OFFSET
0,2
COMMENTS
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), respectively.
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, May 16 2003
For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 4's along the main diagonal, and 2's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 19 2011
Pisano period lengths: 1, 1, 8, 1, 3, 8, 6, 1, 24, 3, 120, 8, 21, 6, 24, 1, 16, 24, 360, 3, ... . - R. J. Mathar, Aug 10 2012
Exponential convolution of Pell numbers (A000129) and companion Pell numbers (A002203), divided by 2 and leading zero dropped. - Vladimir Reshetnikov, Oct 07 2016
LINKS
Indranil Ghosh, Table of n, a(n) for n = 0..1459
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=4, q=4.
Tanya Khovanova, Recursive Sequences.
Wolfdieter 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=4.
Index entries for linear recurrences with constant coefficients, signature (4,4)
FORMULA
a(n) = 4*(a(n-1) + a(n-2)), a(-1)=0, a(0)=1.
G.f.: 1/(1 - 4*x - 4*x^2).
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.
a(n) = Sum_{k=0..n} 3^k*A063967(n,k). - Philippe Deléham, Nov 03 2006
From Johannes W. Meijer, Aug 01 2010: (Start)
a(n) = 4^n*hypergeom([1/2-n/2, -n/2], [-n], -1) for n>=1. - Peter Luschny, Dec 17 2015
MAPLE
A057087 := n -> `if`(n=0, 1, 4^n*hypergeom([1/2-n/2, -n/2], [-n], -1)):
seq(simplify(A057087(n)), n=0..21); # Peter Luschny, Dec 17 2015
MATHEMATICA
Table[Fibonacci[n + 1, 2] 2^n, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 08 2016 *)
LinearRecurrence[{4, 4}, {1, 4}, 30] (* Harvey P. Dale, Aug 17 2017 *)
PROG
(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 */
(PARI) Vec(1/(1-4*x-4*x^2) + O(x^100)) \\ Altug Alkan, Dec 17 2015
(Sage) [lucas_number1(n, 4, -4) for n in range(1, 23)] # Zerinvary Lajos, Apr 23 2009
(Magma) I:=[1, 4]; [n le 2 select I[n] else 4*Self(n-1) + 4*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 16 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Aug 11 2000
STATUS
approved