%I
%S 7,6,15,15,38,39,99,102,259,267,678,699,1775,1830,4647,4791,12166,
%T 12543,31851,32838,83387,85971,218310,225075,571543,589254,1496319,
%U 1542687,3917414,4038807,10255923,10573734,26850355,27682395,70295142,72473451
%N G.f. (7+6*x6*x^23*x^3)/((x^2+x1)*(x^2x1)).
%C One of two sequences involving the Lucas/Fibonacci numbers.
%C This sequence consists of pairs of numbers more or less close to each other with "jumps" in between pairs. "pos((Ex)^n)" sums up over all floretion basis vectors with positive coefficients for each n. The following relations appear to hold: a(2n)  (a(2n1) + a(2n2)) = 2*Luc(2n) a(2n+1)  a(2n) = Fib(2n), apart from initial term a(2n+1)/a(2n1) > 2 + golden ratio phi a(2n)/a(2n2) > 2 + golden ratio phi An identity: (1/2)a(n)  (1/2)A099256(n) = ((1)^n)A000032(n)
%H Harvey P. Dale, <a href="/A099255/b099255.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,3,0,1).
%F a(n) = 2*pos((Ex)^n)
%F a(0) = 7, a(1) = 6, a(2) = a(3) = 15, a(n+4) = 3a(n+2)  a(n).
%F a(2n) = A022097(2n+1), a(2n+1) = A022086(2n+3).
%F a(n)=A061084(n+1)+A013655(n+2). [From _R. J. Mathar_, Nov 30 2008]
%t LinearRecurrence[{0,3,0,1},{7,6,15,15},40] (* _Harvey P. Dale_, Dec 29 2012 *)
%o Floretion Algebra Multiplication Program, FAMP
%Y Cf. A099256, A000032.
%K nonn,easy
%O 0,1
%A _Creighton Dement_, Oct 09 2004
%E More terms from _Creighton Dement_, Apr 19 2005
