A092550 Expansion of -x*(1+x+x^2+x^4)/(-1+2*x^3+x^6).

1, 1, 1, 2, 3, 2, 5, 7, 5, 12, 17, 12, 29, 41, 29, 70, 99, 70, 169, 239, 169, 408, 577, 408, 985, 1393, 985, 2378, 3363, 2378, 5741, 8119, 5741, 13860, 19601, 13860, 33461, 47321, 33461, 80782, 114243, 80782, 195025, 275807, 195025, 470832, 665857

Offset: 1

Author

Roger L. Bagula, Apr 08 2004

Keywords

• nonn

Comments

If prefaced with a 1: denominators in convergents to barover:[1, 0, 1] as follows:

1,....0,....1,....1,....0,....1,....1,....0,....1,....

1/1,..0/1,..1/2,..1/3...1/2...2/5...3/7...2/5...5/12,...;

Gary W. Adamson, Mar 25 2014

For k(n), a term in A249576, k(n+6) mod (k(n+5)) = a(n). - Russell Walsmith, Nov 27 2014

Links

• Marcia Edson, Scott Lewis and Omer Yayenie, The k-periodic Fibonacci sequence and an extended Binet's formula, INTEGERS 11 (2011) #A32.
• D. Panario, M. Sahin, Q. Wang, A family of Fibonacci-like conditional sequences, INTEGERS, Vol. 13, 2013, #A78.
• Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,1).

Crossrefs

Cf. A000045.

Programs

Mathematica

m=3 fib[n_Integer?Positive] :=fib[n] =If[Mod[n, m]==0, fib[n-2], fib[n-1]+fib[n-2]] fib[0]=fib[1] = fib[2] = 1 digits=50 a=Table[fib[n], {n, 1, digits}]
LinearRecurrence[{0,0,2,0,0,1},{1,1,1,2,3,2},50] (* Harvey P. Dale, Jan 13 2015 *)

Formula

a(n) = a(n-2) if 3|n, otherwise a(n)= a(n-1)+a(n-2).

From R. J. Mathar, Dec 08 2010: (Start)

a(n)= +2*a(n-3) +a(n-6).

G.f.: -x*(1+x+x^2+x^4)/(-1+2*x^3+x^6).

a(3n+1) = A000129(n+1). a(3n)=A000129(n). a(3n+2)= A078057(n). (End)

Extensions

Edited, and new name, Joerg Arndt, Sep 17 2013