OFFSET
0,2
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
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
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
EXAMPLE
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.
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 *)
PROG
(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))
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Sarah Arpin, Dec 13 2018
STATUS
approved