A365775 Expansion of g.f. A(x) satisfying A(x) = 1 + x*A(x)/(1 - 5*x*A(x))^2.

1, 1, 11, 106, 1061, 11226, 124026, 1414211, 16515981, 196551736, 2375042076, 29062573926, 359407971786, 4484868410231, 56399986492661, 714067825064426, 9094408567049701, 116436367409647736, 1497734068943432856, 19346547929074098836, 250851388061224003276

Offset

0,3

Comments

Related identities which hold formally for all Maclaurin series F(x):

(1) F(x) = (1/x) * Sum{n>=1} n^(n-1) * x^n * F(x)^n / (1 + n*x*F(x))^(n+1),

(2) F(x) = (2/x) * Sum{n>=1} n*(n+1)^(n-2) * x^n * F(x)^n / (1 + (n+1)*x*F(x))^(n+1),

(3) F(x) = (3/x) * Sum{n>=1} n*(n+2)^(n-2) * x^n * F(x)^n / (1 + (n+2)*x*F(x))^(n+1),

(4) F(x) = (4/x) * Sum{n>=1} n*(n+3)^(n-2) * x^n * F(x)^n / (1 + (n+3)*x*F(x))^(n+1),

(5) F(x) = (k/x) * Sum{n>=1} n*(n+k-1)^(n-2) * x^n * F(x)^n / (1 + (n+k-1)*x*F(x))^(n+1) for all fixed nonzero k.

Links

Paul D. Hanna, Table of n, a(n) for n = 0..400

Formula

a(n) = Sum{k=0..n} binomial(n+1, n-k)/(n+1) * binomial(2*n-k-1, k) * 5^k.

Let A(x)^m = Sum_{n>=0} a(n,m) * x^n then a(n,m) = Sum_{k=0..n} binomial(n+m, n-k)*m/(n+m) * binomial(2*n-k-1, k) * 5^k.

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.

(1) A(x) = 1 + x*A(x)/(1 - 5*x*A(x))^2.

(2) A(x) = (1/x) * Series_Reversion( x/(1 + x/(1 - 5*x)^2) ).

(3) A( x/(1 + x/(1 - 5*x)^2) ) = 1 + x/(1 - 5*x)^2.

(4) A(x) = 1 + (m+1) * Sum{n>=1} n*(n+m)^(n-2) * x^n * A(x)^n / (1 + (n+m-5)*x*A(x))^(n+1) for all fixed nonnegative m.

(4.a) A(x) = 1 + Sum{n>=1} n^(n-1) * x^n * A(x)^n / (1 + (n-5)*x*A(x))^(n+1).

(4.b) A(x) = 1 + 2 * Sum{n>=1} n*(n+1)^(n-2) * x^n * A(x)^n / (1 + (n-4)*x*A(x))^(n+1).

(4.c) A(x) = 1 + 3 * Sum{n>=1} n*(n+2)^(n-2) * x^n * A(x)^n / (1 + (n-3)*x*A(x))^(n+1).

(4.d) A(x) = 1 + 4 * Sum{n>=1} n*(n+3)^(n-2) * x^n * A(x)^n / (1 + (n-2)*x*A(x))^(n+1).

(4.e) A(x) = 1 + 5 * Sum{n>=1} n*(n+4)^(n-2) * x^n * A(x)^n / (1 + (n-1)*x*A(x))^(n+1).

(4.f) A(x) = 1 + 6 * Sum{n>=1} n*(n+5)^(n-2) * x^n * A(x)^n / (1 + n*x*A(x))^(n+1).

a(n) ~ sqrt(3) * 5^(2*n) * (19^(3/2 + n) / (2*sqrt((113 + (28*(47225 + 1083*sqrt(1905))^(1/3))/5^(2/3) - 2632/(5*(47225 + 1083*sqrt(1905)))^(1/3))*Pi) * n^(3/2) * (68 + (2*(-1496331 + 60325*sqrt(1905)))^(1/3)/3^(2/3) - 9214*2^(2/3)/(3*(-1496331 + 60325*sqrt(1905)))^(1/3))^(n + 1/2))). - Vaclav Kotesovec, Oct 06 2023

Example

G.f.: A(x) = 1 + x + 11*x^2 + 106*x^3 + 1061*x^4 + 11226*x^5 + 124026*x^6 + 1414211*x^7 + 16515981*x^8 + 196551736*x^9 + 2375042076*x^10 + ...
where A(x) satisfies A(x) = 1 + x*A(x)/(1 - 5*x*A(x))^2
also
A(x) = 1 + 1^0*x*A(x)/(1 + (-4)*x*A(x))^2 + 2^1*x^2*A(x)^2/(1 + (-3)*x*A(x))^3 + 3^2*x^3*A(x)^3/(1 + (-2)*x*A(x))^4 + 4^3*x^4*A(x)^4/(1 + (-1)*x*A(x))^5 + 5^4*x^5*A(x)^5/(1 + 0*x*A(x))^6 + 6^5*x^6*A(x)^6/(1 + 1*x*A(x))^7 + ...
and
A(x) = 1 + 6*1*6^(-1)*x*A(x)/(1 + 1*x*A(x))^2 + 6*2*7^0*x^2*A(x)^2/(1 + 2*x*A(x))^3 + 6*3*8^1*x^3*A(x)^3/(1 + 3*x*A(x))^4 + 6*4*9^2*x^4*A(x)^4/(1 + 4*x*A(x))^5 + 6*5*10^3*x^5*A(x)^5/(1 + 5*x*A(x))^6 + ...

Prog

(PARI) {a(n) = sum(k=0, n, binomial(n+1, n-k)/(n+1) * binomial(2*n-k-1, k) * 5^k)}
for(n=0, 30, print1(a(n), ", "))
(PARI) {a(n) = my(A = (1/x) * serreverse( x/(1 + x/(1 - 5*x +O(x^(n+2)) )^2) ) ); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

Crossrefs

Cf. A365770, A109081, A365772, A365773, A365774.

Cf. A366235 (dual).

Sequence in context: A250416 A116011 A229070 * A058715 A140617 A224717

Adjacent sequences: A365772 A365773 A365774 * A365776 A365777 A365778

Keyword

nonn

Author

Paul D. Hanna, Oct 04 2023

Status

approved