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A076729 a(n) = A001147(n+1) * Integral_{x=0..1} (1 + x^2)^n dx.

%I A076729 #59 Jan 22 2025 08:52:24
%S A076729 1,4,28,288,3984,70080,1506240,38384640,1133072640,38038533120,
%T A076729 1431213235200,59645279232000,2726781752217600,135661078090137600,
%U A076729 7295806823277772800,421717409630060544000,26071235813929033728000,1716456412254215503872000,119894838461795743137792000
%N A076729 a(n) = A001147(n+1) * Integral_{x=0..1} (1 + x^2)^n dx.
%C A076729 Numerator of the integral where denominator is equal to (2n+1)!! = A001147(n+1).
%C A076729 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.
%C A076729 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
%C A076729 Self-convolution of A001813. - _Vladimir Reshetnikov_, Oct 11 2016
%F A076729 a(n) = 2*n*a(n-1) + (2*n)!/n!.
%F A076729 a(n) = 2^n*Sum_{k=0..n} A001147(k)*A001147(n-k).
%F A076729 a(n) = (2*n+1)!*Sum_{k=0..n} k!*(-2)^k/((2*k+1)!*(n-k)!).
%F A076729 a(n) = (2*n+1)!!*hypergeom([1/2, -n], [3/2], -1). - _Vladeta Jovovic_, Dec 05 2002
%F A076729 E.g.f.: 1/((1-2*x)*sqrt(1-4*x)). - _Vladeta Jovovic_, May 11 2003
%F A076729 G.f.: hypergeom([1,1/2],[],4*x)^2  - _Mark van Hoeij_, May 16 2013
%F A076729 a(n) ~ 2^(2*n+3/2)*n^n/exp(n). - _Vaclav Kotesovec_, Oct 05 2013
%F A076729 a(n) = (2n+1)!!*Sum_{i=0..n} binomial(n,i)/(2i+1). - _John M. Campbell_, Feb 06 2016
%F A076729 From _Vladimir Reshetnikov_, Oct 08 2016: (Start)
%F A076729 a(n) = 2^n*A034430(n) = -(2*n+1)!! * Im(Beta(2, n+1, 1/2))/2.
%F A076729 Recurrence: 2*(3*n+2)*a(n) = a(n+1) + 4*n*(2*n+1)*a(n-1). (End)
%F A076729 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
%F A076729 a(n) = 2^n*n!*JacobiP(n, 1/2, -1 - n, 3). - _Peter Luschny_, Jan 22 2025
%e A076729 For n = 3, (2n+1)!! = 105 and the integral is 96/35 = 288/105, so a(3) = 288.
%p A076729 seq((doublefactorial(2*n+1))*add((binomial(n, i))/(2*i+1), i=0..n), n=0..20) ; # _John M. Campbell_, Feb 06 2016
%p A076729 A076729 := n -> 2^n*n!*JacobiP(n, 1/2, -1 - n, 3):
%p A076729 seq(simplify(A076729(n)), n = 0..18);  # _Peter Luschny_, Jan 22 2025
%t A076729 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 *)
%t A076729 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 *)
%t A076729 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 *)
%o A076729 (PARI) a(n)=if(n<0,0,subst(intformal((1+x^2)^n),x,1)*(2*n+1)!/2^n/n!)
%Y A076729 Cf. A001147, A077595, A077745, A086891, A034430, A001813.
%K A076729 nonn
%O A076729 0,2
%A A076729 Al Hakanson (hawku(AT)hotmail.com), Oct 28 2002