A316987 G.f.: A(x) = Sum_{n>=0} binomial(2*(n+1), n)/(n+1) * x^n / (1+x)^(3*(n+1)).

1, -1, -1, 1, 2, 0, -4, -4, 5, 15, 4, -34, -49, 35, 174, 106, -363, -729, 173, 2311, 2266, -4048, -11573, -2001, 32110, 45110, -41692, -187550, -103297, 445119, 864056, -301392, -3033429, -2820827, 5900903, 16076357, 1596760, -48162378, -64931570, 69966798, 291036757, 143307879, -737616558, -1374497088, 610620123, 5114691723, 4475615447, -10618782829

Offset: 0

Author

Paul D. Hanna, Jul 18 2018

Keywords

• sign

Comments

Note that: binomial(2*(n+1), n)/(n+1) = A000108(n+1) for n >= 0, where C(x) = Sum_{n>=0} A000108(n)*x^n satisfies C(x) = 1 + x*C(x)^2.

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 - x^2 + x^3 + 2*x^4 - 4*x^6 - 4*x^7 + 5*x^8 + 15*x^9 + 4*x^10 - 34*x^11 - 49*x^12 + 35*x^13 + 174*x^14 + 106*x^15 - 363*x^16 + ...
such that
A(x) = 1/(1+x)^3 + 2*x/(1+x)^6 + 5*x^2/(1+x)^9 + 14*x^3/(1+x)^12 + 42*x^4/(1+x)^15 + 132*x^5/(1+x)^18 + ... + A000108(n+1)*x^n/(1+x)^(3*(n+1)) + ...
RELATED SERIES.
Series_Reversion( x*A(x) ) = x + x^2 + 3*x^3 + 9*x^4 + 30*x^5 + 105*x^6 + 382*x^7 + 1429*x^8 + 5463*x^9 + 21248*x^10 + ... + A316371(n)*x^n + ...
which equals the sum:
Sum_{n>=0} binomial(3*(n+1), n)/(n+1) * x^(n+1)/(1+x)^(2*(n+1)).
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..500

Crossrefs

Cf. A316371.

Programs

PARI

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

Formula

G.f. A(x) satisfies:

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

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

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