A334907 Comtet's expansion of the e.g.f. (sqrt(1 + sqrt(8*s)) - sqrt(1 - sqrt(8*s)))/ sqrt(8*s * (1 - 8*s)).

1, 5, 63, 1287, 36465, 1322685, 58503375, 3053876175, 183771489825, 12525477859125, 953725671273375, 80237355387564375, 7391465178302430225, 739967791738943292525, 79993069900054731795375, 9286937373235386442953375, 1152424501315118408602850625

Offset: 0

Author

Petros Hadjicostas, May 15 2020

Keywords

• nonn

Comments

A special case of an integral in Comtet (1967, pp. 85-86) yields

Integral_{t=-oo..oo} dx/(x^2 + t^2)^(2*n) = Pi * a(n-1)/((n-1)! * 2^(3*n - 2) * t^(4*n-1)) for n >= 1 and t > 0. This integral also follows from a theorem in Moll (2002, p. 312, set a=1), but it requires the summation formula for a(n) shown below.

Links

• Louis Comtet, Fonctions génératrices et calcul de certaines intégrales, Publikacije Elektrotechnickog faculteta - Serija Matematika i Fizika, No. 181/196 (1967), 77-87; see pp. 81-83.
• Petros Hadjicostas, Proof of the claim a(n) = n!*A063079(n+1)/A060818(n), 2020.
• V. H. Moll, The evaluation of integrals: a personal story, Notices Amer. Math. Soc., 49 (No. 3, March 2002), 311-317.

Crossrefs

Cf. A001790, A060818, A063079, A223549, A223550.

Formula

a(n) = binomial(4*n+2, 2*n+1)*n!/2^(n+1).

a(n) = n!*A063079(n+1)/A060818(n) = n!*A001790(2*n+1)/A060818(n) (see the link for a proof).

a(n) = n!*Sum_{j=0..n} 2^(n-2*j)*binomial(2*n+1,2*j)*binomial(2*j,j).

a(n) = 2^n*n!*Sum_{k=0..n} A223549(n,k)/A223550(n,k).

E.g.f.: 2/(sqrt(1 - 8*s) * (sqrt(1 + sqrt(8*s)) + sqrt(1 - sqrt(8*s)))).

E.g.f.: sqrt(2/(1 + sqrt(1 - 8*s))/(1 - 8*s)).

D-finite with recurrence (2*n+1)*a(n) -(4*n-1)*(4*n+1)*a(n-1)=0. - R. J. Mathar, May 25 2020