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%I #25 Apr 27 2024 17:21:09
%S 1,2,0,-5,-8,0,21,34,0,-89,-144,0,377,610,0,-1597,-2584,0,6765,10946,
%T 0,-28657,-46368,0,121393,196418,0,-514229,-832040,0,2178309,3524578,
%U 0,-9227465,-14930352,0,39088169,63245986,0,-165580141,-267914296,0
%N a(n+4) = a(n+3) - 2*a(n+2) - a(n+1) - a(n)
%C 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)
%H C. Dement, <a href="http://fumba.eu/sitelayout/Floretion.html">Online Floretion Multiplier</a> [broken link]
%H R. J. Mathar, <a href="/A173343/a173343.pdf">Structure of the Floretion Group</a>
%H R. Munafo, <a href="http://www.mrob.com/pub/math/seq-floretion.html">Sequences Related to Floretions</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,-2,-1,-1).
%F G.f.: (x+1)/(x^4+x^3+2*x^2-x+1).
%F 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]
%t LinearRecurrence[{1,-2,-1,-1},{1,2,0,-5},50] (* _Harvey P. Dale_, Jul 17 2018 *)
%Y Cf. A173344, A117647.
%K easy,sign
%O 0,2
%A _Creighton Dement_, Feb 16 2010