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%I #21 Oct 31 2016 11:27:05
%S 1,5,45,521,6733,92385,1316865,19274925,287694285,4359037985,
%T 66837293545,1034774126325,16149186405025,253737607849445,
%U 4009771017244485,63681603585696321,1015763347140335565,16264070907887454465
%N a(n)= Sum_{i=0..n} C(n,i)*C(2i,i)^2.
%H Seiichi Manyama, <a href="/A248586/b248586.txt">Table of n, a(n) for n = 0..815</a>
%F a(n) = Sum_{i=0..n) A007318(n,i)*A002894(i).
%F Conjecture: n^2*a(n) +(-19*n^2+19*n-5)*a(n-1) +35*(n-1)^2*a(n-2) -17*(n-1)*(n-2)*a(n-3)=0.
%F G.f.: LegendreP(-1/2, (1+15x)/(1-17x)) /[sqrt(1-17x)*sqrt(1-x)]. - Corrected by _Robert Israel_, Oct 28 2016
%F From _Emanuele Munarini_, Oct 28 2016: (Start)
%F a(n) = hypergeometric(1/2,1/2,-n;1,1;-16).
%F G.f.: A(t) = (2/Pi)*(ellipticK(16*t/(1-t))/(1-t)).
%F Diff. eq. satisfied by the g.f.: t*(1-t)*(1-18*t+17*t^2)*A''(t)+(1-t)*(1-37*t+68*t^2)*A'(t)-(34*t^2-35*t+5)*A(t)=0.
%F Remark: the conjectured recurrence for the coefficients a(n) comes from this diff. eq. for A(t).
%F (End)
%F a(n) ~ 17^(n+1)/(16*Pi*n). - _Vaclav Kotesovec_, Oct 30 2016
%t Table[Sum[Binomial[n, k] Binomial[2k, k]^2, {k, 0, n}],{n,0,100}] (* _Emanuele Munarini_, Oct 28 2016 *)
%o (PARI) a(n) = sum(i=0, n, binomial(n,i)*binomial(2*i,i)^2); \\ _Michel Marcus_, Oct 09 2014
%o (Maxima) makelist(sum(binomial(n,k)*binomial(2*k,k)^2,k,0,n),n,0,12); /* _Emanuele Munarini_, Oct 28 2016 */
%Y Cf. A002894 (inverse binomial transform), A002893.
%K nonn,easy
%O 0,2
%A _R. J. Mathar_, Oct 09 2014
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