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%I #45 Nov 04 2016 07:18:52
%S 1,5,41,441,5341,68845,922621,12701245,178338145,2542242545,
%T 36677022081,534311328705,7846771001041,116019251361041,
%U 1725360846921041,25786805857871441,387084441100423541,5832802431123111941
%N Partial sums of binomial(2n,n)^2.
%C Central coefficients of number triangle A115255.
%C p divides all a(n) from a((p-1)/2) to a(p-1) for Gaussian primes p=7,23,31,79,167,431,479,983, ... of the form 4n+3, A002145(n) and for primes of the form 8n+7, A007522(n). - _Alexander Adamchuk_, Jul 05 2006
%C Conjecture: For any positive integer n, the polynomials Sum_{k=0}^n binomial(2k,k)^2*x^k and Sum_{k=0}^n binomial(2k,k)^2*x^k/(k+1) are irreducible over the field of rational numbers. - _Zhi-Wei Sun_, Mar 23 2013
%H Vincenzo Librandi, <a href="/A115257/b115257.txt">Table of n, a(n) for n = 0..200</a>
%H DLMF Digital Library of Mathematical Functions, <a href="http://dlmf.nist.gov/19.2#E4">Elliptic Integrals</a>, NIST, 2016.
%F a(n) = Sum_{k=0..n} C(2k, k)^2. a(n) = A115255(2n, n).
%F a(n) = C(2n,n)^2 + C(2n-2,n-1)^2 + ... + C(2k,k)^2 + ... + C(2,1)^2 + C(0,0)^2, where C(2k,k) = (2k)!/(k!)^2 are the central binomial coefficients A000984(k). - _Alexander Adamchuk_, Jul 05 2006
%F a(n) = Sum_{k=0..n} ((2k)!/(k!)^2)^2. a(n) = Sum_{k=0..n} A000984[k]^2. - _Alexander Adamchuk_, Jul 05 2006
%F Recurrence: n^2*a(n) = (17*n^2-16*n+4)*a(n-1) - 4*(2*n-1)^2*a(n-2). - _Vaclav Kotesovec_, Oct 19 2012
%F a(n) ~ 16^(n+1)/(15*Pi*n). - _Vaclav Kotesovec_, Oct 19 2012
%F From _Emanuele Munarini_, Oct 28 2016: (Start)
%F Let K(x) be the complete elliptic integral of the first kind as defined in [DLMF, 19.2.4] for phi = Pi/2.
%F a(n) = (2/Pi)*K(16)-((16^(n+1)*Gamma(n+3/2)^2)/(Pi*Gamma(n+2)^2))*hypergeometric (1,n+3/2,n+3/2;n+2,n+2;16).
%F G.f.: A(t) = (2/Pi)*(K(16*t)/(1-t)).
%F Diff. eq. satisfied by the g.f. t*(1-17*t+16*t^2)*A''(t)+(1-35*t+64*t^2)*A'(t)-(5-36*t)*A(t)=0. (End)
%p series( 2*EllipticK(4*x^(1/2))/(Pi*(1-x)) ,x=0,20); # _Mark van Hoeij_, Apr 06 2013
%t Table[Sum[((2k)!/(k!)^2)^2,{k,0,n}], {n,0,40}] (* _Alexander Adamchuk_, Jul 05 2006 *)
%t Accumulate[(Binomial[2#,#])^2&/@Range[0,20]] (* _Harvey P. Dale_, Mar 04 2011 *)
%o (Maxima) makelist(sum(binomial(2*k,k)^2,k,0,n),n,0,12); /* _Emanuele Munarini_, Oct 28 2016 */
%o (PARI) a(n) = sum(k=0, n, binomial(2*k, k)^2); \\ _Michel Marcus_, Oct 30 2016
%Y Cf. A000984, A002145, A007522, A115255, A228002.
%K easy,nonn
%O 0,2
%A _Paul Barry_, Jan 18 2006
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