cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A188680 Alternate partial sums of binomial(3n,n)^2.

Original entry on oeis.org

1, 8, 217, 6839, 238186, 8779823, 335842273, 13185196127, 527732395714, 21438596184911, 881264330165314, 36575197658193086, 1530121867019096914, 64443673226319500222, 2729760145163758146178, 116203781083772019594878
Offset: 0

Views

Author

Emanuele Munarini, Apr 08 2011

Keywords

Crossrefs

Cf. Alternate partial sums of binomial(k*n,n)^2: A228002 (k=2), this sequence (k=3).

Programs

  • Mathematica
    Table[Sum[Binomial[3k,k]^2(-1)^(n-k),{k,0,n}],{n,0,20}]
  • Maxima
    makelist(sum(binomial(3*k,k)^2*(-1)^(n-k),k,0,n),n,0,20);
    
  • PARI
    a(n)=my(t=1); sum(k=1,n, t*=(27*k^2 - 27*k + 6)/(4*k^2 - 2*k); (-1)^(n-k)*t^2)+(-1)^n \\ Charles R Greathouse IV, Nov 02 2016

Formula

a(n) = sum(C(3k,k)^2*(-1)^(n-k), k=0..n).
Recurrence: 4*(2*n^2+7*n+6)^2 * a(n+2) -(713*n^4+4262*n^3+9509*n^2 +9384*n+3456) * a(n+1) -9*(9*n^2+27*n+20)^2 * a(n) = 0.
G.f.: (1+x)^(-1)*F(1/3,1/3,2/3,2/3;1/2,1/2,1;729*x/16), where F(a1,a2,a3,a4;b1,b2,b3;z) is a hypergeometric series.
a(n) ~ 3^(6*n+7)/(745*Pi*n*2^(4*n+2)). - Vaclav Kotesovec, Aug 06 2013

A115257 Partial sums of binomial(2n,n)^2.

Original entry on oeis.org

1, 5, 41, 441, 5341, 68845, 922621, 12701245, 178338145, 2542242545, 36677022081, 534311328705, 7846771001041, 116019251361041, 1725360846921041, 25786805857871441, 387084441100423541, 5832802431123111941
Offset: 0

Views

Author

Paul Barry, Jan 18 2006

Keywords

Comments

Central coefficients of number triangle A115255.
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
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

Crossrefs

Programs

  • Maple
    series( 2*EllipticK(4*x^(1/2))/(Pi*(1-x)) ,x=0,20); # Mark van Hoeij, Apr 06 2013
  • Mathematica
    Table[Sum[((2k)!/(k!)^2)^2,{k,0,n}], {n,0,40}] (* Alexander Adamchuk, Jul 05 2006 *)
    Accumulate[(Binomial[2#,#])^2&/@Range[0,20]]  (* Harvey P. Dale, Mar 04 2011 *)
  • Maxima
    makelist(sum(binomial(2*k,k)^2,k,0,n),n,0,12); /* Emanuele Munarini, Oct 28 2016 */
    
  • PARI
    a(n) = sum(k=0, n, binomial(2*k, k)^2); \\ Michel Marcus, Oct 30 2016

Formula

a(n) = Sum_{k=0..n} C(2k, k)^2. a(n) = A115255(2n, n).
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
a(n) = Sum_{k=0..n} ((2k)!/(k!)^2)^2. a(n) = Sum_{k=0..n} A000984[k]^2. - Alexander Adamchuk, Jul 05 2006
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
a(n) ~ 16^(n+1)/(15*Pi*n). - Vaclav Kotesovec, Oct 19 2012
From Emanuele Munarini, Oct 28 2016: (Start)
Let K(x) be the complete elliptic integral of the first kind as defined in [DLMF, 19.2.4] for phi = Pi/2.
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).
G.f.: A(t) = (2/Pi)*(K(16*t)/(1-t)).
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)
Showing 1-2 of 2 results.