A173343 a(n+4) = a(n+3) - 2*a(n+2) - a(n+1) - a(n).

1, 2, 0, -5, -8, 0, 21, 34, 0, -89, -144, 0, 377, 610, 0, -1597, -2584, 0, 6765, 10946, 0, -28657, -46368, 0, 121393, 196418, 0, -514229, -832040, 0, 2178309, 3524578, 0, -9227465, -14930352, 0, 39088169, 63245986, 0, -165580141, -267914296, 0

Offset: 0

Author

Creighton Dement, Feb 16 2010

Keywords

• easy
• sign

Comments

Sequence appears to give signed Fibonacci numbers, where those Fibonacci numbers "missing" are in A173344. A117647 gives a nonnegative version without zeros. (a(n)) = kjbseq(X) with X = -0.25'i + 0.5'j + 0.5'k + 0.25'i + j' + 0.5k' - 0.25ii - 0.25'jj' - 0.25'kk' + 0.5'ij' + 0.5'ik' - 0.5'ji' -0.25'jk' + 0.25'kj' + 0.25'ee' (see Munafo link for definitions)

Links

• C. Dement, Online Floretion Multiplier [broken link]
• R. J. Mathar, Structure of the Floretion Group
• R. Munafo, Sequences Related to Floretions
• Index entries for linear recurrences with constant coefficients, signature (1,-2,-1,-1).

Crossrefs

Cf. A173344, A117647.

Programs

Mathematica

LinearRecurrence[{1,-2,-1,-1},{1,2,0,-5},50] (* Harvey P. Dale, Jul 17 2018 *)

Formula

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

a(n) = b(n)+b(n-1) where b(3n) = b(3n+1) = -b(3n+2) = (-1)^n*A001076(n+1). [From R. J. Mathar, Apr 01 2010]