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A092550 Expansion of -x*(1+x+x^2+x^4)/(-1+2*x^3+x^6). 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,4
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.
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)
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 *)
CROSSREFS
Cf. A000045.
Sequence in context: A347000 A354271 A344482 * A058977 A337246 A347104
KEYWORD
nonn
AUTHOR
Roger L. Bagula, Apr 08 2004
EXTENSIONS
Edited, and new name, Joerg Arndt, Sep 17 2013
STATUS
approved

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Last modified April 26 21:53 EDT 2024. Contains 372004 sequences. (Running on oeis4.)