A112029 a(n) = Sum_{k=0..n} binomial(n+k, k)^2.
1, 5, 46, 517, 6376, 82994, 1119210, 15475205, 217994860, 3115374880, 45035696036, 657153097330, 9663914317396, 143050882063262, 2129448324373546, 31853280798384645, 478503774600509620, 7215090439396842572, 109154411037070011504, 1656268648035559711392
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
- F. Baldassarri, S. Bosch, B. Dwork, (eds), p-adic Analysis. Lecture Notes in Mathematics, vol. 1454, pp. 194 - 204, Springer, Berlin, Heidelberg.
- Matthijs J. Coster, Supercongruences.
- C. Elsner, On recurrence formulas for sums involving binomial coefficients, Fib. Q., 43,1 (2005), 31-45.
- Vaclav Kotesovec, Asymptotic of generalized Apery sequences with powers of binomial coefficients, Nov 04 2012
- Pedro J. Miana, Hideyuki Ohtsuka, and Natalia Romero, Sums of powers of Catalan triangle numbers, arXiv:1602.04347 [math.NT], 2016.
Programs
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Magma
[(&+[Binomial(n+j, j)^2: j in [0..n]]): n in [0..20]]; // G. C. Greubel, Jul 06 2021
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Maple
f := 64*x^2/(16*x-1); S := sqrt(x)*sqrt(4-x); H := ((10*x-5/8)*hypergeom([1/4,1/4],[1],f)-(21*x-21/8)*hypergeom([1/4,5/4],[1],f))/(S*(1-16*x)^(5/4)); ord := 30; ogf := series(int(series(H,x=0,ord),x)/S,x=0,ord); # Mark van Hoeij, Mar 27 2013
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Mathematica
Table[Sum[Binomial[n+k,k]^2, {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Nov 23 2012 *) -
PARI
a(n) = sum(k=0, n, binomial(n+k, k)^2); \\ Michel Marcus, Jul 07 2021
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Sage
[sum(binomial(n+j, j)^2 for j in (0..n)) for n in (0..20)] # G. C. Greubel, Jul 06 2021
Formula
a(n) ~ 2^(4*n+2)/(3*Pi*n). - Vaclav Kotesovec, Nov 23 2012
Recurrence: 2*(2*n+1)*(21*n-13)*n^2*a(n) = (1365*n^4 - 1517*n^3 + 240*n^2 + 216*n - 64)*a(n-1) - 4*(n-1)*(2*n-1)^2*(21*n+8)*a(n-2). - Vaclav Kotesovec, Nov 23 2012
G.f.: see Maple code. - Mark van Hoeij, Mar 27 2013
a(p-1) == 1 (mod p^3) for all primes p >= 5. See the comments in A173774. - Peter Bala, Jul 12 2024
a(n-1) = 1/(4*n) * binomial(2*n, n)^2 * ( 1 + 3*((n - 1)/(n + 1))^3 + 5*((n - 1)*(n - 2)/((n + 1)*(n + 2)))^3 + 7*((n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)))^3 + ... ) for n >= 1. - Peter Bala, Jul 22 2024
a(m*p^r - 1) == a(m*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and positive integers m and r. See Coster, Theorem 4. - Peter Bala, Nov 29 2024
a(n) = A110197(2n,n). - Alois P. Heinz, Mar 21 2025