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A322518 Binomial transform of the Apéry numbers (A005259).

%I A322518 #28 Jan 06 2020 17:29:33
%S A322518 1,6,84,1680,39240,999216,26899896,752939424,21691531800,638947312080,
%T A322518 19155738105504,582589712312064,17930566188602136,557417298916695600,
%U A322518 17477836958370383280,552090876791399769600,17552554240486710112920,561230779055361080132880
%N A322518 Binomial transform of the Apéry numbers (A005259).
%C A322518 Starting with the a(3) term, each term is divisible by 8. (Empirical observation.)
%C A322518 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
%H A322518 Jackson Earles, Justin Ford, Poramate Nakkirt, Marlo Terr, Dr. Ilia Mishev, Sarah Arpin, <a href="https://www.colorado.edu/math/binomial-transforms-sequences-fall-2018">Binomial Transforms of Sequences</a>, Fall 2018.
%H A322518 N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>
%F A322518 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
%F A322518 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
%e A322518 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:
%e A322518 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.
%t A322518 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 *)
%o A322518 (Sage)
%o A322518 def OEISbinomial_transform(N, seq):
%o A322518     BT = [seq[0]]
%o A322518     k = 1
%o A322518     while k< N:
%o A322518         next = 0
%o A322518         j = 0
%o A322518         while j <=k:
%o A322518             next = next + ((binomial(k,j))*seq[j])
%o A322518             j = j+1
%o A322518         BT.append(next)
%o A322518         k = k+1
%o A322518     return BT
%o A322518 Apery = oeis('A005259')
%o A322518 OEISBinom = OEISbinomial_transform(18,Apery.first_terms(20))
%o A322518 (Julia)
%o A322518 function BinomialTransform(seq)
%o A322518     N = length(seq)
%o A322518     bt = Array{BigInt,1}(undef,N)
%o A322518     bt[1] = seq[1]
%o A322518     for k in 1:N-1
%o A322518         next = BigInt(0)
%o A322518         for j in 0:k next += binomial(k, j)*seq[j+1] end
%o A322518         bt[k+1] = next
%o A322518     end
%o A322518 bt end
%o A322518 BinomialTransform([A005259(n) for n in 0:18]) |> println # _Peter Luschny_, Jan 06 2020
%Y A322518 Cf. A005259, A322519.
%K A322518 nonn,easy
%O A322518 0,2
%A A322518 _Sarah Arpin_, Dec 13 2018