A098617 G.f. A(x) satisfies: A(x*G(x)) = G(x), where G(x) is the g.f. for A098616(n) = Pell(n+1)*Catalan(n).

1, 2, 6, 16, 46, 128, 364, 1024, 2902, 8192, 23188, 65536, 185420, 524288, 1483096, 4194304, 11863910, 33554432, 94908420, 268435456, 759257636, 2147483648, 6074027496, 17179869184, 48592102396, 137438953472, 388736403144

Offset

0,2

Comments

G.f. satisfies: A(x) = x/Series_Reversion(x*G(x)), where G(x) is the g.f. for A098616 = {1*1, 2*1, 5*2, 12*5, 29*14, 70*42, 169*132, ...}.

Hankel transform is 2^n. - Paul Barry Jan 19 2011

Links

Fung Lam, Table of n, a(n) for n = 0..1000

Formula

G.f.: (sqrt(1-4*x^2) + 2*x)/(1-8*x^2).

a(2*n+1) = 2*8^n.

a(n) = sum{k=0..floor((n+1)/2), (C(n,k)-C(n,k-1))*A000129(n-2k+1)}. - Paul Barry Jan 19 2011

a(n) = 2^n*sum(j=0..n/2, binomial((n-1)/2,j)). - Vladimir Kruchinin, May 18 2011

a(n) = Sum_{k, 0<=k<=n} A201093(n,k)*2^k. - Philippe Deléham, Nov 27 2011

G.f.: 1/(1-2x/(1-x/(1+x/(1+x/(1-x/(1-x/(1+x/(1+x/(1-x/(1-... (continued fraction). - Philippe Deléham, Nov 27 2011

Recurrence: (n+6)*a(n)=256*(n+1)*a(n-6)-128*(n+3)*a(n-4)+4*(5*n+23)*a(n-2), for even n. - Fung Lam, Mar 31 2014

Recurrence: n*a(n) = 12*(n-1)*a(n-2) - 32*(n-3)*a(n-4). - Vaclav Kotesovec, Mar 31 2014

Asymptotic approximation: a(n) ~ (4/sqrt(2))^n/sqrt(2)+2^(n+1)/sqrt(2*Pi*n^3), for even n. - Fung Lam, Mar 31 2014

0 = a(n) * (+64*a(n+1) - 8*a(n+3)) + a(n+2) * (-8*a(n+1) + a(n+3)) if n>=0. - Michael Somos, Apr 07 2014

Example

G.f. = 1 + 2*x + 6*x^2 + 16*x^3 + 46*x^4 + 128*x^5 + 364*x^6 + 1024*x^7 + ...

Mathematica

CoefficientList[Series[(Sqrt[1-4*x^2] + 2*x)/(1-8*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 31 2014 *)

Prog

(PARI) a(n)=polcoeff((sqrt(1-4*x^2+x^2*O(x^n))+2*x)/(1-8*x^2), n)
(Maxima)
a(n):=2^n*sum(binomial((n-1)/2, j), j, 0, n/2); (* Vladimir Kruchinin, May 18 2011

Crossrefs

Cf. A098616, A098615, A000129, A000108.

Sequence in context: A148440 A148441 A307606 * A291036 A092687 A094039

Adjacent sequences: A098614 A098615 A098616 * A098618 A098619 A098620

Keyword

nonn

Author

Paul D. Hanna, Oct 14 2004

Status

approved