(phi(m)+1) divides m

[FRANCISCO SALINAS]

Franklin wrote

>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?

Cohen and Segal (1) showed that in case there were other solutions to this 
problem (which appeared to be posed by Schinzel), then they should have at 
least 15 distinct prime factors. Moreover, there is a connection with the 
Lehmer's totient problem (2) which asks whether
there is a composite n such that phi(n)|(n-1). If no such composite exist 
then p and 2p are the only members for Leroy's sequence.

(1) G. L Cohen; S. L Segal, A note concerning those n for which phi(n)+1 
divides n. Fibonacci Quarterly 27 (1989), no. 3, pp. 285-286.
(2) http://mathworld.wolfram.com/LehmersTotientProblem.html





>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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