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Floretion with FAMP code 1teszapsumseq[C*B] with C = - .5'j + .5'k - .5j' + .5k' - 'ii' - .5'ij' - .5'ik' - .5'ji' - .5'ki' and B = - .5'i + .5'j - .5i' + .5j' - 'kk' - .5'ik' - .5'jk' - .5'ki' - .5'kj'.
2

%I #26 May 05 2025 03:01:23

%S 1,0,0,1,-1,2,0,1,0,0,1,-1,2,-2,1,-2,0,-1,0,0,-1,1,-2,2,-3,4,-2,3,-2,

%T 2,-1,2,0,1,0,0,1,-1,2,-2,3,-3,4,-4,5,-4,4,-3,4,-2,3,-2,2,-1,2,0,1,0,

%U 0,1,-1,2,-2,3,-3,4,-4,5,-5,6,-6,5,-6,4,-5,4,-4,3,-4,2,-3,2,-2

%N Floretion with FAMP code 1teszapsumseq[C*B] with C = - .5'j + .5'k - .5j' + .5k' - 'ii' - .5'ij' - .5'ik' - .5'ji' - .5'ki' and B = - .5'i + .5'j - .5i' + .5j' - 'kk' - .5'ik' - .5'jk' - .5'ki' - .5'kj'.

%C The following parity property of this sequence seems interesting: Assume 0 is both positive and negative and let sgn(a(n)) be the sign of a(n). Then apparently sgn(a(n)) = -sgn(a(n+1)) = sgn(a(n+2)) for all n. But sgn(a(2n)) != sgn(a(2m)) for all n,m.

%H Creighton Dement, <a href="/A117154/b117154.txt">First 821 terms</a> [after this it alternates 39, -39, 39, -39 forever]

%H Creighton Dement, <a href="/A126626/a126626.txt">Notes on A126626 and A117154</a>

%H Creighton Dement, <a href="https://www.mrob.com/pub/seq/CKD-2010.pdf">Floretions 2009, DRAFT</a>. See section 4.1 Algorithms.

%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (-1).

%K sign,easy,hear

%O 1,6

%A _Creighton Dement_, Apr 21 2006