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

[Leroy Quet]

I wrote (in part):

>....
>Is there an easy and direct (unrecursive) way to determine the sign 
>before each 1?
>
>...
>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=b4be2fdf.03
03281748.762fa20d%40posting.google.com&rnum=3&prev=
>
>http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&threadm=b4be2fdf.03
03251950.25d8d0e5%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
>
>....
>Thanks,
>Leroy Quet



I believe (but am not absolutely certain) that a(m-1) = -(-1)^c(m),

where c(m) = 

the number of e()'s where:

e(e(e(...e(m)...))) = 0,

and e(m) is a nonnegative integer such that:
2^e(m) is the highest power of 2 dividing m.

I discuss this c-sequence in the first link I give above.

(Is there anything related to these sequences yet in the EIS?)

thanks,
Leroy
 Quet