A088327 G.f.: exp(Sum_{k>=1} B(x^k)/k), where B(x) = x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + ... = (C(x)-1)/x and C is the g.f. for the Catalan numbers A000108.
1, 1, 3, 8, 25, 77, 256, 854, 2940, 10229, 36124, 128745, 463137, 1677816, 6118165, 22432778, 82660369, 305916561, 1136621136, 4238006039, 15852603939, 59471304434, 223704813807, 843547443903, 3188064830876, 12074092672950, 45816941923597, 174173975322767
Offset: 0
Keywords
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
Programs
-
Magma
m:=35; f:= func< x | (&*[1/(1-x^j)^Catalan(j): j in [1..m+2]]) >; R
:=PowerSeriesRing(Rationals(), m); Coefficients(R!( f(x) )); // G. C. Greubel, Dec 12 2022 -
Maple
with(numtheory): a:= proc(n) option remember; `if`(n=0, 1, add(add(d* binomial(2*d, d)/(d+1), d=divisors(j))*a(n-j), j=1..n)/n) end: seq(a(n), n=0..40); # Alois P. Heinz, Sep 10 2012
-
Mathematica
With[{nn=35}, CoefficientList[Series[Product[1/(1-x^i)^CatalanNumber[i], {i,nn}], {x,0,nn}], x]] (* Geoffrey Critzer, Feb 26 2013 *)
-
SageMath
# uses[EulerTransform from A166861] b = EulerTransform(lambda n: binomial(2*n, n)/(n+1)) print([b(n) for n in range(30)]) # Peter Luschny, Nov 11 2020
Formula
Euler transform of Catalan numbers (A000108). - Franklin T. Adams-Watters, Mar 01 2006
a(n) ~ c * 4^n / n^(3/2), where c = exp(Sum_{k>=1} (-2 + 4^k - 4^k*sqrt(1 - 4^(1-k)))/(2*k) ) / sqrt(Pi) = 1.60022306097485382475864802335610662545... - Vaclav Kotesovec, Mar 21 2021
Comments