Perhaps, this is interesting (?)

[creigh at o2online.de]

The following result should be of interest, however, I am not an expert
so this may have been pointed out before:

3^2     = 1^2 + 2^2 + 2^2
9^2     = 1^2 + 4^2 + 8^2
33^2   = 1^2 + 8^2 + 32^2  
129^2 = 1^2 + 16^2 + 128^2
...
->
[A087289(n)]^2  = 1^2 + [2^(n+1)]^2 + [2^(2n+1)]^2  
-> 
[A087289(n)]^2  = 1^2 + 2^(2n+2) + 2^(4n+2)  

This result comes at the same time I realized a major point had been
overlooked in the "Floretion" paper at http://www.crowdog.de .  I mentioned 
that x +1 = 'i - 'k - 'jj' -'ji' 'jk' + 1 and x -1, etc. 
form "division groups" (chapter "Chung-shu Deciphers Floret's star"). 
However, because of a mistaken glance at FAMP's result on the 
computer screen 4 months ago, I'd thought x^2 = 0 
until today when I finally realized that  x^2 = 1. In a way, this changes
much and almost immediately led to the above result. 

Sincerely,
Creighton