A333592 a(n) = Sum_{k = 0..n} binomial(n+k-1, k)^2.

1, 2, 14, 146, 1742, 22252, 296438, 4063866, 56884430, 808970960, 11649069764, 169444272692, 2485268015414, 36707034407396, 545386280953262, 8144809577111146, 122177689609022670, 1839933272106181720, 27804610617723365072, 421476329309967621504, 6406685024966332359492

Offset: 0

Author

Peter Bala, Mar 27 2020

Keywords

• nonn
• easy

Comments

Compare with the closed-form evaluation Sum_{k = 0..n} C(n+k-1,k) = C(2*n,n) = A000984(n) (see the first comment in A001700 for a proof).

It is well-known that Sum_{k = 0..n} C(n,k)^2 = C(2*n,n). Here, we consider by analogy Sum_{k = 0..n} C(-n,k)^2, where C(-n,k) = (-1)^k*C(n+k-1,k) for integer n and nonnegative integer k.

The sequence b(n) = C(2*n,n) of central binomial coefficients satisfies the supercongruences b(n*p^k) = b(n*p^(k-1)) ( mod p^(3*k) ) for all primes p >= 5 and positive integers n and k - see Meštrović. We conjecture that the present sequence also satisfies these congruences. Some examples of the congruences are given below.

More generally, calculation suggests that for positive integer A and integer B, the sequence a(A,B;n) := Sum_{k = 0..A*n} C(B*n+k-1,k)^2 may satisfy the same congruences.

The sequence (a(p) - 2)/(2*p^3) for prime p >= 5 begins [89, 5924, 63652995, 8353899501, 187251503369243, 30724327840061789, 937835335872800013431, ...]. Cf. A034602.

Examples

			Examples of supercongruences:
a(11) - a(1) = 169444272692 - 2 = 2*(3^2)*5*7*(11^3)*397*509 == 0 ( mod 11^3 ).
a(2*7) - a(2) = 545386280953262 - 14 = (2^5)*(3^2)*(7^4)*788714021 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 5375188503768783714940459752 - 22252 = (2^2)*(5^6)*(31^2)* 89493252924350197127 == 0 ( mod 5^6 ).
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..500
• Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Crossrefs

Cf. A000984, A034602, A333593.

Programs

Maple

seq( add( binomial(n+k-1,k)^2, k = 0..n ), n = 0..25);

Mathematica

Table[Binomial[2*n-1, n]^2 * HypergeometricPFQ[{1, -n, -n}, {1 - 2*n, 1 - 2*n}, 1], {n, 1, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

PARI

a(n) = sum(k=0, n, binomial(n+k-1, k)^2); \\ Michel Marcus, Mar 29 2020.

Python

from math import comb
def A333592(n): return sum(comb(n+k-1,k)**2 for k in range(n+1)) if n else 1 # Chai Wah Wu, Oct 28 2022

Formula

a(n) ~ 2^(4*n) / (3*Pi*n). - Vaclav Kotesovec, Mar 28 2020

For n >= 1, a(n) = 2 * Sum_{k = 0..n-1} binomial(n+k, k)*binomial(n+k-1, k) = 2 * Sum_{k = 0..n-1} (n + k)/n *binomial(n+k-1, k)^2. - Peter Bala, Nov 02 2024