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Old 2017-01-26, 01:29   #4
Batalov
 
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"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

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P.S. (For "and" case) Interestingly, n cannot be a multiple of 21 (or of 2, or of 9, 25... or more generally, must be square-free. This is easily proven.)

More generally, if n = p*q*... and p | phi(q), then n is not in the sequence.

The case of n = 2*a is a bit trickier. Either 3 | n and then 2 | phi(3) and n is not in the sequence or 3 doesn't divide n, but then 3 divides either n-phi(n) or either n+phi(n).

Last fiddled with by Batalov on 2017-01-26 at 01:48
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