A217596 Expansion of g.f. x / reversion(x - x^2 - x^3).

1, -1, -2, -5, -17, -64, -259, -1098, -4815, -21659, -99385, -463385, -2189070, -10455340, -50402858, -244929608, -1198504743, -5900360016, -29204546125, -145244328630, -725451444795, -3637422742470, -18301949731665, -92380935149100, -467659449093330

Offset: 0

Author

Vladimir Kruchinin, Mar 19 2013

Keywords

• sign
• easy

Comments

For n>0, -a(n) is the number of times a convex polygon with n + 2 sides can be divided into triangles and quadrilaterals by non-intersecting diagonals such that none of the dividing diagonals passes through a chosen vertex. - Muhammed Sefa Saydam, Jan 27 2025

Examples

			G.f.: A(x) = 1 - x - 2*x^2 - 5*x^3 - 17*x^4 - 64*x^5 - 259*x^6 - 1098*x^7 - ...
		

References

• H. S. Wilf, Generatingfunctionology, Academic Press, NY, 1990.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..600

Crossrefs

Cf. A001002, A007440.

Programs

Mathematica

CoefficientList[x/InverseSeries[Series[x-x^2-x^3, {x, 0, 20}], x],x] (* Vaclav Kotesovec, Feb 15 2015 *)

Maxima

a(n):=if n=0 then 1 else if n=1 then -1 else -sum(binomial(i,n-i)*binomial(n+i-2,n-2),i,ceiling(n/2),n)/(n-1);

PARI

{a(n)=polcoeff(x/serreverse(x-x^2-x^3+x^2*O(x^n)),n)} \\ Paul D. Hanna, Mar 19 2013

PARI

/* Using Vladimir Kruchinin's binomial sum: */
{a(n)=if(n==0,1,if(n==1,-1,-sum(i=n\2,n,binomial(i,n-i)*binomial(n+i-2,n-2))/(n-1)))} \\ Paul D. Hanna, Mar 19 2013

PARI

{a(n)=local(A=1);for(i=1,n,A=1-x/A-x^2/A^2+x*O(x^n));polcoeff(A,n)} \\ Paul D. Hanna, Mar 19 2013

Formula

a(n) = -Sum_{i=ceiling(n/2)..n} binomial(i,n-i)*binomial(n+i-2,n-2)/(n-1), n>1, a(0)=1, a(1)=-1.

a(n) = -Sum_{i=0..n-1} A001002(n-i)*a(i), a(0)=1. [corrected by Muhammed Sefa Saydam, Jul 23 2025]

From Paul D. Hanna, Mar 19 2013: (Start)

G.f. satisfies:

(1) A(x) = 1 - x/A(x) - x^2/A(x)^2.

(2) A(x - x^2 - x^3) = 1 - x - x^2.

(3) [x^n] A(x)^n = -A000204(n), where A000204 is the Lucas numbers.

(4) [x^n] A(x)^(n+1) = 0 for n>2. (End)

Conjecture: 25*n*(n-1)*a(n) - 5*(n-1)*(25*n-42)*a(n-1) + 3*(-23*n^2 + 59*n + 4)*a(n-2) + 9*(3*n-10)*(3*n-11)*a(n-3)=0. - R. J. Mathar, May 23 2014

Maple's sumrecursion command gives the second-order recurrence equation: 5*n*(n - 1)*(4*n - 9)*a(n) = 2*(n - 1)*(44*n^2 - 165*n + 150)*a(n-1) + 3*(4*n - 5)(3*n - 7)(3*n - 8)*a(n-2) with initial conditions a(1) = -1 and a(2) = -2. Mathar's conjectured third-order recurrence follows from this. - Peter Bala, Feb 15 2015

a(n) = -A001002(n-1) - Sum_{i=0..n-2} A001002(i) * A001002(n-i-2) for n >= 2. - Muhammed Sefa Saydam, Jul 25 2025