%I #26 Mar 18 2024 13:05:01

%S -1,1,2,1,1,4,7,9,14,25,41,64,103,169,274,441,713,1156,1871,3025,4894,

%T 7921,12817,20736,33551,54289,87842,142129,229969,372100,602071,

%U 974169,1576238,2550409,4126649,6677056,10803703,17480761,28284466,45765225

%N a(n) = a(n-1) + a(n-3) + a(n-4) for n > 3, a(0) = -1, a(1) = 1, a(2) = 2, a(3) = 1.

%C In reference to the program code given, 4*tesseq[A*H] = A001638 (a Fielder sequence) where A001638(2n) = L(n)^2. Here we have: a(2n+1) = A007598(n+1) = Fibonacci(n+1)^2.

%C Floretion Algebra Multiplication Program, FAMP Code: 4kbaseiseq[B+H] with B = - .25'i + .25'j - .25i' + .25j' + k' - .5'kk' - .25'ik' - .25'jk' - .25'ki' - .25'kj' - .5e and H = + .75'ii' + .75'jj' + .75'kk' + .75e

%C First bisection is A260259 (see previous comment for the second bisection). [_Bruno Berselli_, Nov 02 2015]

%H Daniel C. Fielder, <a href="http://www.fq.math.ca/Scanned/6-3/fielder.pdf">Special integer sequences controlled by three parameters</a>, Fibonacci Quarterly 6, 1968, 64-70.

%H Robert 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, 0, 1, 1).

%F G.f.: (1-2*x-x^2)/((x^2+x-1)*(1+x^2)).

%F a(n) = 2*A056594(n+3)/5 - 6*A056594(n)/5 + A000032(n+1)/5. [_R. J. Mathar_, Nov 12 2009]

%Y Cf. A000045, A001638, A007598, A111570, A111571, A111572, A111573, A260259.

%K easy,sign

%O 0,3

%A _Creighton Dement_, Aug 07 2005