OFFSET
0,2
COMMENTS
Numerator of the integral where denominator is equal to (2n+1)!! = A001147(n+1).
Also numerator of the integral (1-x^2)^-(n+1/2) for x from 0 to sqrt(1/2). Here the sequence starts at n=1; at n=2 the function is 4.
a(n) = Integral_{x=0..log(1+sqrt(2))} cosh(x)^(2*n-1) dx where the denominators are b(n) = (2*n)!/(n!*2^n). E.g., a(3)=28 and b(3)=15; both offsets are 1. - Al Hakanson (hawkuu(AT)excite.com), Mar 02 2004
Self-convolution of A001813. - Vladimir Reshetnikov, Oct 11 2016
FORMULA
a(n) = 2*n*a(n-1) + (2*n)!/n!.
a(n) = (2*n+1)!*Sum_{k=0..n} k!*(-2)^k/((2*k+1)!*(n-k)!).
a(n) = (2*n+1)!!*hypergeom([1/2, -n], [3/2], -1). - Vladeta Jovovic, Dec 05 2002
E.g.f.: 1/((1-2*x)*sqrt(1-4*x)). - Vladeta Jovovic, May 11 2003
G.f.: hypergeom([1,1/2],[],4*x)^2 - Mark van Hoeij, May 16 2013
a(n) ~ 2^(2*n+3/2)*n^n/exp(n). - Vaclav Kotesovec, Oct 05 2013
a(n) = (2n+1)!!*Sum_{i=0..n} binomial(n,i)/(2i+1). - John M. Campbell, Feb 06 2016
From Vladimir Reshetnikov, Oct 08 2016: (Start)
a(n) = 2^n*A034430(n) = -(2*n+1)!! * Im(Beta(2, n+1, 1/2))/2.
Recurrence: 2*(3*n+2)*a(n) = a(n+1) + 4*n*(2*n+1)*a(n-1). (End)
Expansion of square of continued fraction 1/(1 - 2*x/(1 - 4*x/(1 - 6*x/(1 - 8*x/(1 - 10*x/(1 - ...)))))). - Ilya Gutkovskiy, Apr 19 2017
EXAMPLE
For n=3, (2n+1)!!=105 and the integral is 96/35 = 288/105, so a(3) = 288.
MAPLE
seq((doublefactorial(2*n+1))*sum((binomial(n, i))/(2*i+1), i=0..n), n=0..20) ; # John M. Campbell, Feb 06 2016
MATHEMATICA
a[n_] := (2n + 1)!!*Integrate[(1 + x^2)^n, {x, 0, 1}]; Table[ a[n], {n, 0, 16}] (* Robert G. Wilson v, Feb 27 2004 *)
Round@Table[-(2 n + 1)!! Im[Beta[2, n + 1, 1/2]]/2, {n, 0, 20}] (* Round is equivalent to FullSimplify here, but is much faster - Vladimir Reshetnikov, Oct 08 2016 *)
nxt[{n_, a_}]:={n+1, 2a(n+1)+(2(n+1))!/(n+1)!}; NestList[nxt, {0, 1}, 20][[All, 2]] (* Harvey P. Dale, Feb 04 2023 *)
PROG
(PARI) a(n)=if(n<0, 0, subst(intformal((1+x^2)^n), x, 1)*(2*n+1)!/2^n/n!)
CROSSREFS
KEYWORD
nonn
AUTHOR
Al Hakanson (hawku(AT)hotmail.com), Oct 28 2002
STATUS
approved