OFFSET
0,2
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..790
Elżbieta Liszewska, Wojciech Młotkowski, Some relatives of the Catalan sequence, arXiv:1907.10725 [math.CO], 2019.
FORMULA
Let G(x) = (1-x - sqrt(1 - 10*x + x^2)) / (4*x), then g.f. A(x) satisfies:
(1) A(x) = (1/x)*Series_Reversion(x/G(x)),
(2) A(x) = G(x*A(x)) and G(x) = A(x/G(x)),
where G(x) is the g.f. of A103210.
Recurrence: 4*n*(2*n+1)*(19*n-26)*a(n) = (2717*n^3 - 6435*n^2 + 4342*n - 840)*a(n-1) + 2*(n-2)*(2*n-3)*(19*n-7)*a(n-2). - Vaclav Kotesovec, Dec 28 2013
a(n) ~ (3/19)^(1/4) * (5+sqrt(57)) * ((143 + 19*sqrt(57))/16)^n / (16*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Dec 28 2013
From Seiichi Manyama, Jul 26 2020: (Start)
a(n) = Sum_{k=0..n} 2^k * binomial(n,k) * binomial(2*n+k+1,n)/(2*n+k+1).
a(n) = (1/(2*n+1)) * Sum_{k=0..n} 2^(n-k) * binomial(2*n+1,k) * binomial(3*n-k,n-k). (End)
From Seiichi Manyama, Aug 10 2023: (Start)
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * 3^(n-k) * binomial(n,k) * binomial(3*n-k,n-1-k) for n > 0.
a(n) = (1/n) * Sum_{k=1..n} 3^k * 2^(n-k) * binomial(n,k) * binomial(2*n,k-1) for n > 0. (End)
a(n) = (-1)^(n+1) * (3/n) * Jacobi_P(n-1, 1, n+1, -5) for n >= 1. - Peter Bala, Sep 08 2024
EXAMPLE
G.f.: A(x) = 1 + 3*x + 24*x^2 + 255*x^3 + 3102*x^4 + 40854*x^5 +...
Related expansions:
A(x)^2 = 1 + 6*x + 57*x^2 + 654*x^3 + 8310*x^4 + 112560*x^5 +...
A(x)^3 = 1 + 9*x + 99*x^2 + 1224*x^3 + 16272*x^4 + 227187*x^5 +...
The g.f. satisfies A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where
G(x) = 1 + 3*x + 15*x^2 + 93*x^3 + 645*x^4 + 4791*x^5 +...+ A103210(n)*x^n +...
MATHEMATICA
CoefficientList[1/x*InverseSeries[Series[4*x^2/(1-x-Sqrt[1-10*x+x^2]), {x, 0, 20}], x], x] (* Vaclav Kotesovec, Dec 28 2013 *)
PROG
(PARI) /* Formula A(x) = 1 + x*(A(x)^2 + 2*A(x)^3): */
{a(n)=my(A=1); for(i=1, n, A=1+x*(A^2+2*A^3) +x*O(x^n)); polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))
(PARI) /* Formula using Series Reversion: */
{a(n)=my(A=1, G=(1-x-sqrt(1-10*x+x^2+x^3*O(x^n)))/(4*x)); A=(1/x)*serreverse(x/G); polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))
(PARI) a(n) = sum(k=0, n, 2^k*binomial(n, k)*binomial(2*n+k+1, n)/(2*n+k+1)); \\ Seiichi Manyama, Jul 26 2020
(PARI) a(n) = sum(k=0, n, 2^(n-k)*binomial(2*n+1, k)*binomial(3*n-k, n-k))/(2*n+1); \\ Seiichi Manyama, Jul 26 2020
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul D. Hanna, Nov 21 2012
STATUS
approved