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A098616
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Product of Pell and Catalan numbers: a(n) = A000129(n+1)*A000108(n).
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12
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1, 2, 10, 60, 406, 2940, 22308, 175032, 1408550, 11561836, 96425836, 814773960, 6960289532, 60012947800, 521582661000, 4564643261040, 40190674554630, 355772529165900, 3164408450118300, 28266363849505320, 253466716153665300, 2280803103062033160, 20588945107316958840
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OFFSET
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0,2
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COMMENTS
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Radius of convergence: r = (sqrt(2)-1)/4, where A(r) = sqrt(2+sqrt(2)).
More generally, given {S} such that: S(n) = b*S(n-1) + c*S(n-2), |b|>0, |c|>0, then Sum_{n>=0} S(n)*Catalan(n)*x^n = sqrt( (1-2*b*x - sqrt(1-4*b*x-16*c*x^2))/(2*b^2+8*c) )/x.
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LINKS
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FORMULA
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G.f.: A(x) = sqrt( (1-4*x - sqrt(1-8*x-16*x^2))/16 )/x.
Run lengths of zeros (mod 10) equal (5^k - (-1)^k)/2 - 1 starting at index (5^k + (-1)^k)/2:
a(n) == 0 (mod 10) for n = (5^k + (-1)^k)/2 through n = 5^k - 1 when k>=1.
a(n) ~ 2^(2*n-3/2) * (1+sqrt(2))^(n+1) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, May 09 2014
A(-x) = 1/x * series reversion( x*(2*x + sqrt(1 - 4*x^2)) ). Compare with the o.g.f. B(x) of the central binomial numbers A000984, which satisfies B(-x) = 1/x * series reversion( x*(2*x + sqrt(1 + 4*x^2)) ). See also A214377. - Peter Bala, Oct 19 2015
n*(n+1)*a(n) -4*n*(2*n-1)*a(n-1) -4*(2*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 17 2018
Sum_{n>=0} a(n)/16^n = 2*sqrt(3-sqrt(7)). - Amiram Eldar, May 05 2023
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EXAMPLE
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Sequence begins: [1*1, 2*1, 5*2, 12*5, 29*14, 70*42, 169*132, 408*429,...].
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MATHEMATICA
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With[{nn=30}, Times@@@Thread[{LinearRecurrence[{2, 1}, {1, 2}, nn], CatalanNumber[ Range[0, nn-1]]}]] (* Harvey P. Dale, Jan 04 2012 *)
a[n_] := Fibonacci[n + 1, 2] * CatalanNumber[n]; Array[a, 25, 0] (* Amiram Eldar, May 05 2023 *)
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PROG
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(PARI) a(n)=binomial(2*n, n)/(n+1)*round(((1+sqrt(2))^(n+1)-(1-sqrt(2))^(n+1))/(2*sqrt(2)))
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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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