0,2

a(n) gives the length of the word obtained after n steps with the substitution rule 0->11111, 1->111110, starting from 0. The number of 1's and 0's of this word is 5*a(n-1) and 5*a(n-2), resp.

A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case n->n+1, a=0,b=1; p=5, q=5.

W. Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Eqs.(39) and (45),rhs, m=5.

Tanya Khovanova, Recursive Sequences

Index entries for sequences related to Chebyshev polynomials.

a(n) = 5*(a(n-1)+a(n-2)), a(-1)=0, a(0)=1.

a(n)= S(n, i*sqrt(5))*(-i*sqrt(5))^n with S(n, x) := U(n, x/2), Chebyshev's polynomials of the 2nd kind, A049310.

G.f.: 1/(1-5*x-5*x^2).

a(n)=(1/3)*sum(k=0, n, binomial(n, k)*Fibonacci(k)*3^k) - Benoit Cloitre (benoit7848c(AT)orange.fr), Oct 25 2003

a(n)=((5+3sqrt(5))/2)^n(1/2+sqrt(5)/6)+(1/2-sqrt(5)/6)((5-3sqrt(5))/2)^n - Paul Barry (pbarry(AT)wit.ie), Sep 22 2004

(a(n)) appears to be given by the floretion - 0.75'i - 0.5'j + 'k - 0.75i' + 0.5j' + 0.5k' + 1.75'ii' - 1.25'jj' + 1.75'kk' - 'ij' - 0.5'ji' - 0.75'jk' - 0.75'kj' - 1.25e ("jes") - Creighton Dement (Smith(AT)xxx.yyy.com), Nov 28 2004

a(n)=Sum_{k, 0<=k<=n}4^k*A063967(n,k) . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Nov 03 2006

Adjacent sequences: A057085 A057086 A057087 this_sequence A057089 A057090 A057091

Sequence in context: A094972 A084158 A111469 this_sequence A105481 A051738 A094167

nonn,easy

Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de) Aug 11 2000