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.

1, 3, 13, 79, 641, 6579, 81677, 1187039, 19728193, 368562723, 7639512013, 173893382575, 4310656806977, 115569893763411, 3331588687405133, 102751933334045375, 3375782951798785921, 117693183724386637635

Offset: 1

Author

Paul D. Hanna, Oct 09 2005

Keywords

• nonn

Examples

			log(1+x + 2*x*[x + 3*x^2 + 13*x^3 + 79*x^4 + 641*x^5 +...])
= x + 3/2*x^2 + 13/3*x^3 + 79/4*x^4 + 641/5*x^5 +...
		

Crossrefs

Cf. A001147, A112934; A112936, A112937, A112938, A112939, A112940, A112941, A112942, A112943.

Programs

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))}

Formula

G.f.: log(1 + x + 2*x*[Sum_{n>=1} a(n)*x^n]) = Sum_{k>=1} a(n)/n*x^n.

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

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