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A050474
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Solutions to 2*phi(x) = x+1.
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1
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OFFSET
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1,2
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COMMENTS
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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 independent terms and next term of this sequence if it exists is the third such term. - Farideh Firoozbakht, May 01 2007
The next term, if it exists, has at least 7 distinct prime factors (see Beiler, p. 92). - Jud McCranie, Dec 13 2012
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REFERENCES
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A. H. Beiler, Recreations in the Theory of Numbers, page 92.
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LINKS
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Table of n, a(n) for n=1..8.
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FORMULA
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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, Jan 26 2006
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EXAMPLE
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2*phi(15)=2*8=15+1, so 15 is a member of the sequence.
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CROSSREFS
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Cf. A000010.
Cf. A129613, A129614, A129615, A203966.
Sequence in context: A173146 A139289 A116518 * A051179 A122591 A120607
Adjacent sequences: A050471 A050472 A050473 * A050475 A050476 A050477
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KEYWORD
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hard,nonn
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AUTHOR
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Jud McCranie, Dec 24 1999
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STATUS
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approved
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