A322518 Binomial transform of the Apéry numbers (A005259).

1, 6, 84, 1680, 39240, 999216, 26899896, 752939424, 21691531800, 638947312080, 19155738105504, 582589712312064, 17930566188602136, 557417298916695600, 17477836958370383280, 552090876791399769600, 17552554240486710112920, 561230779055361080132880

Offset: 0

Author

Sarah Arpin, Dec 13 2018

Keywords

• nonn
• easy

Comments

Starting with the a(3) term, each term is divisible by 8. (Empirical observation.)

The above is true and follows easily from the pair of known congruences for the Apéry numbers A(n): A(2*n) == 1 (mod 8) and A(2n+1) == 5 (mod 8). - Peter Bala, Jan 06 2020

Examples

			a(2) = binomial(2,0)*A(0) + binomial(2,1)*A(1) + binomial(2,2)*A(2), where A(k) denotes the k-th Apéry number. Using this definition:
a(2) = binomial(2,0)*(binomial(0,0)*binomial(0,0))^2 + binomial(2,1)*((binomial(1,0)*binomial(1,0))^2 + (binomial(1,1)*binomial(2,1))^2) + binomial(2,2)*((binomial(2,0)*binomial(2,0))^2 + (binomial(2,1)*binomial(3,1))^2 + (binomial(2,2)*binomial(4,2))^2) = 84.
		

Links

• Jackson Earles, Justin Ford, Poramate Nakkirt, Marlo Terr, Dr. Ilia Mishev, Sarah Arpin, Binomial Transforms of Sequences, Fall 2018.
• N. J. A. Sloane, Transforms

Crossrefs

Cf. A005259, A322519.

Programs

Julia

function BinomialTransform(seq)
    N = length(seq)
    bt = Array{BigInt,1}(undef,N)
    bt[1] = seq[1]
    for k in 1:N-1
        next = BigInt(0)
        for j in 0:k next += binomial(k, j)*seq[j+1] end
        bt[k+1] = next
    end
bt end
BinomialTransform([A005259(n) for n in 0:18]) |> println # Peter Luschny, Jan 06 2020

Mathematica

a[n_] := Sum[Binomial[n, k] * Sum[(Binomial[k, j] * Binomial[k+j, j])^2, {j, 0, k}], {k, 0, n}]; Array[a, 20, 0] (* Amiram Eldar, Dec 13 2018 *)

Sage

def OEISbinomial_transform(N, seq):
    BT = [seq[0]]
    k = 1
    while k< N:
        next = 0
        j = 0
        while j <=k:
            next = next + ((binomial(k,j))*seq[j])
            j = j+1
        BT.append(next)
        k = k+1
    return BT
Apery = oeis('A005259')
OEISBinom = OEISbinomial_transform(18,Apery.first_terms(20))

Formula

a(n) ~ 2^(n - 3/4) * 3^(n + 3/2) * (1 + sqrt(2))^(2*n - 1) / (Pi*n)^(3/2). - Vaclav Kotesovec, Dec 17 2018

The Gauss congruences hold: a(n*p^k) == a(n*p^(k-1)) (mod p^k) for all primes p and n a positive integer. - Peter Bala, Jan 06 2020