A333593 a(n) = Sum_{k = 0..n} (-1)^(n + k)*binomial(n + k - 1, k)^2.

1, 0, 6, 72, 910, 12000, 163086, 2266544, 32043726, 459167040, 6651400756, 97214919648, 1431514320886, 21213380196736, 316072831033350, 4731683468079072, 71128104013487310, 1073134004384407680, 16243463355081280080, 246585461357885877920

Offset: 0

Author

Peter Bala, Mar 27 2020

Keywords

• nonn
• easy

Comments

It is known that Sum_{k = 0..2*n} (-1)^(n+k)*C(2*n,k)^2 = C(2*n,n). Here, we consider by analogy Sum_{k = 0..n} (-1)^(n+k)*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 prime p >= 5 and any positive integers n and k - see Mestrovic. We conjecture that the present sequence also satisfies these congruences. Some examples of the congruences are given below. For a proof of the particular case of these congruences, a(p) == 0 ( mod p^3 ) for prime p >= 5, see the Bala link. The sequence a(p)/(2*p^3) for prime p >= 5 begins [48, 304, 36519504, 4827806144, 109213719151680, 17975321574419440, ...]. Cf. A034602.

More generally, calculation suggests that for positive integer A and integer B, the sequence a(A,B;n) := Sum_{k = 0..A*n} (-1)^(n+k)* C(-B*n,k)^2 = Sum_{k = 0..A*n} (-1)^(n+k)*C(B*n+k-1,k)^2 may also satisfy the above congruences for all prime p >= 5.

Examples

			Examples of congruences:
a(11)  = 97214919648 = (2^5)*3*(7^2)*(11^3)*15527 == 0 ( mod 11^3 ).
a(2*7) - a(2) = 316072831033350 - 6 = (2^13)*3*(7^3)*11*691*4933 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 3164395891098711251676512000 - 12000 = (2^5)*(5^6)*29* 124891891*1747384859327 == 0 ( mod 5^6 ).
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..500
• Peter Bala, Notes on A333593
• R. Mestrovic, 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, A333592.

Programs

Maple

seq( add( (-1)^(n+k)*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, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

PARI

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

Formula

a(n) ~ 2^(4*n) / (5*Pi*n). - Vaclav Kotesovec, Mar 28 2020