(phi(m)+1) divides m

[Franklin T. Adams-Watters]

This isn't a complete answer, but:

The sequence includes p and 2p for p an odd prime.
Other than 2, no other prime powers are included.
If n is the product of 2 distinct odd primes, phi(n) > n/2, so n is not
in the set.

Question: are there any members other than p and 2p?

Leroy Quet <qq-quet at mindspring.com> wrote:

>Let phi(m) 
>be the number of positive integers <= m and relatively prime with m.
>
>I get, by hand, that the sequence of m's where
>
>(phi(m)+1)|m
>
>begins
>
>2, 3, 5, 6, 7, 10, 11, 13, 14, 17, 19,..
>
>(Not in the EIS, apparently.)
>
>What can be said about this sequence, such as its asymptotics, the best 
>way to determine if an integer is in it, etc ?
>

-- 
Franklin T. Adams-Watters
16 W. Michigan Ave.
Palatine, IL 60067
847-776-7645


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