Divisors of n within binary n

[franktaw at netscape.net]

You can get a more than that trivially.  If n is in this sequence, so 
is 2n.  So the real question is, are there infinitely many members of 
the sequence which are not twice a previous element?  So far, you've 
got 1, 3, and 10.

In fact, any number of the form 2^(2n-1) + 2^(n-1) is in the sequence.  
This sequence is A007582:

1,3,10,36,136,528,...

Suppose that n has 3 or more 1's in its binary representation.  Take 
the substring from the first through the next to last 1 (or the second 
through the last); call this m.  Then n can be represented as 2^u*m + 
2^v, and m is an odd number greater than 1.  It follows that m does not 
divide n.

So any member of Leroy's sequence can have at most 2 1's in its binary 
representation.  For numbers with 2 1's, there must clearly be at least 
as many zeros at the end as between the two 1's, so the solutions are 
precisely the numbers A007582(k)*2^u.  The sequence can alternatively 
be described as all numbers of the form 2^a + 2^b, where 0 <= b <= a <= 
2b+1.

Franklin T. Adams-Watters


-----Original Message-----
From: Leroy Quet <qq-quet at mindspring.com>

>%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.

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.

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