A316371 G.f.: A(x) = Sum_{n>=0} binomial(3*(n+1), n)/(n+1) * x^n / (1+x)^(2*(n+1)).

1, 1, 3, 9, 30, 105, 382, 1429, 5463, 21248, 83813, 334485, 1348102, 5479366, 22433934, 92437445, 383026643, 1595053047, 6672007021, 28020635404, 118106170713, 499454897337, 2118477808719, 9010443044061, 38420834606794, 164210479691902, 703352241046710, 3018668702116310, 12979807315841432, 55908387904005714

Offset: 0

Author

Paul D. Hanna, Jul 02 2018

Keywords

• nonn

Comments

Note that: binomial(3*(n+1), n)/(n+1) = A001764(n+1) for n >= 0, where F(x) = Sum_{n>=0} A001764(n)*x^n satisfies F(x) = 1 + x*F(x)^3.

Compare the g.f. to:

(C1) M(x) = Sum_{n>=0} binomial(2*(n+1), n)/(n+1) * x^n / (1+x)^(n+1) where M(x) = 1 + M(x) + M(x)^2 is the g.f. of Motzkin numbers (A001006).

(C2) 1 = Sum_{n>=0} binomial(m*(n+1), n)/(n+1) * x^n / (1+x)^(m*(n+1)) holds for fixed m.

(C3) If S(x,p,q) = Sum_{n>=0} binomial(p*(n+1),n)/(n+1) * x^n/(1+x)^(q*(n+1)), then Series_Reversion ( x*S(x,p,q) ) = x*S(x,q,p) holds for fixed p and q.

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 9*x^3 + 30*x^4 + 105*x^5 + 382*x^6 + 1429*x^7 + 5463*x^8 + 21248*x^9 + 83813*x^10 + 334485*x^11 + 1348102*x^12 + ...
such that
A(x) = 1/(1+x)^2 + 3*x/(1+x)^4 + 12*x^2/(1+x)^6 + 55*x^3/(1+x)^8 + 273*x^4/(1+x)^10 + 1428*x^5/(1+x)^12 + ... + A001764(n+1)*x^n/(1+x)^(2*(n+1)) + ...
RELATED SERIES.
(E1) 1 - 1/A(x) = x + 2*x^2 + 4*x^3 + 11*x^4 + 34*x^5 + 114*x^6 + 402*x^7 + 1470*x^8 + 5522*x^9 + 21181*x^10 + 82610*x^11 + 326611*x^12 + ...
where A(1/A(x) - 1) = 1/(1+x).
(E2) Series_Reversion( x*A(x) ) = x - x^2 - x^3 + x^4 + 2*x^5 - 4*x^7 - 4*x^8 + 5*x^9 + 15*x^10 + 4*x^11 - 34*x^12 - 49*x^13 + 35*x^14 + 174*x^15 + ...
which equals the sum:
Sum_{n>=0} binomial(2*(n+1), n)/(n+1) * x^(n+1)/(1+x)^(3*(n+1)).
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..400

Crossrefs

Cf. A316987, A001764.

Programs

PARI

{a(n) = my(A = sum(m=0,n, binomial(3*(m+1), m)/(m+1) * x^m / (1+x +x*O(x^n))^(2*(m+1)))); polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

Formula

G.f. A(x) satisfies:

(1) A( 1/A(x) - 1 ) = 1/(1+x).

(2) A(x) = (1 + 3*x^2*A(x)^2 + x^3*A(x)^3) / (1 - x + x^2).

(3) A(x) = (1/x) * Series_Reversion( (1 + x + 3*x^2 + x^3 - (1+x)*sqrt(1 + 2*x^2 + 4*x^3 + x^4))/(2*x) ).

a(n) ~ 33^(1/4) * (19 + 3*sqrt(33))^(n+1) / (sqrt(3*Pi) * n^(3/2) * 2^(3*n + 5)). - Vaclav Kotesovec, Jul 06 2018