A357559 a(n) = Sum_{k = 0..n} (-1)^(n+k)*k^3*binomial(n,k)*binomial(n+k,k)^2.

0, 4, 270, 8448, 192000, 3669300, 62952162, 1003770880, 15182515584, 220700443500, 3110529630450, 42769154678976, 576313309494000, 7636526099508852, 99765264496070250, 1287663145631539200, 16446680778536421888, 208154776511034178380, 2613380452317012835386

Offset

0,2

Comments

Define S_m(n) = Sum_{k = 0..n} (-1)^(n+k)*k^m*binomial(n,k)*binomial(n+k,k)^2, so that S_0(n) = A005258(n), one type of Apéry numbers. The present sequence is the case m = 3. See A357558 for the case m = 1.

It is known that S_0(2*n) - 1 is divisible by (2*n + 1)^3 provided 2*n + 1 is a prime greater than 3 (see, for example, Straub, Example 3.4).

Let C_m = Product_{odd primes p <= m} p.

Conjectures:

1) S_2(2*n) is divisible by n^2*(2*n + 1)^2.

2) for even m >= 4, C_m * S_m(2*n) is divisible by n^3*(2*n + 1)^2.

3) for odd m >= 5, C_m * S_m(2*n) is divisible by n^2*(2*n + 1)^4.

Links

Table of n, a(n) for n=0..18.

A. Straub, Multivariate Apéry numbers and supercongruences of rational functions, arXiv:1401.0854 [math.NT] (2014).

Formula

Conjecture: 3*a(2*n) == 0 ( mod n^2*(2*n + 1)^3 ).

Recurrence a(0) = 0, a(1) = 4, and for n >= 2, n*(n + 3)*(5*n - 6)*(n - 1)^4*a(n) = (n - 1)*(55*n^4 - 66*n^3 - 22*n^2 + 15*n + 6)*(n + 1)^2*a(n-1) + n^4*(5*n - 1)*(n + 1)^2*a(n-2).

a(n) ~ n^2 * phi^(5*n + 11/2) / (2*Pi*5^(1/4)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Oct 07 2022

Maple

seq( add( (-1)^(n+k) * k^3 * binomial(n, k) * binomial(n+k, k)^2, k = 0..n ), n = 0..20 );

Crossrefs

Cf. A005258, A357510, A357512, A357558, A357560, A357561.

Sequence in context: A052136 A089667 A119008 * A108134 A221081 A340916

Adjacent sequences: A357556 A357557 A357558 * A357560 A357561 A357562

Keyword

nonn,easy

Author

Peter Bala, Oct 04 2022

Status

approved