A213507 E.g.f.: exp( Sum_{n>=1} A000108(n)*x^n/n ), where A000108(n) = binomial(2*n,n)/(n+1) forms the Catalan numbers.

1, 1, 3, 17, 149, 1809, 28399, 550297, 12732873, 343231361, 10576764251, 367054970721, 14173669352413, 602974492511377, 28027436035348359, 1413479599558432169, 76879014760731439889, 4486205132570631391617, 279595430611791210216883, 18536284947404377562405041

Offset

0,3

Comments

Compare to: exp( Sum_{n>=1} (n+1)/2*A000108(n)*x^n/n ) = Sum_{n>=0} A000108(n)*x^n.

Conjecture: a(n) is not divisible by {2,5,13,29,31,37,41,43,53,71,97}.

Links

Table of n, a(n) for n=0..19.

Formula

E.g.f.: (4*exp(1-2/(sqrt(1-4*x)+1)))/(sqrt(1-4*x)+1)^2. - Vladimir Kruchinin, Mar 22 2013

a(n) ~ 2^(2*n+5/2)*n^(n-2)/exp(n+1). - Vaclav Kotesovec, Sep 24 2013

From Peter Bala, Nov 21 2016: (Start)

Conjectural e.g.f.: 1/x * Series_Reversion ( x/(1 + F(x)) ), where F(x) = x + Sum_{n >= 2} (n - 1)^(n - 2)*x^n/n! = x + x^2/2! + 2*x^3/3! + 3^2*x^4/4! + 4^3*x^5/5! + 5^4*x^6/6! + ...; that is, dF/dx = 1 - LambertW(-x) = 1 + Euler's tree function T(x). See A000169.

The conjecture is equivalent to the result: Catalan(n) = [x^n] (1 + F(x))^n = [x^n] (2*x + x*T(x) + x/T(x))^n. (End)

Example

E.g.f.: A(x) = 1 + x + 3*x^2/2! + 17*x^3/3! + 149*x^4/4! + 1809*x^5/5! + ...
such that
log(A(x)) = x + 2*x^2/2 + 5*x^3/3 + 14*x^4/4 + 42*x^5/5 + 132*x^6/6 + 429*x^7/7 + 1430*x^8/8 + ... + A000108(n)*x^n/n + ...

Mathematica

With[{nn=20}, CoefficientList[Series[(4Exp[1-2/(Sqrt[1-4x]+1)])/ (Sqrt[ 1-4x]+1)^2, {x, 0, nn}], x] Range[0, nn]!] (* Harvey P. Dale, Jan 18 2016 *)

Prog

(PARI) {a(n)=n!*polcoeff(exp(sum(m=1, n, binomial(2*m, m)/(m+1)*x^m/m)+x*O(x^n)), n)}
for(n=0, 30, print1(a(n), ", "))

Crossrefs

Cf. A000108, A000169, A243953.

Sequence in context: A234289 A009813 A319946 * A305471 A368597 A135750

Adjacent sequences: A213504 A213505 A213506 * A213508 A213509 A213510

Keyword

nonn,easy

Author

Paul D. Hanna, Jun 12 2012

Status

approved