cp's OEIS Frontend

A112935 Logarithmic derivative of A112934 such that a(n)=(1/2)*A112934(n+1) for n>0, where A112934 equals the INVERT transform of double factorials A001147.

%I A112935 #14 Feb 17 2017 07:47:52
%S A112935 1,3,13,79,641,6579,81677,1187039,19728193,368562723,7639512013,
%T A112935 173893382575,4310656806977,115569893763411,3331588687405133,
%U A112935 102751933334045375,3375782951798785921,117693183724386637635
%N A112935 Logarithmic derivative of A112934 such that a(n)=(1/2)*A112934(n+1) for n>0, where A112934 equals the INVERT transform of double factorials A001147.
%F A112935 G.f.: log(1 + x + 2*x*[Sum_{n>=1} a(n)*x^n]) = Sum_{k>=1} a(n)/n*x^n.
%F A112935 G.f.: (1 - 1/Q(0))/x where Q(k) = 1 - x*(2*k-1)/(1 - x*(2*k+4)/Q(k+1) ); (continued fraction). - _Sergei N. Gladkovskii_, Mar 19 2013
%F A112935 G.f.: 1/(x*G(0)) - 1/(2*x), where G(k)= 1 + 1/(1 - 2*x*(2*k+2)/(2*x*(2*k+2) - 1 + 2*x*(2*k+1)/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, May 31 2013
%e A112935 log(1+x + 2*x*[x + 3*x^2 + 13*x^3 + 79*x^4 + 641*x^5 +...])
%e A112935 = x + 3/2*x^2 + 13/3*x^3 + 79/4*x^4 + 641/5*x^5 +...
%o A112935 (PARI) {a(n)=local(F=1+x+x*O(x^n));for(i=1,n,F=1+x+2*x^2*deriv(F)/F); return(n*polcoeff(log(F),n,x))}
%Y A112935 Cf. A001147, A112934; A112936, A112937, A112938, A112939, A112940, A112941, A112942, A112943.
%K A112935 nonn
%O A112935 1,2
%A A112935 _Paul D. Hanna_, Oct 09 2005