A108628 n-th term of the crystal ball sequence for A_{n+1} lattice for n >= 0.

1, 7, 55, 471, 4251, 39733, 380731, 3716695, 36808723, 368750757, 3728940249, 38003358693, 389866749975, 4022124746409, 41697566691555, 434124925278807, 4536783726146499, 47569453938399445, 500266519237489357, 5275183203229043221, 55760274296452936741

Offset

0,2

Comments

Equals the secondary diagonal of square array A108625, in which row n equals the crystal ball sequence for A_n lattice. Main diagonal of square array A108625 equals the Apery numbers (A005258).

Links

G. C. Greubel, Table of n, a(n) for n = 0..950

Armin Straub, Multivariate Apéry numbers and supercongruences of rational functions, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; arXiv preprint, arXiv:1401.0854 [math.NT], 2014.

Formula

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

a(n) = A108625(n+1, n).

exp( Sum_{n >= 1} a(n-1)*x^n/n ) = 1 + x + 4*x^2 + 22*x^3 + 144*x^4 + 1048*x^5 + 8189*x^6 + 67325*x^7 + 574999*x^8 + ... appears to have integer coefficients. Cf. A208675. - Peter Bala, Jan 12 2016

Recurrence: (n+1)^2*(5*n^2 - 6*n + 2)*a(n) = (55*n^4 - 11*n^3 - 26*n^2 + 5*n + 5)*a(n-1) + (n-1)^2*(5*n^2 + 4*n + 1)*a(n-2). - Vaclav Kotesovec, Jan 13 2016

a(n) ~ sqrt(89/8 + 199/(8*sqrt(5))) * ((1+sqrt(5))/2)^(5*n) / (Pi*n). - Vaclav Kotesovec, Jan 13 2016

Equivalently, a(n) ~ phi^(5*n + 11/2) / (2*5^(1/4)*Pi*n), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 06 2021

From Peter Bala, Mar 24 2022: (Start)

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

Using binomial(-n,k) = (-1)^k*binomial(n+k-1,k) for nonnegative k, we have a(-n) = Sum_{k} binomial(-n,k)*binomial(-n+1,k)*binomial(-n+k+1,k) = Sum_{k} (-1)^k* binomial(n+k-1,k)*binomial(n+k-2,k)*binomial(n-2,k) = (-1)^n*A208675(n-1) for n >= 2.

a(n-1) = (1/2)*Sum_{k = 0..floor(n/2)} binomial(n,k)^2 * binomial(3*n-2*k-1,n-2*k) for n >= 1. Cf. A103882.

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

Equivalently, a(n) = [(x*z)^(n+1)*y^n] ( (x + y + z)^(n+1) * (x + y)^n * (y + z)^(n+1) ).

a(n) = (1/5)*( 2*A005258(n+1) - A005258(n) ).

a(n) = hypergeometric3F2([n + 2, -n, -n - 1], [1, 1], 1).

a(n) = (-1)^n*(n+1)*hypergeom([n + 2, n + 2, -n ], [1, 2], 1).

a(n) = [x^n] 1/(1 - x)*P(n+1,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Compare with A208675(n) = [x^n] 1/(1 - x)*P(n-1,(1 + x)/(1 - x)) for n >= 1 and A005258(n) = [x^n] 1/(1 - x)*P(n,(1 + x)/(1 - x)).

a(n) = B(n+1,n,n+1) in the notation of Straub, equation 24. Hence

a(n) = [(x*z)^(n+1)*y^n] 1/(1 - x - y - z + x*z + y*z - x*y*z).

The following takes the sequence offset to be 1: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and all positive integers n and k.

Conjectures: a((p-1)/2) == 0 (mod p) and a((p^3-1)/2) == 0 (mod p^3) for primes p of the form 4*m + 1; a((p^3-1)/2) == 0 (mod p^2) for primes p of the form 4*m + 3; a((p^2-1)/2) == 0 (mod p^2) for primes p >= 5. (End)

Maple

seq(add(binomial(n, k)*binomial(n+1, k)*binomial(n+k+1, k), k = 0..n), n = 0..20); # Peter Bala, Apr 14 2022

Mathematica

Table[Sum[Binomial[n+1, k]^2 Binomial[n+k, k-1], {k, 0, n+1}], {n, 0, 20}] (* Harvey P. Dale, Apr 01 2013 *)

Prog

(PARI) a(n)=sum(k=0, n+1, binomial(n+1, k)^2*binomial(n+k, k-1))
(Python)
def A108628(n):
    m, g = 1, 0
    for k in range(n+1):
        g += m
        m *= (n+k+2)*(n-k)*(n-k+1)
        m //= (k+1)**3
    return g # Chai Wah Wu, Oct 03 2022
(Magma)
A108628:= func< n | (&+[Binomial(n+1, k)^2*Binomial(n+k, k-1): k in [0..n+1]]) >;
[A108628(n): n in [0..30]]; // G. C. Greubel, Oct 06 2023
(SageMath)
def A108628(n): return sum(binomial(n+1, k)^2*binomial(n+k, k-1) for k in range(n+2))
[A108628(n) for n in range(31)] # G. C. Greubel, Oct 06 2023

Crossrefs

Cf. A005258, A103882, A108625, A208675.

Sequence in context: A152262 A078018 A357207 * A116862 A096307 A199564

Adjacent sequences: A108625 A108626 A108627 * A108629 A108630 A108631

Keyword

nonn,easy

Author

Paul D. Hanna, Jun 14 2005

Status

approved