A098616 - OEIS

%I #34 May 05 2023 01:37:26

%S 1,2,10,60,406,2940,22308,175032,1408550,11561836,96425836,814773960,

%T 6960289532,60012947800,521582661000,4564643261040,40190674554630,

%U 355772529165900,3164408450118300,28266363849505320,253466716153665300,2280803103062033160,20588945107316958840

%N Product of Pell and Catalan numbers: a(n) = A000129(n+1)*A000108(n).

%C Radius of convergence: r = (sqrt(2)-1)/4, where A(r) = sqrt(2+sqrt(2)).

%C 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.

%H Harvey P. Dale, <a href="/A098616/b098616.txt">Table of n, a(n) for n = 0..1000</a>

%F G.f.: A(x) = sqrt( (1-4*x - sqrt(1-8*x-16*x^2))/16 )/x.

%F Run lengths of zeros (mod 10) equal (5^k - (-1)^k)/2 - 1 starting at index (5^k + (-1)^k)/2:

%F a(n) == 0 (mod 10) for n = (5^k + (-1)^k)/2 through n = 5^k - 1 when k>=1.

%F a(n) ~ 2^(2*n-3/2) * (1+sqrt(2))^(n+1) / (sqrt(Pi) * n^(3/2)). - _Vaclav Kotesovec_, May 09 2014

%F 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

%F 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

%F Sum_{n>=0} a(n)/16^n = 2*sqrt(3-sqrt(7)). - _Amiram Eldar_, May 05 2023

%e Sequence begins: [1*1, 2*1, 5*2, 12*5, 29*14, 70*42, 169*132, 408*429,...].

%t With[{nn=30},Times@@@Thread[{LinearRecurrence[{2,1},{1,2},nn], CatalanNumber[ Range[0,nn-1]]}]] (* _Harvey P. Dale_, Jan 04 2012 *)

%t a[n_] := Fibonacci[n + 1, 2] * CatalanNumber[n]; Array[a, 25, 0] (* _Amiram Eldar_, May 05 2023 *)

%o (PARI) a(n)=binomial(2*n,n)/(n+1)*round(((1+sqrt(2))^(n+1)-(1-sqrt(2))^(n+1))/(2*sqrt(2)))

%Y Cf. A000129, A000108, A098614, A098617, A098618, A000984, A214377.

%K nonn,easy

%O 0,2

%A _Paul D. Hanna_, Oct 09 2004