1,1

This sequence is infinite, in fact 3*2^n is a subsequence because if m=3*2^n then phi(m-phi(m))=phi(3*2^n-2^n)=2^n=phi(m). Also, if p is a Sophie Germain prime greater than 3 then for each natural number n, 2^n*3*p^2 is in the sequence. Note that there exist terms of this sequence like 750 or 2310 that they aren't of either of these forms. - Farideh Firoozbakht (mymontain(AT)yahoo.com), Jun 19 2005

If n is an even term greater than 2 in this sequence then 2n is also in the sequence. Because for even numbers m we have phi(2m)=2*phi(m) so phi(2n)=2*phi(n)=2*phi(n-phi(n)) and since n is an even number greater than 2 n-phi(n) is even so 2*phi(n-phi(n))=phi(2n-2*phi(n))=phi(2n-phi(2n)) hence phi(2n)=phi(2n-phi(2n)) and 2n is in the sequence. Conjecture: All terms of this sequence are even. - Farideh Firoozbakht (mymontain(AT)yahoo.com), Jul 04 2005

If n is in the sequence and the natural number m divides gcd(n,phi(n)) then m*n is in the sequence. The facts that I have found about this sequence earlier (Jun 19 2005 and Jul 04 2005) are consequences of this. If p is a Sophie Germain prime greater than 3, k>1 and k & n are natural numbers then 2^n*3*p^k are in the sequence. - Farideh Firoozbakht (mymontain(AT)yahoo.com), Dec 10 2005

Numbers n such that phi(n) = phi(n + phi(n)) includes all n = 2^k. - Jonathan Vos Post (jvospost3(AT)gmail.com), Oct 25 2007

R. K. Guy, Unsolved Problems in Number Theory B42.

T. D. Noe, Table of n, a(n) for n=1..1000

Select[Range[11700], EulerPhi[ # ] == EulerPhi[ # - EulerPhi[ # ]] &] (Firoozbakht)

Cf. A000027, A005384, A051488.

Sequence in context: A054061 A118224 A003680 this_sequence A111286 A058295 A132176

Adjacent sequences: A051484 A051485 A051486 this_sequence A051488 A051489 A051490

nonn,nice,easy

R. K. Guy (rkg(AT)cpsc.ucalgary.ca)

More terms from James A. Sellers (sellersj(AT)math.psu.edu)