Divisors of n within binary n

[Leroy Quet]

(Reinhard Zumkeller: the referenced sequences, A078822 and A093640, are 
your sequences.)


I just submitted the following:

>%S A000001 
>0,1,0,1,2,1,3,0,1,2,4,0,1,2,5,0,1,2,3,6,1,3,7,0,1,2,4,8,0,1,2,4,9,0,1,2,5,
>10,0,1,2,3,5,11,0,1,2,3,4,6,12,0,1,2,3,5,6,13,0,1,2,3,6,7,14,1,3,7,15,0,1,
>2,4,8,16,0,1,2,4,8,17
>%N A000001 Table where nth row (of A078822(n) terms) contains the distinct 
>nonnegative integers which, when written in binary, are substrings of n 
>written in binary.
>%e A000001 12 in binary is 1100. Within this binary representation there 
>is 0 (occurring twice), 1 (occurring twice), 10 (= 2 in decimal), 11 (= 3 
>in decimal), 100 (= 4 in decimal), 110 (= 6 in decimal), and 1100 (= 12 in 
>decimal).
>So row 12 = (0,1,2,3,4,6,12).
>%Y A000001 A078822
>%O A000001 0
>%K A000001 ,easy,nonn,

(Hopefully I did not make an error with the submitted numbers.)

a(n) of sequence A093640 is the number of such integers, k, in row n 
(above) such that k divides n.
Now I wonder about the integers n where, except for the 0, the integers 
in row n above are precisely those positive integers which divide n. (ie 
A093640(n) = A000005(n).)

I have this sequence beginning:  1,2,3,4,6,8,10,12,16,20,...
I know this sequence is infinite, because every power of 2 is in it.
But I wonder, without thinking about this much myself (so this may be a 
trivial question) if there are an infinite number of terms which are not 
powers of 2.

thanks,
Leroy Quet