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A050474 Solutions to 2*phi(x)=x+1. +0
1
1, 3, 15, 255, 65535, 83623935, 4294967295, 6992962672132095 (list; graph; listen)
OFFSET

1,2

COMMENT

The next term, if it exists, has at least 7 distinct prime factors.

If n is in the sequence and n+2 is prime then m=n*(n+2) is in the sequence because 2*phi(m)=2*phi(n*(n+2))=2*phi(n)*(n+1)=(n+1)^2= m+1. We can obtain the terms 3, 15, 255, 65535 & 4294967295 from 1(the first term) in this way. Also since 83623935 is a term and 83623935+2 is prime 83623935*(83623935+2)=6992962672132095 is in the sequence. So 1 and 83623935 are the only known indepedent terms and next term of this sequence if it exists is the third such term. - Farideh Firoozbakht (mymontain(AT)yahoo.com), May 01 2007

REFERENCES

A. H. Beiler, Recreations in the Theory of Numbers, page 92.

FORMULA

n is in the sequence iff phi(n^2)=1+2+3+...+n because n is in the sequence <=> 2*phi(n)=n+1 <=> n*phi(n)=n*(n+1)/2 <=> phi(n^2)=1+2+3++...+n. For n=1,2,...,5 a(n)=2^2^(n-1)-1. - Farideh Firoozbakht (mymontain(AT)yahoo.com), Jan 26 2006

EXAMPLE

2*phi(15)=2*8=15+1, so 15 is a member of the sequence.

CROSSREFS

Cf. A000010.

Cf. A129613, A129614, A129615.

Sequence in context: A114735 A139289 A116518 this_sequence A051179 A122591 A120607

Adjacent sequences: A050471 A050472 A050473 this_sequence A050475 A050476 A050477

KEYWORD

hard,nonn

AUTHOR

Jud McCranie (j.mccranie(AT)comcast.net), Dec 24 1999

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Last modified December 15 00:47 EST 2009. Contains 170825 sequences.


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