A249925 G.f. satisfies: A(x) = 1 + 2*x*A(x) + 5*x^2*A(x)^2.

1, 2, 9, 38, 186, 932, 4889, 26238, 143966, 802652, 4536874, 25932348, 149650516, 870675912, 5101656889, 30078478318, 178309845686, 1062198928812, 6355149937934, 38172142221748, 230094601968876, 1391444403490552, 8439240940653834, 51323083138005388, 312896262064813036, 1911980839096481432

Offset: 0

Author

Paul D. Hanna, Nov 22 2014

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + 2*x + 9*x^2 + 38*x^3 + 186*x^4 + 932*x^5 + 4889*x^6 +...
where the square-root of the g.f. yields
sqrt(A(x)) = 1 + x + 4*x^2 + 15*x^3 + 70*x^4 + 336*x^5 + 1716*x^6 + 9009*x^7 + 48620*x^8 +...+ Fibonacci(n+1)*A000108(n)*x^n + +...
Related expansions.
A(x)^2 = 1 + 4*x + 22*x^2 + 112*x^3 + 605*x^4 + 3292*x^5 + 18298*x^6 +...
which obeys A(x) = 1 + 2*x*A(x) + 5*x^2*A(x)^2.
Given series bisections A(x) = B0(x^2) + x*B1(x^2),
B0(x) = 1 + 9*x + 186*x^2 + 4889*x^3 + 143966*x^4 + 4536874*x^5 +...
B1(x) = 2 + 38*x + 932*x^2 + 26238*x^3 + 802652*x^4 + 25932348*x^5 +...
then B1(x)/B0(x) = 2 + 10*x*B1(x):
B1(x)/B0(x) = 2 + 20*x + 380*x^2 + 9320*x^3 + 262380*x^4 + 8026520*x^5 +...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Paul Barry, On the inversion of Riordan arrays, arXiv:2101.06713 [math.CO], 2021.

Crossrefs

Cf. A098614, A000045, A000108.

Programs

Mathematica

CoefficientList[Series[(1-2*x-Sqrt[1-4*x-16*x^2]) / (10*x^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Nov 29 2014 *)

PARI

{a(n)=local(X=x+O(x^(n+3)),A); A = (1-2*x - sqrt(1-4*X-16*x^2)) / (10*x^2); polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

PARI

{a(n) = sum(k=0,n,fibonacci(n-k+1)*fibonacci(k+1)*binomial(2*(n-k),n-k)*binomial(2*k,k)/((n-k+1)*(k+1)))}
for(n=0,30,print1(a(n),", "))

Formula

G.f.: (1-2*x - sqrt(1-4*x-16*x^2)) / (10*x^2).

Self-convolution square of A098614, where A098614(n) = A000045(n+1)*A000108(n), the term-wise product of Fibonacci and Catalan numbers.

a(n) = Sum_{k=0..n} A000045(k+1)*A000045(n-k+1) * A000108(k)*A000108(n-k).

a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*Fibonacci(k+1) * C(2*(n-k),n-k)*C(2*k,k) / ((n-k+1)*(k+1)).

a(n) == 1 (mod 2) iff n = 2*(2^k - 1) for k>=0.

Given series bisections B0 and B1 such that A(x) = B0(x^2) + x*B1(x^2), then B1(x)/B0(x) = 2 + 10*x*B1(x), thus B1(x) = 2*B0(x)/(1 - 10*x*B0(x)).

a(n) ~ sqrt(5+2*sqrt(5)) * 2^(n+2) * (1+sqrt(5))^n / (5 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Nov 29 2014. Equivalently, a(n) ~ 5^(1/4) * 2^(2*n+2) * phi^(n + 3/2) / (5 * sqrt(Pi) * n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 06 2021

Recurrence: (n+2)*a(n) = 2*(2*n+1)*a(n-1) + 16*(n-1)*a(n-2). - Vaclav Kotesovec, Nov 29 2014