OFFSET
1,1
COMMENTS
Most of numbers n in this sequence are divisible by 2, and it appears that n/2 belongs to A179188. The other ones are listed in sequence A217141.
Proof of the comment: If n is even and not a multiple of 4 then phi(n)=phi(n/2). If n is a multiple of 4 then phi(n)=2 * phi(n/2). So when k is a multiple of 4 and phi(n)=phi(n+k), then phi(n/2)=phi(n/2+k/2). QED. This also applies to A179186, A179202. - Jud McCranie, Dec 30 2012
LINKS
Jud McCranie, Table of n, a(n) for n = 1..10000
Kevin Ford, Solutions of phi(n)=phi(n+k) and sigma(n)=sigma(n+k), arXiv:2002.12155 [math.NT], 2020.
MATHEMATICA
Select[Range[1, 5000], EulerPhi[#] == EulerPhi[# + 12] &] (* Vincenzo Librandi, Jun 24 2014 *)
PROG
(PARI) {op=vector(N=12); for( n=1, 1e4, if( op[n%N+1]+0==op[n%N+1]=eulerphi(n), print1(n-N, ", ")))}
(Magma) [n: n in [1..3000] | EulerPhi(n) eq EulerPhi(n+12)]; // Vincenzo Librandi, Sep 08 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Michel Marcus, Sep 27 2012
STATUS
approved