0,2

a(n) gives the length of the word obtained after n steps with the substitution rule 0->1111, 1->11110, starting from 0. The number of 1's and 0's of this word is 4*a(n-1) and 4*a(n-2), resp.

Inverse binomial transform of odd Pell bisection A001653. With a leading zero, inverse binomial transform of even Pell bisection A001542, divided by 2. - Paul Barry (pbarry(AT)wit.ie), May 16 2003

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=4, q=4.

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=4.

Index entries for sequences related to linear recurrences with constant coefficients

Tanya Khovanova, Recursive Sequences

Index entries for sequences related to Chebyshev polynomials.

a(n) = 4*(a(n-1)+a(n-2)), a(-1)=0, a(0)=1.

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

G.f.: 1/(1-4*x-4*x^2).

a(n)=Sum_{k, 0<=k<=n}3^k*A063967(n,k) . - Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Nov 03 2006

a(n)=-(1/8)*sqrt(2)*[2-2*sqrt(2)]^(n-1)+(1/8)*[2+2*sqrt(2)]^(n-1)*sqrt(2), with n>=0 [From Paolo P. Lava (ppl(AT)spl.at), Nov 20 2008]

((2+sqrt8)^n-(2-sqrt8)^n)/sqrt32. Offset 1. a(3)=20. [From Al Hakanson (hawkuu(AT)gmail.com), Jan 07 2009]

a(n)=A000129(n+1)*A000079(n). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 08 2009]

(PARI) a(n)=if(n<0, 0, (2*I)^n*subst(I*poltchebi(n+1)+poltchebi(n), 'x, -I)/2) /* Michael Somos Sep 16 2005 */

(Other) sage: [lucas_number1(n, 4, -4) for n in xrange(1, 23)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 23 2009]

Pairwise sums are in A086347.

Sequence in context: A094971 A099025 A008353 this_sequence A151254 A098225 A073532

Adjacent sequences: A057084 A057085 A057086 this_sequence A057088 A057089 A057090

nonn,easy

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