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%I #34 Oct 27 2023 10:18:59
%S 3,4,6,12,60,1020,262140,334495740,17179869180,27971850688528380
%N Numbers n such that phi(n) - 1 divides n, where phi is Euler's totient function.
%C The sequence b(n) = 4*A050474(n) is a subsequence of this sequence, and comprises solutions of n/(phi(n) - 1) = 4, accounting for all terms up to a(9) except a(1) and a(3). Proof: suppose n/(phi(n) - 1) = 4. With n = 4*x, x/(phi(4*x) - 1) = 1, or phi(4*x) = x + 1. Since phi(k) is even for k > 2, x is odd, and phi(4*x) = 2*phi(x) = x + 1, the definition of A050474. It follows that 4*A050474(8) = 27971850688528380 is a term of this sequence. - _Chris Boyd_, Mar 22 2015
%C Similarly, the terms with n/(phi(n) - 1) = 3 are given by 3 * terms of A050474 coprime to 3; n/(phi(n) - 1) = 6 are given by 6 * terms of A050474 coprime to 6. Also, the terms of n/(phi(n) - 1) = 5 are given by 5 * terms t of A203966 coprime to 5 and having (t+1)/phi(t) = 4. Note that n/(phi(n) - 1) = 2 is impossible. - _Max Alekseyev_, Oct 26 2023
%t Select[1 + Range[1000000], Divisible[#, EulerPhi[#] - 1] &]
%o (PARI) for(n=3,1e7,if(n%(eulerphi(n)-1)==0,print1(n", "))) \\ _Charles R Greathouse IV_, Dec 26 2011
%Y Cf. A000010, A050474, A203966.
%K nonn,more
%O 1,1
%A _José María Grau Ribas_, Dec 25 2011
%E a(8) from _Charles R Greathouse IV_, Dec 27 2011
%E a(9) from _Donovan Johnson_, Dec 29 2011
%E a(10) from _Chris Boyd_ confirmed by _Max Alekseyev_, Oct 26 2023