A219536 - OEIS

%I #35 Sep 09 2024 09:35:02

%S 1,3,24,255,3102,40854,566934,8164263,120864390,1827982362,

%T 28122626760,438720097638,6923868098820,110346550539780,

%U 1773394661610258,28707809007278775,467677404522668742,7661583171651546786,126137791939032756960,2085923447593966281378

%N G.f. satisfies A(x) = 1 + x*(A(x)^2 + 2*A(x)^3).

%H G. C. Greubel, <a href="/A219536/b219536.txt">Table of n, a(n) for n = 0..790</a>

%H Elżbieta Liszewska, Wojciech Młotkowski, <a href="https://arxiv.org/abs/1907.10725">Some relatives of the Catalan sequence</a>, arXiv:1907.10725 [math.CO], 2019.

%F Let G(x) = (1-x - sqrt(1 - 10*x + x^2)) / (4*x), then g.f. A(x) satisfies:

%F (1) A(x) = (1/x)*Series_Reversion(x/G(x)),

%F (2) A(x) = G(x*A(x)) and G(x) = A(x/G(x)),

%F where G(x) is the g.f. of A103210.

%F 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

%F 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

%F From _Seiichi Manyama_, Jul 26 2020: (Start)

%F a(n) = Sum_{k=0..n} 2^k * binomial(n,k) * binomial(2*n+k+1,n)/(2*n+k+1).

%F 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)

%F From _Seiichi Manyama_, Aug 10 2023: (Start)

%F 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.

%F 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)

%F a(n) = (-1)^(n+1) * (3/n) * Jacobi_P(n-1, 1, n+1, -5) for n >= 1. - _Peter Bala_, Sep 08 2024

%e G.f.: A(x) = 1 + 3*x + 24*x^2 + 255*x^3 + 3102*x^4 + 40854*x^5 +...

%e Related expansions:

%e A(x)^2 = 1 + 6*x + 57*x^2 + 654*x^3 + 8310*x^4 + 112560*x^5 +...

%e A(x)^3 = 1 + 9*x + 99*x^2 + 1224*x^3 + 16272*x^4 + 227187*x^5 +...

%e The g.f. satisfies A(x) = G(x*A(x)) and G(x) = A(x/G(x)) where

%e G(x) = 1 + 3*x + 15*x^2 + 93*x^3 + 645*x^4 + 4791*x^5 +...+ A103210(n)*x^n +...

%t 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 *)

%o (PARI) /* Formula A(x) = 1 + x*(A(x)^2 + 2*A(x)^3): */

%o {a(n)=my(A=1);for(i=1,n,A=1+x*(A^2+2*A^3) +x*O(x^n));polcoeff(A,n)}

%o for(n=0,25,print1(a(n),", "))

%o (PARI) /* Formula using Series Reversion: */

%o {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)}

%o for(n=0,25,print1(a(n),", "))

%o (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

%o (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

%Y Column k=2 of A336574.

%Y Cf. A003168, A103210, A219534, A219535.

%K nonn,easy

%O 0,2

%A _Paul D. Hanna_, Nov 21 2012