Fibonacci Polynomials

[Creighton Dement]

Dear Seqfans, 

This is something I noticed while working on my "Lucas Strings" file
which seems quite interesting but is probably way out of the range of an
amature's grasp...

The Fibonacci polynomials start out:

    F_1(y)=1 
    F_2(y)=y 
    F_3(y)=y^2+1 
    F_4(y)=y^3+2y 
    F_5(y)=y^4+3y^2+1 
    F_6(y)=y^5+4y^3+3y 

A comment to A000129 by T. D. Noe states that a(n)=F(n,2), the nth
Fibonacci polynomial evaluated at y=2. 

As mentioned before, for prime p, I get A = A001333 with y = 2 and A =
Lucas with y = 1 where A is defined as  

A(1) = 1
A(2) = 3
A(3) = y^(3-1) + 3*1
A(5) = y^(5-1) + 5*(1 + y^2)
A(7) = y^(7-1) + 7*(1 + 2*y^2 + y^4)
A(11) = y^(11-1) + 11*(1 + 5*y^2 + 7*y^4 + 4*y^6 + y^8)

(See http://www.crowdog.de/LucSt2.html )

By the circumstances under which the polynomials originally occured, it
seemed  more or less apparent to me that the polynomials appearing on
the right side of the above equations are somehow of "fundamental
nature".  

We know that A000129(n) + A000129(n+1) = A001333(n+1). 

By now at the latest, it would seem both sets of polynomials are
related. 

Ex. for n = 5 (a prime number), we have  

41 = A001333(5) = A000129(4) + A000129(5) = (2^3+2*2) + (2^4+3*2^2+1)
[[Fibonacci polynomial formula]] =  2^(5-1) + 5*(1 + 2^2) [[Necklace
formula]]

Or stated a bit differently...  the rational polynomial function 
f_5(y) = (y^4+3y^2+1 + y^3+2y - y^(5-1))/(1 + y^2) = (3y^2+1 +
y^3+2y)/(1 + y^2) is a prime number for f_5(2). 

Do others see similar relations which might be of interest?

Sincerely, 
Creighton 


-It's a shame when the girl of your dreams would still rather be with
someone else when you're actually in a dream.