A213684 Logarithmic derivative of A001002.

1, 5, 22, 105, 511, 2534, 12720, 64449, 328900, 1688115, 8705060, 45064110, 234054198, 1219053680, 6364813192, 33302104593, 174570695175, 916628799380, 4820160541350, 25381091113455, 133808636072595, 706211862466500, 3730964595817680, 19729042153581150

Offset: 1

Author

Paul D. Hanna, Jun 22 2012

Keywords

• nonn
• easy

Comments

A001002(n) is the number of dissections of a convex (n+2)-gon into triangles and quadrilaterals by nonintersecting diagonals.

The g.f. of A001002 satisfies: G(x) = 1 + x*G(x)^2 + x^2*G(x)^3.

Central terms in A155161: a(n) = A155161(2*n,n). - Reinhard Zumkeller, Apr 17 2013

a(n) is the 2n-th term of the n-fold self-convolution of the Fibonacci numbers. - Alois P. Heinz, Feb 07 2021

Examples

			L.g.f.: L(x) = x + 5*x^2/2 + 22*x^3/3 + 105*x^4/4 + 511*x^5/5 +...
such that
L(x) = x*(1+x) + d/dx x^3*(1+x)^2/2! + d^2/dx^2 x^5*(1+x)^3/3! + d^3/dx^3 x^7*(1+x)^4/4! +...
The g.f. of A001002 begins:
exp(L(x)) = 1 + x + 3*x^2 + 10*x^3 + 38*x^4 + 154*x^5 + 654*x^6 +...
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
• Dmitry V. Kruchinin and Vladimir V. Kruchinin, A Generating Function for the Diagonal T2n,n in Triangles, Journal of Integer Sequence, Vol. 18 (2015), article 15.4.6.

Crossrefs

Cf. A000045, A001002, A037027, A155161, A298610, A350383.

Programs

Haskell

a213684 n = a155161 (2*n) n  -- Reinhard Zumkeller, Apr 17 2013

Maple

with(orthopoly): seq(add(i, i in [seq((-1)^iquo(n-k,2)*coeff(G(n,n,x/2), x, k), k=0..n)]), n=1..24); # Peter Luschny, Jan 26 2018

Mathematica

Table[n*Sum[Binomial[k+n-1,n]*Binomial[k,n-k]/k,{k,1,n}],{n,1,20}] (* Vaclav Kotesovec, Oct 20 2012 *)

PARI

{a(n)=n*sum(r=1,n,binomial(r+n-1,n)*binomial(r,n-r)/r)}
for(n=1, 30, print1(a(n), ", "))

PARI

{Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
{a(n)=local(A=1); A=(sum(m=1, n+1, Dx(m-1, x^(2*m-1)*(1+x)^m/m!)+x*O(x^n))); n*polcoeff(A, n)}
for(n=1, 30, print1(a(n), ", "))

Formula

a(n) = n * Sum_{r=1..n} binomial(r+n-1,n) * binomial(r,n-r) / r.

L.g.f.: Sum_{n>=1} d^(n-1)/dx^(n-1) x^(2*n-1)*(1+x)^n/n! = Sum_{n>=1} a(n)*x^n/n.

Recurrence: 75*(n-1)*n*a(n) = 5*(n-1)*(59*n-12)*a(n-1) + (559*n^2-1503*n+1100)* a(n-2) + 21*(3*n-8)*(3*n-7)*a(n-3). - Vaclav Kotesovec, Oct 20 2012

a(n) ~ 3^(3*n)/(2*5^(n-1/2)*sqrt(6*Pi*n)). - Vaclav Kotesovec, Oct 20 2012

a(n) = A037027(2*n-1,n-1). - Vladimir Kruchinin, Feb 28 2013

a(n) = Sum_{k=0..n} (-1)^floor((n-k)/2) [x^k] G(n,n,x/2), where G(n,a,x) denotes the n-th Gegenbauer polynomial; row sums of A298610. - Peter Luschny, Jan 26 2018

a(n) = [x^n] (1/(1-x-x^2))^n. - Alois P. Heinz, Feb 07 2021

From Peter Bala, Mar 11 2025: (Start)

a(n) = Sum_{k = 0..n} n/(2*n-k) * binomial(2*n-k, k)*binomial(2*n-2*k, n).

a(n) = (1/2)*binomial(2*n, n)*hypergeom([-n/2, (-n+1)/2], [-2*n+1], -4). Cf. A350383.

Second-order recurrence: 5*n*(n-1)*(8*n-13)*a(n) = 2*(n-1)*(88*n^2-187*n+75)*a(n-1) + 3*(8*n-5)*(3*n-4)*(3*n-5)*a(n-2) with a(1) = 1 and a(2) = 5. (End)