A200031 G.f. satisfies: A(x) = 1 + x + 3*x*A(x) + x*A(x)^2.

1, 5, 25, 150, 1000, 7125, 53125, 409375, 3234375, 26059375, 213296875, 1768625000, 14825156250, 125419296875, 1069473046875, 9182583593750, 79319843750000, 688837802734375, 6010580419921875, 52670308222656250, 463321803125000000, 4089876834521484375, 36217014743896484375

Offset

0,2

Comments

Counts colored Motzkin paths starting with a level step H(1,0), where H(1,0) and U(1,1) each have 5 colors and D(1,-1) has one color. - Alexander Burstein, May 27 2021

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group, arXiv:2112.11595 [math.CO], 2021.

Formula

G.f.: A(x) = (1-3*x - sqrt(1 - 10*x + 5*x^2))/(2*x).

Recurrence: (n+1)*a(n) = 5*(2*n-1)*a(n-1) - 5*(n-2)*a(n-2). - Vaclav Kotesovec, Oct 20 2012

a(n) ~ sqrt(10+5*sqrt(5))*(5+2*sqrt(5))^n/(2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 20 2012. Equivalently, a(n) ~ 5^((n+1)/2) * phi^(3*n + 3/2) / (2*sqrt(Pi)*n^(3/2)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021

a(n) = Sum_{k=0..floor((n-1)/2)}{binomial(n-1,2k)*A000108(k)*5^{n-k}} for n>=1. - Alexander Burstein, May 24 2021

G.f.: Let B(x) = 2 + A(x) and let G(x) be the g.f. for A344623. Then G(x) = 1 + x*G(x)*B(x^2*G(x)). - Alexander Burstein, May 25 2021

a(n) = 5*A107265(n-1) for n >= 1. - Alexander Burstein, May 27 2021

Example

G.f.: A(x) = 1 + 5*x + 25*x^2 + 150*x^3 + 1000*x^4 + 7125*x^5 + 53125*x^6 + ...
The g.f. satisfies A(x) = 1 + x*(1 + 3*A(x) + A(x)^2) where:
A(x)^2 = 1 + 10*x + 75*x^2 + 550*x^3 + 4125*x^4 + 31750*x^5 + 250000*x^6 + ...

Mathematica

CoefficientList[Series[(1-3*x-Sqrt[1-10*x+5*x^2])/(2*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)

Prog

(PARI) {a(n)=polcoeff((1-3*x - sqrt(1-10*x+5*x^2 +x^2*O(x^n)))/(2*x), n)}

Crossrefs

Cf. A107265.

Sequence in context: A002050 A047782 A106565 * A216689 A297589 A092166

Adjacent sequences: A200028 A200029 A200030 * A200032 A200033 A200034

Keyword

nonn

Author

Paul D. Hanna, Nov 12 2011

Status

approved