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%I #45 Feb 18 2024 08:17:54
%S 1,3,15,255,65535,83623935,4294967295,6992962672132095
%N Solutions to 2*phi(x) = x+1.
%C 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
%C The next term, if it exists, has at least 7 distinct prime factors (see Beiler, p. 92). - _Jud McCranie_, Dec 13 2012
%C From _Chris Boyd_, Mar 22 2015: (Start)
%C Solutions to k*phi(x) = x + 1, including a(1) - a(8), were published in 1932 by D. H. Lehmer. In the paper's summing up, "3*5*353*929" (= 4919055) was printed in error; it should have read "3*5*17*353*929" (= 83623935), i.e., a(6). This error has been propagated in several subsequent texts, including Wong's thesis.
%C Lehmer identified solutions where x has fewer than 7 distinct prime factors. Wong showed that no additional solutions exist unless x has at least 8 distinct prime factors. It appears not to be excluded by either author that an unidentified solution < a(8) with 8 or more distinct prime factors may exist. (End)
%C There are no other terms below 10^17. - _Max Alekseyev_, Oct 30 2021
%D A. H. Beiler, Recreations in the Theory of Numbers, page 92.
%H D. H. Lehmer, <a href="http://dx.doi.org/10.1090/s0002-9904-1932-05521-5">On Euler's totient function</a>, Bulletin of the American Mathematical Society, 38 (1932), 745-751.
%H E. Wong, <a href="http://www.collectionscanada.gc.ca/obj/s4/f2/dsk2/ftp04/mq24272.pdf">Computations on normal families of primes</a>, Simon Fraser University, 1997, MSc thesis.
%F A number 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
%e 2*phi(15) = 2*8 = 15 + 1, so 15 is a member of the sequence.
%t Select[Range[700000], (# + 1)== 2 EulerPhi[#] &] (* _Vincenzo Librandi_, Mar 22 2015 *)
%o (PARI) is_A050474(n)=if(2*eulerphi(n)==n+1,1,0) \\ _Chris Boyd_, Mar 22 2015
%o (Magma) [n: n in [1..2*10^6] | 2*EulerPhi(n) eq (n+1)]; // _Vincenzo Librandi_, Mar 22 2015
%Y Cf. A000010, A129613, A129614, A129615, A202855, A203966.
%K hard,nonn
%O 1,2
%A _Jud McCranie_, Dec 24 1999
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