%I #23 Mar 23 2024 10:50:23
%S 1,3,15,85,526,3438,23358,163306,1167235,8490513,62648451,467769217,
%T 3527692298,26832220834,205601792340,1585604105312,12297768490441,
%U 95861469636203,750611119223931,5901214027721577,46564408929573723,368644188180241449,2927350250765841801,23310167641788680947,186089697960587977233,1489085453187335910243
%N G.f.: Sum_{n>=0} binomial(4*(n+1), n)/(n+1) * x^n / (1+x)^(n+1).
%C Note that: binomial(4*(n+1), n)/(n+1) = A002293(n+1) for n >= 0, where F(x) = Sum_{n>=0} A002293(n)*x^n satisfies F(x) = 1 + x*F(x)^4.
%C Compare the g.f. to:
%C (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 (C2) 1 = Sum_{n>=0} binomial(m*(n+1), n)/(n+1) * x^n / (1+x)^(m*(n+1)) holds for fixed m.
%C (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.
%F G.f. A(x) satisfies:
%F (1) A(x) = (1 + x*A(x))^4 / (1+x).
%F (2) A(x) = (1/x) * Series_Reversion( x/((1+x)^4 - x) ).
%F (3) A(x) = Sum_{n>=0} binomial(4*(n+1), n)/(n+1) * x^n / (1+x)^(n+1).
%F a(n) ~ 229^(n + 3/2) / (sqrt(Pi) * 2^(7/2) * n^(3/2) * 3^(3*n + 9/2)). - _Vaclav Kotesovec_, Jul 22 2018
%F a(n) = (1/(n+1)) * Sum_{k=0..n} (-1)^k * binomial(n+1,k) * binomial(4*n-4*k+4,n-k). - _Seiichi Manyama_, Mar 23 2024
%e G.f.: A(x) = 1 + 3*x + 15*x^2 + 85*x^3 + 526*x^4 + 3438*x^5 + 23358*x^6 + 163306*x^7 + 1167235*x^8 + 8490513*x^9 + 62648451*x^10 + ...
%e such that
%e A(x) = 1/(1+x) + 4*x/(1+x)^2 + 22*x^2/(1+x)^3 + 140*x^3/(1+x)^4 + 969*x^4/(1+x)^5 + 7084*x^5/(1+x)^6 + ... + A002293(n+1)*x^n/(1+x)^(n+1) + ...
%e RELATED SERIES.
%e Series_Reversion( x*A(x) ) = x/((1+x)^4 - x) = x - 3*x^2 + 3*x^3 + 5*x^4 - 22*x^5 + 27*x^6 + 28*x^7 - 163*x^8 + 235*x^9 + 134*x^10 + ...
%e which equals the sum:
%e Sum_{n>=0} binomial(n+1, n)/(n+1) * x^(n+1)/(1+x)^(4*(n+1)).
%t Rest[CoefficientList[InverseSeries[Series[x/((1 + x)^4 - x), {x, 0, 20}], x], x]] (* _Vaclav Kotesovec_, Jul 22 2018 *)
%o (PARI) {a(n) = my(A = sum(m=0, n, binomial(4*(m+1), m)/(m+1) * x^m / (1+x +x*O(x^n))^(1*(m+1)))); polcoeff(A, n)}
%o for(n=0, 30, print1(a(n), ", "))
%o (PARI) {a(n) = my(A = (1/x) * serreverse( x/((1+x)^4 - x +x*O(x^n)) ) ); polcoeff(A, n)}
%o for(n=0, 30, print1(a(n), ", "))
%Y Cf. A316371, A127897, A317134, A349361, A349362, A349363, A349364.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Jul 21 2018
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