A200031 - OEIS

%I #33 Dec 23 2021 03:30:07

%S 1,5,25,150,1000,7125,53125,409375,3234375,26059375,213296875,

%T 1768625000,14825156250,125419296875,1069473046875,9182583593750,

%U 79319843750000,688837802734375,6010580419921875,52670308222656250,463321803125000000,4089876834521484375,36217014743896484375

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

%C 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

%H Vincenzo Librandi, <a href="/A200031/b200031.txt">Table of n, a(n) for n = 0..200</a>

%H Alexander Burstein and Louis W. Shapiro, <a href="https://arxiv.org/abs/2112.11595">Pseudo-involutions in the Riordan group</a>, arXiv:2112.11595 [math.CO], 2021.

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

%F Recurrence: (n+1)*a(n) = 5*(2*n-1)*a(n-1) - 5*(n-2)*a(n-2). - _Vaclav Kotesovec_, Oct 20 2012

%F 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

%F 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

%F 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

%F a(n) = 5*A107265(n-1) for n >= 1. - _Alexander Burstein_, May 27 2021

%e G.f.: A(x) = 1 + 5*x + 25*x^2 + 150*x^3 + 1000*x^4 + 7125*x^5 + 53125*x^6 + ...

%e The g.f. satisfies A(x) = 1 + x*(1 + 3*A(x) + A(x)^2) where:

%e A(x)^2 = 1 + 10*x + 75*x^2 + 550*x^3 + 4125*x^4 + 31750*x^5 + 250000*x^6 + ...

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

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

%Y Cf. A107265.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Nov 12 2011