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(n) / a(n-1) converges to (5 + (3 * sqrt(5))) / 2 as n approaches infinity. (5 + (3 * sqrt(5))) / 2 can also be written as phi^2 + (2 * phi), phi^3 + phi, phi + sqrt(5) + 2, (3 * phi) + 1, (3 * phi^2) - 2, phi^4 - 1 and (5 + (3 * (L(n) / F(n)))) / 2, where L(n) is the n-th Lucas number and F(n) is the n-th Fibonacci number as n approaches infinity. -Ross La Haye,Aug 18 2003, on another version

Pisano period lengths: 1, 3, 3, 6, 1, 3, 24, 12, 9, 3, 10, 6, 56, 24, 3, 24,288, 9, 18, 6,... -R. J. Mathar,Aug 10 2012

Indranil Ghosh,Table of n, a(n) for n = 0..1300

Martin Burtscher, Igor Szczyrba, Rafał Szczyrba,Analytic Representations of the n-anacci Constants and Generalizations Thereof,Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.

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.

Tanya Khovanova,Recursive Sequences

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

Eric Weisstein's World of Mathematics,Horadam Sequence

Index entries for linear recurrences with constant coefficients,signature (5,5)

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,Oct 25 2003

a(n) = ((5 + 3*sqrt(5))/2)^n(1/2 + sqrt(5)/6) + (1/2 - sqrt(5)/6)((5 - 3*sqrt(5))/2)^n. -Paul Barry,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,Nov 28 2004

a(n) = Sum_{k=0..n} 4^k*A063967(n,k). -Philippe Deléham,Nov 03 2006

G.f.: G(0)/(2-5*x), where G(k)= 1 + 1/(1 - x*(9*k-5)/(x*(9*k+4) - 2/G(k+1))); (continued fraction). -Sergei N. Gladkovskii,Jun 17 2013

FromEhren Metcalfe,Nov 18 2017: (Start)

With F(n) =A000045(n), L(n) =A000032(n), beta = (1-sqrt(5))/2:

a(2*n-1) = 5^n*F(4*n)/3 = (5^(n-1/2)*L(4*n) - 2*5^(n-1/2)*beta^(4*n))/3.

a(2*n) = 5^n*L(4*n+2)/3 = (5^(n+1/2)*F(4*n+2) + 2*5^n*beta^(4*n+2))/3.

a(n) = round 5^((n+1)/2)*F(2*(n+1))/3.

a(n) = round 5^(n/2)*L(2*(n+1))/3. (End)

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=5*a[n-1]+5*a[n-2]od: seq(a[n], n=1..33); #Zerinvary Lajos,Dec 14 2008

LinearRecurrence[{5, 5}, {1, 5}, 30] (*G. C. Greubel,Jan 16 2018 *)

(Sage) [lucas_number1(n, 5, -5) for n in range(1, 22)] #Zerinvary Lajos,Apr 24 2009

(PARI) x='x+O('x^30); Vec(1/(1 - 5*x - 5*x^2)) \\G. C. Greubel,Jan 16 2018

(Magma) I:=[1, 5]; [n le 2 select I[n] else 5*Self(n-1) + 5*Self(n-2): n in [0..30]]; //G. C. Greubel,Jan 16 2018

Cf.A001076,A006190,A007482,A015520,A015521,A015523,A015524,A015525,A015528,A015529,A015530,A015531,A015532,A015533,A015534,A015535,A015536,A015537,A015443,A015447,A030195,A053404,A057087,A083858,A085939,A090017,A091914,A099012,A180222,A180226.

Sequence in context:A241588A229246A276598*A156195A105481A242157

Adjacent sequences:A057085A057086A057087*A057089A057090A057091

nonn,easy

Wolfdieter Lang,Aug 11 2000

approved