests that for integer h and a nonzero integer k the expansion of exp ( Sum_{n >= 1} (2*k)^(2*n-1)*E(2*n-1,h/(2*k)) )*x^n/n has integer coefficients. This is the case h = 1 and k = 2.

%H G. C. Greubel, <a href="/A255900/b255900.txt">Table of n, a(n) for n = 0..200</a>

%F O.g.f.: exp( x + 11*x^2/2 + 361*x^3/3 + 24611*x^4/4 + ... ) = 1 + x + 6*x^2 + 126*x^3 + 6291*x^4 + ....

%F a(0) = 1 and a(n) = 1/n*Sum_{k = 1..n} (-1)^k*4^(2*k-1)*E(2*k-1,1/4)*a(n-k) for n >= 1.

%p #A255900

%p k := 2:

%p exp(add((2*k)^(2*n-1)*euler(2*n-1, 1/(2*k))*(-x)^n/n, n = 1 .. 15)): seq(coeftayl(%, x = 0, n), n = 0 .. 15);

%t A000464:= With[{nn = 200}, Take[CoefficientList[Series[Sin[x]/Cos[2 x], {x, 0, nn}], x]*Range[0, nn-1]!, {2, -1, 2}]]; a:= With[{nmax = 50}, CoefficientList[Series[Exp[x + Sum[A000464[[k]]*x^(k)/(k), {k, 2, 75}]], {x, 0, nmax}], x]]; Table[a[[n]], {n, 1, 50}] (* _G. C. Greubel_, Aug 26 2018 *)

%Y Cf. A000464, A188514, A255881.

%K nonn,easy

%O 0,3

%A _Peter Bala_, Mar 10 2015