example of using famp to arrive at a result of Generalized Gaussian Fibonacci integers

[creigh at o2online.de]

Dear Seqfans, 

This page was originally written an example of using the identities
listed at http://www.crowdog.de/20801/20828.html - a page which is
still under construction. [As far as the transformations mentioned in my 
last posting are concerned- I'm getting fairly excited about them and will 
try to discuss them in more detail soon- perhaps using "nicer" examples 
than last time] 

Example:  Using Famp to arrive at 
(b(n+1))^2 + 3(b(n))^2 - 2*b(n+1)*b(n) = 3^n
b(n) = http://www.research.att.com/projects/OEIS?Anum=A088137

The floretion + 'i + 'j + i' + j' + 'ii' + 'jj' + 'ij' + 'ji' + e 
generates the following set of sequences (among many others). 

1mixtesseq: 3, 9, 27, 81, 243, 729, 2187, 6561, 19683, 59049,
1dia[I]tesseq: 4, 1, 16, 121, 100, 169, 3136, 5329, 484, 69169,
1dia[J]tesseq: 4, 1, 16, 121, 100, 169, 3136, 5329, 484, 69169, 
1dia[K]tesseq: 1, 1, 25, 49, 1, 529, 1849, 289, 9025, 58081, 38809, 
1vesseq: 9, 9, 9, 225, 441, 9, 4761, 16641, 2601, 81225, 522729, 349281, 
974169, 
1tesseq: 1, 1, 25, 49, 1, 529, 1849, 289, 9025, 58081, 38809, 108241, 1560001

Using the identities 
dia[I] + dia[J] + dia[K] + mix + tes  = ves
and symtes = sym +  tes

we have

dia[I]tes + dia[J]tes + dia[K]tes + mix + tes  = ves + 3tes
-> mix + tes = mixtes = ves + 3tes - (dia[I]tes + dia[J]tes + dia[K]tes)

Thus, 
[3, 9, 27, 81, 243, 729, 2187]  = 
[ 9, 9, 9, 225, 441, 9, 4761] + 3* [1, 1, 25, 49, 1, 529, 1849,]  - 
(2*[4, 1, 16, 121, 100, 169, 3136,] + [1, 1, 25, 49, 1, 529, 1849,])   

upon "sym"plifying...

[3, 9, 27, 81, 243, 729, 2187]  = [ 9, 9, 9, 225, 441, 9, 4761] + 
2*[1, 1, 25, 49, 1, 529, 1849,] - 2*[4, 1, 16, 121, 100, 169, 3136,]
= 9*[1, 1, 1,25, 49, 1, 529, 1849,]  + 2*[1, 1, 25, 49, 1, 529, 1849,]  
- 2*[4, 1, 16, 121, 100, 169, 3136,]  

Finally 
 9*(A087455(n))^2 + 2*(A087455(n+1))^2 - 2*(A088137(n+2))^2 = 3^(n+1)  

At this point we could also make use of another formula 
A087455(n+1) = A088137(n+2) - A088137(n+1)
(if I remember correctly, this was originally derived a few months ago 
with the help of a second floretion).  
then, setting A087455 = a, A088137 = b :
 9*(a(n))^2 + 2*(a(n+1))^2 - 2*(b(n+2))^2 = 3^(n+1)  and 
(a(n+1))^2 = (b(n+2) - b(n+1))^2 = (b(n+2))^2 + (b(n+1))^2 - 2*b(n+2)*b(n+1)

->

9*[ (b(n+1))^2 + (b(n))^2 - 2*b(n+1)*b(n) ] + 
2*[ (b(n+2))^2 + (b(n+1))^2 - 2*b(n+2)*b(n+1) ] -
2*(b(n+2))^2 = 3^(n+1), n > 0
->
11*(b(n+1))^2 + 9(b(n))^2 - 18*b(n+1)*b(n)) - 4*b(n+2)*b(n+1)) = 3^(n+1), n 
> 0
-> (using the recursion formula given)
11*(b(n+1))^2 + 9(b(n))^2 - 18*b(n+1)*b(n)) - 4(2*b(n+1) - 3*b(n))*b(n+1) 
= 3^(n+1), n > 0
->
3*(b(n+1))^2 + 9(b(n))^2 - 6*b(n+1)*b(n) = 3^(n+1),  n > 0
->
(b(n+1))^2 + 3(b(n))^2 - 2*b(n+1)*b(n) = 3^n,  n > 0
 
Note that we also have Paul Barry's neat formula 
b(n)=((1+isqrt(2))^n-(1-isqrt(2))^n)/(2isqrt(2)) to go along with 
the above relation. 

Sincerely, 
Creighton