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

%I A316987 #7 Jul 19 2018 11:48:40
%S A316987 1,-1,-1,1,2,0,-4,-4,5,15,4,-34,-49,35,174,106,-363,-729,173,2311,
%T A316987 2266,-4048,-11573,-2001,32110,45110,-41692,-187550,-103297,445119,
%U A316987 864056,-301392,-3033429,-2820827,5900903,16076357,1596760,-48162378,-64931570,69966798,291036757,143307879,-737616558,-1374497088,610620123,5114691723,4475615447,-10618782829
%N A316987 G.f.: A(x) = Sum_{n>=0} binomial(2*(n+1), n)/(n+1) * x^n / (1+x)^(3*(n+1)).
%C A316987 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.
%C A316987 Compare the g.f. to:
%C A316987 (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).
%C A316987 (C2) 1 = Sum_{n>=0} binomial(m*(n+1), n)/(n+1) * x^n / (1+x)^(m*(n+1)) holds for fixed m.
%C A316987 (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.
%H A316987 Paul D. Hanna, <a href="/A316987/b316987.txt">Table of n, a(n) for n = 0..500</a>
%F A316987 G.f. A(x) satisfies:
%F A316987 (1) A(x) = Sum_{n>=0} binomial(2*(n+1), n)/(n+1) * x^n / (1+x)^(3*(n+1)).
%F A316987 (2) A(x) = (1 + x^2*A(x)^2) / (1 + x + 3*x^2 + x^3).
%F A316987 (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).
%e A316987 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 + ...
%e A316987 such that
%e A316987 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)) + ...
%e A316987 RELATED SERIES.
%e A316987 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 + ...
%e A316987 which equals the sum:
%e A316987 Sum_{n>=0} binomial(3*(n+1), n)/(n+1) * x^(n+1)/(1+x)^(2*(n+1)).
%o A316987 (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)}
%o A316987 for(n=0, 30, print1(a(n), ", "))
%Y A316987 Cf. A316371.
%K A316987 sign
%O A316987 0,5
%A A316987 _Paul D. Hanna_, Jul 18 2018