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A322518
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Binomial transform of the Apéry numbers (A005259).
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1
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1, 6, 84, 1680, 39240, 999216, 26899896, 752939424, 21691531800, 638947312080, 19155738105504, 582589712312064, 17930566188602136, 557417298916695600, 17477836958370383280, 552090876791399769600, 17552554240486710112920, 561230779055361080132880
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OFFSET
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0,2
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COMMENTS
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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
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LINKS
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FORMULA
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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
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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