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 A322518 Binomial transform of the Apéry numbers (A005259). 1
 1, 6, 84, 1680, 39240, 999216, 26899896, 752939424, 21691531800, 638947312080, 19155738105504, 582589712312064, 17930566188602136, 557417298916695600, 17477836958370383280, 552090876791399769600, 17552554240486710112920, 561230779055361080132880 (list; graph; refs; listen; history; text; internal format)
 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 Cf. A005259, A322519. Sequence in context: A306244 A277304 A128575 * A014062 A147626 A123312 Adjacent sequences:  A322515 A322516 A322517 * A322519 A322520 A322521 KEYWORD nonn,easy AUTHOR Sarah Arpin, Dec 13 2018 STATUS approved

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Last modified November 28 12:19 EST 2020. Contains 338720 sequences. (Running on oeis4.)