New method and two new sequences relating to Fib

[creigh at o2online.de]

I would like to introduce two new sequences resulting from 
"dynamic symmetries". Both sequences are extracted from  
a single floretion ( "Ex", below).  For the sake of completeness, 
I also give all sequences arising from static symmetries
for this floretion. 

Take:
E = ( 'i + i' + 'ii' + 'jj' + 'kk' + 'jk' + kj' + 1 )/4 
x = 'i + i' + 'ji' + 'ki' + 1 

All formulas below are of the form:
(a(n)) = sym((Ex)^n) = symseq(Ex) = (a(0), a(1), ...)
where "sym" is some choice of basis vectors whose coefficients
are to be summed up. As the choice is usually based on some symmetric 
property, I write "sym".

Static Symmetries:
("static" meaning loosely that the choice of basis vectors
is independant of n as well as of the floretion itself) 

(ves((Ex)^n)) = vesseq(Ex) = ((-1)^n)*Luc   

4*tesseq(Ex) = (-1, 3, -4, 7, -11, 18, -19) = ((-1)^(n+1))*(Fib at 1 3)

(4/9)*lesseq(Ex) = (1, -1, 2, -3, 5, -8, 11, -13) = ((-1)^n)*(Fib at 1 1)

2*jesseq(Ex) = (0, 1, -1, 2, -3, 5, -8, 11) = ((-1)^(n+1))*Fib 

emseq(Ex) = vesseq(Ex) = ((-1)^n)*Luc   

(4/7)*chuseq(Ex) = (1, -1, 2, -3, 5, -8, 11, -13) = ((-1)^n)*(Fib at 1 1)

2*chutesseq(Ex) = (3, -2, 5, -7, 12, -19)  = ((-1)^n)*(Fib at 3 2)

4*chu'(Ex) = (1, 3, -2, 5, 7, -12, 19)  // similar to chutesseq

2*achu(Ex) = (1, 0, 1, -1, 2, -3, 5, -8, 11, -13) // similar to Fib 

ves identities:
jes + les + tes = chu + chu' = chu + achu + tes = ves

Dynamic Symmetries:

"pos((Ex)^n)" sums up over all basis vectors with positive coefficients for 
each n. 

2*posseq(Ex) = (7, 6, 15, 15, 38, 39, 99, 102, 259, 267, 678, 699, 1775, 1830)
comments: notice that the sequence consists of pairs of numbers more or 
less close to each other with "jumps" inbetween pairs. 

Adding up each pair and subtracting it from the next highest number, i.e. 
a(n+2) - (a(n+1) + a(n)), gives (third term is 22 since 22 = 99 - (38 + 
39) )  (2, 8, 22, 58, 152, ) = 2*Luc(2n)

Subracting each pair gives:
(1, 0, 1, 3, 8, 21, 55, ) // = Fib(2n) apart from initial term

2*negseq(Ex) = (3, 8, 9, 23, 24, 61, 63, 160, 165, 419, 432, 1097, 1131)

Subtracting each pair (starting at 8) gives:
(1, 1, 2, 5, 13, 34, ) // =  F(2n+1), apart from initial term

ves identity:
pos - neg = ves


Sincerely, 
Creighton