A119020 Eigenvector of triangle A055151 of Motzkin polynomial coefficients, where A055151(n,k) = n!/((n-2k)!*k!*(k+1)!) for 0<=k<=[n/2], n>=0.

1, 1, 2, 4, 11, 31, 96, 302, 1023, 3607, 13318, 50348, 195361, 772565, 3112630, 12715692, 52648847, 220705119, 937145214, 4028239116, 17522172021, 77071709841, 342583183572, 1537550150766, 6961838925069, 31774593686661

Offset: 0

Author

Paul D. Hanna, May 09 2006

Keywords

• nonn

Comments

Binomial transform is A119021. Inverse binomial transform is A119022.

Examples

			A(x) = 1 + x + 2*x^2 + 4*x^3 + 11*x^4 + 31*x^5 + 96*x^6 +...
A(x/(1+x))/(1+x) = 1 + x^2 + 2*2*x^4 + 4*5*x^6 + 11*14*x^8 +...+ a(n)*A000108(n)*x^(2n) +...
		

Crossrefs

Cf. A055151 (Motzkin polynomials), A119021 (binomial), A119022 (inverse binomial).

Programs

PARI

{a(n)=if(n==0,1,sum(k=0,n\2,n!/((n-2*k)!*k!*(k+1)!)*a(k)))}

Formula

Eigenvector: a(n) = Sum_{k=0..[n/2]} n!/((n-2k)!*k!*(k+1)!)*a(k), for n>=0, with a(0)=1.

G.f. satisfies: A(x) = A(-x/(1-2*x))/(1-2*x); i.e., 2nd inverse binomial transform equals A(-x).

G.f. satisfies: A(x/(1-x))/(1-x) = A(-x/(1-3*x))/(1-3*x).

G.f. of inverse binomial transform: A(x/(1+x))/(1+x) = B(x^2) where [x^n] B(x) = a(n)*C(2*n,n)/(n+1) = a(n)*A000108(n) and A000108=Catalan.