'Super-Palindromic' Sequence Of +-1's

[Leroy Quet]

I posted the below to sci.math. But in regards to that group, I was more 
interested in the generating function for the sequence.

Here, however, I am more interested in the sequence itself.

Is there an easy and direct (unrecursive) way to determine the sign 
before each 1?

And, this brings up a good point regarding the EIS too.
Is there a way to do a search where ONLY sequences with the properly 
signed integers appear? 
Sequences like the one below get a lot of matches, but (in my opinion) 
needlessly.
(Did not we recently discuss this topic on seq.fan?)

thanks,
Leroy Quet


the post:
----

This is a little like what I discuss at these threads:
http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&threadm=b4be2fd

f.0303281748.762fa20d%40posting.google.com&rnum=3&prev=

http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&threadm=b4be2fd

f.0303251950.25d8d0e5%40posting.google.com&rnum=7&prev=



Let A(0) = {1} = {a(0)};

Let A(n+1) = {A(n),-a(n),A(n)},

for all n >= 0,

where a(n) is the n_th term in each sequence A where a(n) is defined.

(the A's are each a sequence formed by concatenation.)

So, we have:

A(4) =

1, -1, 1, 1, 1, -1, 1, -1, 1, -1, 1, 1, 1, -1, 1, -1, 1, -1, 1, 1, 1, -1, 
1, -1, 1, -1, 1, 1, 1, -1, 1

(For each m, A(m) has 2^(m+1)-1 terms, a(0) to a(2^(m+1)-2).)

So, if we let 

F_m(y) = sum{k=0 to 2^(m+1)-2} a(k) y^k

(and F(y) = limit{m-> oo} F_m(y)),

then:

F_m(y) = 

F_{m-1}(y) (1 + y^(2^m)) - a(m-1) y^(2^m -1),


and so

F(y) - 1 =

sum{k=1 to oo} (F_{k-1}(y) y  - a(k-1)) y^(2^k -1),


and therefore

F_{2^m-1}(1) + F{m-1}(1) - 1 =

sum{k=0 to 2^m-2) F_k(1).

(all of above right?)

So, what exactly is the closed-form for F(y)??

Thanks,
Leroy Quet