OFFSET
0,3
COMMENTS
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.
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
First bisection is A260259 (see previous comment for the second bisection). [Bruno Berselli, Nov 02 2015]
LINKS
Daniel C. Fielder, Special integer sequences controlled by three parameters, Fibonacci Quarterly 6, 1968, 64-70.
Robert Munafo, Sequences Related to Floretions.
Index entries for linear recurrences with constant coefficients, signature (1, 0, 1, 1).
FORMULA
G.f.: (1-2*x-x^2)/((x^2+x-1)*(1+x^2)).
CROSSREFS
KEYWORD
easy,sign
AUTHOR
Creighton Dement, Aug 07 2005
STATUS
approved