A217139 - OEIS

%I #29 Sep 08 2022 08:46:04

%S 48,68,72,78,86,88,114,143,144,156,157,164,168,186,192,203,216,222,

%T 247,273,292,356,402,432,444,450,452,456,612,654,728,732,762,798,834,

%U 864,876,884,932,942,964,1032,1054,1080,1086,1124,1147,1152,1194,1209,1220

%N Numbers n such that phi(n) = phi(n+12), with Euler's totient function phi = A000010.

%C 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.

%C 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

%H Jud McCranie, <a href="/A217139/b217139.txt">Table of n, a(n) for n = 1..10000</a>

%H Kevin Ford, <a href="https://arxiv.org/abs/2002.12155">Solutions of phi(n)=phi(n+k) and sigma(n)=sigma(n+k)</a>, arXiv:2002.12155 [math.NT], 2020.

%t Select[Range[1, 5000], EulerPhi[#] == EulerPhi[# + 12] &] (* _Vincenzo Librandi_, Jun 24 2014 *)

%o (PARI) {op=vector(N=12); for( n=1, 1e4, if( op[n%N+1]+0==op[n%N+1]=eulerphi(n), print1(n-N, ", ")))}

%o (Magma) [n: n in [1..3000] | EulerPhi(n) eq EulerPhi(n+12)]; // _Vincenzo Librandi_, Sep 08 2016

%Y Cf. A000010, A179188, A217141, A007015.

%K nonn,easy

%O 1,1

%A _Michel Marcus_, Sep 27 2012