|
%I #17 Mar 01 2014 05:00:56
%S 1,5,27,195,1785,19845,259875,3918915,66891825,1274998725,26843892075,
%T 618718975875,15495473018025,419010239773125,12167108660581875,
%U 377607284571391875,12473420957563190625,436953531082636693125,16179945969799083346875,631461179013117650071875
%N Numerators of sequence of fractions with e.g.f. (1+x)/(1-x)^(3/2).
%C Denominators are successive powers of 2.
%H Vincenzo Librandi, <a href="/A126119/b126119.txt">Table of n, a(n) for n = 0..200</a>
%F E.g.f.: (1+2*x)/(1-2*x)^(3/2)
%F From _Sergei N. Gladkovskii_, Jul 23 2012: (Start)
%F a(n) = (4*n+1)*((2*n)!)/( n!*(2^n) ).
%F G.f.: 3*Q(0), where Q(k)= 1 - (2*k+2)/(3*(2*k+1) - 9*x*(2*k+1)^2*(2*k+3)/(3*x*(2*k+1)*(2*k+3) - (2*k+2)/Q(k+1))); (continued fraction, 3rd kind, 3-step).
%F (End).
%F a(n) ~ 2^(n+5/2) * n^(n+1) / exp(n). - _Vaclav Kotesovec_, Feb 25 2014
%e The fractions are 1, 5/2, 27/4, 195/8, 1785/16, 19845/32, 259875/64, ...
%t CoefficientList[Series[(1+2*x)/(1-2*x)^(3/2), {x, 0, 20}], x] * Range[0, 20]! (* _Vaclav Kotesovec_, Feb 25 2014 *)
%o (PARI) {a(n)= if(n<0, 0, n!* polcoeff( (1+2*x)/ (1-2*x +x*O(x^n))^(3/2), n))} /* _Michael Somos_, Apr 08 2007 */
%Y Cf. A001174, A126115.
%K nonn,frac
%O 0,2
%A _N. J. A. Sloane_, Mar 22 2007
|
