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a(n) = (4*n)! / ((2*n)!*n!^2).
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%I #121 Oct 16 2024 09:23:15

%S 1,12,420,18480,900900,46558512,2498640144,137680171200,7735904619300,

%T 441233078286000,25467973278667920,1484298740174927040,

%U 87202550985276963600,5157850293780050462400,306839461354466267304000,18344908596179023234548480

%N a(n) = (4*n)! / ((2*n)!*n!^2).

%C Appears in Ramanujan's theory of elliptic functions of signature 4.

%C H. A. Verrill proves that a(n) = Sum_{p + q + r = 3n} w^(p-q) * {(3n)!/(p! q! r!)}^2, with p, q, r >= 0 and w = primitive 3rd root of unity.

%C The family of elliptic curves "x=2*H1=p^2+q^2-(1/4)*q^4, 0<x<1" generates these a_n as the coefficients of the period-energy function "T(x)=2*Pi*2F1(1/4,3/4;1;x)". Applying complex transformation "q->sqrt(-1)*q" to H1 produces "x=2*H2=p^2-q^2-(1/4)*q^4, 0<x<1", with "T(x)=sqrt(2)*Pi*2F1(1/4,3/4;1;1-x)". This explains the appearance of factor sqrt(2)/2 in Ramanujan's nome q_1. - _Bradley Klee_, Feb 25 2018

%C Even-order terms in the diagonal of rational function 1/(1 - (x^2 + y^2 + z)). - _Gheorghe Coserea_, Aug 09 2018

%D E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, 1975, p. 96.

%H Vincenzo Librandi, <a href="/A000897/b000897.txt">Table of n, a(n) for n = 0..200</a>

%H Alin Bostan, Armin Straub, and Sergey Yurkevich, <a href="https://arxiv.org/abs/2212.10116">On the representability of sequences as constant terms</a>, arXiv:2212.10116 [math.NT], 2022.

%H H. J. Brothers, <a href="http://www.brotherstechnology.com/docs/Pascal&#39;s_Prism_(supplement).pdf">Pascal's Prism: Supplementary Material</a>

%H B. Klee, <a href="http://list.seqfan.eu/oldermail/seqfan/2017-December/018186.html">Geometric G.F. for Ramanujan Periods</a>, seqfan mailing list, 2017.

%H S. Ramanujan, <a href="http://ramanujan.sirinudi.org/Volumes/published/ram06.pdf">Modular Equations and Approximations to Pi</a>, Quarterly Journal of Mathematics, XLV (1914), 350-372.

%H L. C. Shen, <a href="https://doi.org/10.1007/s11139-011-9360-8">A note on Ramanujan’s identities involving the hypergeometric function 2F1(1/6,5/6;1;z)</a>, The Ramanujan Journal, 30.2 (2013), 211-222.

%H H. A. Verrill, <a href="https://arxiv.org/abs/math/0407327">Sums of squares of binomial coefficients, with applications to Picard-Fuchs equations</a>, arXiv:math/0407327v1 [math.CO], 2004.

%F E.g.f.: Sum_{k>=0} (-1)^k * a(k) * x^(4*k) / (4*k)! = BesselI(0, 2x) * BesselJ(0, 2x).

%F G.f.: F(1/4, 3/4; 1; 64*x). - _Michael Somos_, Oct 31 2005

%F a(n) = A008977(n)/A000984(n) - _Zerinvary Lajos_, Jun 28 2007

%F Sum_{k>=0} a(k) * x^(3k)/(3k)!^2 = f(x)*f(x*w)*f(x/w) where f(x) = BesselI(0, 2*sqrt(x)) and w = primitive 3rd root of unity. - _Michael Somos_, Jul 25 2007

%F In general, for (BesselI(b, 2x))*(BesselJ(b, 2x))=((x^(2*b))/((GAMMA(b+1))^2)*(1-(x^4)/(Q(0)+(x^4))); Q(k)=(k+1)*(k+b+1)*(2*k+b+1)*(2*k+b+2)-(x^4)+(x^4)*(k+1)*(k+b+1)*(2*k+b+1)*(2*k+b+2)/Q(k+1)) ; (continued fraction). - _Sergei N. Gladkovskii_, Nov 24 2011

%F D-finite with recurrence 0 = a(n)*4*(4*n + 1)*(4*n + 3) - a(n+1)*(n + 1)^2 for all n in Z. - _Michael Somos_, Aug 12 2014

%F 0 = a(n)*(-4026531840*a(n+2) +2005401600*a(n+3) -103896576*a(n+4) +1251948*a(n+5)) + a(n+1)*(+41418752*a(n+2) -30435328*a(n+3) +1863228*a(n+4) -24604*a(n+5)) + a(n+2)*(-16896*a(n+2) +75608*a(n+3) -6740*a(n+4) +105*a(n+5)) for all n in Z. - _Michael Somos_, Aug 12 2014

%F From _Peter Bala_, Jul 12 2016: (Start)

%F a(n) = binomial(3*n,n)*binomial(4*n,n) = A005809(n)*A005810(n) = ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) = [x^n](F(x)^(12*n)), where F(x) = 1 + x + 6*x^2 + 105*x^3 + 2448*x^4 + 67043*x^5 + 2028307*x^6 + ... appears to have integer coefficients. Cf. A002894, A002897, A006480, A008977, A186420 and A188662. (End)

%F a(n) ~ 2^(6*n-1/2)/(Pi*n). - _Ilya Gutkovskiy_, Jul 12 2016

%F G.f.: 2*EllipticK(sqrt((sqrt(1-64*x)-1)/(2*sqrt(1-64*x))))/(Pi*(1-64*x)^(1/4)) where EllipticK is the complete elliptic integral of the first kind (in Maple's notation). - _Robert Israel_, Jul 12 2016

%F a(n) = Sum_{k = 0..3*n} (-1)^k*C(3*n,k)*C(6*n-k,3*n)*C(2*k,k). - _Peter Bala_, Feb 10 2018

%F From _Bradley Klee_, Feb 27 2018: (Start)

%F a(n) = A000984(n)*A001448(n).

%F G.f.: (1/(sqrt(2)*Pi))*Integral_{q=-oo..oo} 1/sqrt(q^2+(1/4)*q^4+(1-64*x)) dq.

%F G.f.: (1/(2*Pi))*Integral_{phi=0..2*Pi} 1/sqrt(1-64*x*sin^4(phi)) dphi. (End)

%F From _Peter Bala_, Mar 20 2022: (Start)

%F Right-hand side of the following identities valid for n >= 1:

%F Sum_{k = 0..2*n} 2*n*(2*n+k-1)!/(k!*n!^2) = (4*n)!/((2*n)!*n!^2;

%F (3/2)*Sum_{k = 0..n} 2*n*(3*n+k-1)!/(k!*n!*(2*n)!) = (4*n)!/((2*n)!*n!^2).

%F Cf. A001451. (End)

%F a(n) = (4^n/n!^2)*Product_{k = 0..2*n-1} (2*k + 1). - _Peter Bala_, Feb 26 2023

%F a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * A108625(3*n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). - _Peter Bala_, Oct 15 2024

%e G.f.: 1 + 12*x + 420*x^2 + 18480*x^3 + 900900*x^4 + 46558512*x^5 + 2498640144*x^6 + ...

%p seq((4*n)!/(n!)^4/binomial(2*n,n), n=0..14); # _Zerinvary Lajos_, Jun 28 2007

%t Table[(4n)!/((2n)! n!^2), {n, 0, 30}] (* _Stefan Steinerberger_, Apr 14 2006 *)

%t a[ n_] := Binomial[ 4 n, 2 n] Binomial[ 2 n, n]; (* _Michael Somos_, Mar 24 2013 *)

%t a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/4, 3/4, 1, 64 x], {x, 0, n}]; (* _Michael Somos_, Mar 24 2013 *)

%t a[ n_] := If[ n < 0, 0, With[{m = 4 n}, (-1)^n m! SeriesCoefficient[ BesselI[ 0, 2 x] BesselJ[ 0, 2 x], {x, 0, m}]]]; (* _Michael Somos_, Aug 12 2014 *)

%t a[ n_] := 64^n Pochhammer[1/4, n] Pochhammer[3/4, n] / n!^2; (* _Michael Somos_, Aug 12 2014 *)

%o (PARI) {a(n) = if( n<0, 0, (4*n)! / ((2*n)! * n!^2))}; /* _Michael Somos_, Oct 31 2005 */

%o (GAP) a:=n->Sum([0..3*n],k->(-1)^k*Binomial(3*n,k)*Binomial(6*n-k,3*n)*

%o Binomial(2*k,k));;

%o A000897:=List([0..14],n->a(n)); # _Muniru A Asiru_, Feb 11 2018

%Y Cf. A002897, A008977, A186420, A188662. Elliptic Integrals: A002894, A113424, A006480. Factors: A005809, A005810, A000984, A001448.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_