|
| |
|
|
A050474
|
|
Solutions to 2*phi(x)=x+1.
|
|
1
| | |
|
|
|
OFFSET
| 1,2
|
|
|
COMMENTS
| 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: A173146 A139289 A116518 * A051179 A122591 A120607
Adjacent sequences: A050471 A050472 A050473 * A050475 A050476 A050477
|
|
|
KEYWORD
| hard,nonn
|
|
|
AUTHOR
| Jud McCranie (JudMcCranie(AT)ugaalum.uga.edu), Dec 24 1999
|
| |
|
|
