Lucas and A001333 / Lucas Strings

[Creighton Dement]

Dear Seqfans, 

As mentioned in a previous posting, I find comparisons between A001333
and Lucas numbers worth looking into. For prime indices p, they appear
to be "particularly easy" to compare (for non-primes there are similar
formula but starting with the prime case seemed easiest):

A = A001333: Numerators of continued fraction convergents to sqrt(2). 

A(1) = 2^(1-1) 
A(2) = 2^(2-1) + 2*(1/2) = 3
A(3) = 2^(3-1) + 3*2*(1/2) = 7
A(5) = 2^(5-1) + 5*(2*(1/2) + 2^2) = 41
A(7) = 2^(7-1) + 7*(2*(1/2) + 2*2^2 + 2^4) = 239
A(11) = 2^(11-1) + 11*(2*(1/2) + 5*2^2 + 7*2^4 + 4*2^6 + 2^8) = 8119

Next, replace 2's, above, with x as such:

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


Then x = 2 ->  A(p) = A001333(p) and x = 1 -> A(p) = Lucas(p)

The last statement is a reformulation of 
Lucas(p) = 1 + p*A006206(p)
A006206: Number of aperiodic binary necklaces with no subsequence 00,
excluding the necklace "0"

Specifically, it can be shown that for prime p, A001333 has the form
2tes[(a + b)^p] = 2^{p-1} + p*Q(p)$ where 
1. $Q(2) = 1/2$ is not an integer / it is an odd number for all odd
prime p (my paper needs to be updated, but some of this is given under
http://www.crowdog.de/Flointseq.pdf ) 
2. It can be written as the sum over A006206(p) terms of multiples of
powers of 2.
3. It appears to be the sequence A106514(p) (Convolution of A000079
-powers of 2- and A001333).

My Question: Is there a formula to determine "all the cofficients of the
powers of 2" such that we may give an explicit formulate using A006206-
not only for Lucas numbers, but also A001333? 
 
*****

Concerning the previously discussed "Lucas string" sequence, 4 more
terms were calculated. For some reason, I can't get my computer to
calculate a fifth (waited over 5 hours and eventually turned the machine
off).

 (a(n)) = (0, 1, 1, 3, 3, 8, 8, 17, 23, 41, 55, 102, 144, 247, 387, 631,
987, 1636, 2584, 4233, 6787, 11011 )
 http://www.research.att.com/~njas/nsequences/A113166
 
 The conjecture: a(p) = Fib(p-1) for prime p is now confirmed up to p <=
19.   
 http://www.research.att.com/~njas/nsequences/A113166

Also, Fib - (a(n)) = (0, 0, 0, 1, 0, 3, 0, 4, 2, 7, 0, 13, 0, 14, 10,
21, 0, 39, 0, 52, 22, 65, ) 

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.