A352654 a(0) = 0, a(n) = [x^n] P(n-1,(1 + x)/(1 - x)) for n >= 1, where P(n,x) denotes the n-th Legendre polynomial.

0, 0, 2, 18, 162, 1500, 14240, 137886, 1356194, 13507416, 135916002, 1379301990, 14097919968, 144977296932, 1498750896708, 15564971674518, 162298598439330, 1698353774374704, 17828728267167326, 187693442844729174, 1981038544180652162, 20957881615473229620

Offset

0,3

Comments

Compare with A103882(n) = [x^n] P(n,(1 + x)/(1 - x)).

Using binomial(-n,k) = (-1)^k*binomial(n+k-1,k), valid for nonnegative k, we can extend the binomial sum a(n) = Sum_{k} binomial(n-1,k)*binomial(n-1,k-1)*binomial(n+k-1,k) to negative values of n to find a(-n) = Sum_{k} binomial(-n-1,k)*binomial(-n-1,k-1)* binomial(-n+k-1,k) = Sum_{k} (-1)^(k-1)*binomial(n+k,k)*binomial(n+k-1,k-1)*binomial(n,k) = (-1)^(n+1)*A103882(n) (n != 0).

Links

Table of n, a(n) for n=0..21.

Formula

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

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

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

a(n) = (1/5)*(A005258(n) - 3*A005258(n-1)) for n >= 1. It follows that the supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) hold for primes p >= 5 and positive integers n and k.

a(n) = A108628(n-1) - A208675(n) for n >= 1.

a(n) ~ (1/10)*5^(3/4)*(sqrt(5) + 1)/(Pi*sqrt(22 + 10*sqrt(5))*n)*((1/2)*(11 + 5*sqrt(5)))^n.

n^2*(5*n - 7)*(n - 2)*a(n) = (n - 1)*(55*n^3 - 187*n^2 + 190*n - 48)*a(n-1) + (n - 2)^2*(5*n -2 )*(n - 1)*a(n-2) with a(1) = 0 and a(2) = 2.

Conjecture: a(n) = [(x*y)^(n-1)*z^n] 1/(1 - x - x*y - y*z - x*z - x*y*z) = [(x*z)^(n-1))*y*n] 1/(1 - x - x*y - y*z - x*z - x*y*z) for n >= 1.

a(n) = n*(n-1)*hypergeom([2-n, 1-n, 1+n], [2, 2], 1). - Peter Luschny, Jan 03 2024

Maple

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

Prog

(PARI) a(n) = polcoef(pollegendre(n-1, (1 + x)/(1 - x)) + O(x^(n+1)), n); \\ Michel Marcus, Apr 17 2022
(Python)
def A352654(n):
    if n == 0: return 0
    m, g = 1, 0
    for k in range(n+1):
        g += -m*n//(n+k) if (n+k)&1 else m*n//(n+k)
        m *= (n+k+1)*(n-k)*(n+k)
        m //= (k+1)**3
    return g # Chai Wah Wu, Oct 03 2022
(SageMath)
def a(n): return n * (n-1) * hypergeometric([2-n, 1-n, 1+n], [2, 2], 1)
print([simplify(a(n)) for n in range(22)]) # Peter Luschny, Jan 03 2024

Crossrefs

Cf. A005258, A103882, A108628, A208675.

Sequence in context: A108550 A322282 A270369 * A144513 A037518 A037721

Adjacent sequences: A352651 A352652 A352653 * A352655 A352656 A352657

Keyword

nonn,easy

Author

Peter Bala, Apr 14 2022

Status

approved