# Divisible by a square

CRGreathouse
Homework Helper
The sequence includes all odd multiples of {11, 21, 39, 136, 171, 202, 243, 253, 292, 406, 548, 1081, 1711, 1751, 1830, 1958, 2667, 3165, 3197, 3615, 3934, 4656, 5648, 5886, 6123, 6275, 6328, 7184, 8515, 9653, 10256, 11026, 11167, 13546, 13861, 15931, 16290, 18205, 18528, 20242, 21389, 22544, 24753, 26106, 27028, 27148, 29470, 32415, 32896, 34453, 34689, 34732, 35410, 35651, 36046, 40100, 41718, 45023, 45455, ...}.

For odd n,
$$11^2 | 10^{11n}$$
$$7^2 | 10^{21n}$$
$$13^2 | 10^{39n}$$
$$17^2 | 10^{136n}$$
$$19^2 | 10^{171n}$$
$$101^2 | 10^{202n}$$
and so on.

Is there a number in A086982 such that:

- It is not an odd multiple of a previous member, and
- Any of its odd multiples is not in A086982?

(Actually only the second condition is relevant, the first is rather cosmetic. For example, all the odd multiples of 33 are in the sequence, since all the odd multiples of 11 are.)

If there is none, it would appear that the list in post #26 (the non-zero, sorted values of A086981, I presume you mean) is more fundamental than, and in some sense "generates", A086982.

CRGreathouse
Homework Helper
Is A086982 precisely the odd multiples of the nonzero values of A086981? Yes.

But careful. I don't know that my list is the start of the sorted list of the nonzero members of A086981 (though I think it is). There could be a prime square dividing 10^k+1 for a fairly small k. I've only searched for divisibility by primes up to 40,000 or so (thus prime squares up to 1.6 billion) so technically I only know that 1 to 9 are not on the list.

Hmm. 10^k+1 can't be a prime square by Mihăilescu's theorem (otherwise 10^k and 10^k+1 would be consecutive powers). It can't be 2, 3, or 5 times a prime square since it's 1 mod 2, 2 mod 3, and 1 mod 5. So at worst it is 7 times a prime square.

Actually, since primes (greater than 5) are 1, 3, 7, or 9 mod 10, prime squares are 1 or 9 mod 10. Thus 7 times a prime square would be 3 or 7 mod 10, which is right out. 11 times a prime square, though... still a possibility, for all I know. So I haven't missed anything up to 17.6 billion, which is more than 10^10, so I know that 11 is the first member of the list.

Of course for the small entries we could just factor 10^k+1 directly. It might be practical to go to k = 200 or so, proving the first five members of the sequence. (I think I did this already.) But proving the first ten seems far out of reach with current SNFS technology. msieve has factored 260-digit numbers of special form, but 405-digit numbers are about 12,000 times harder.

Last edited: