|
%I #15 Apr 14 2018 02:37:34
%S 3,4,6,8,14,18,20,32,62,90,108,128,522,608,1280,2204,2282,3218,4254,
%T 4424,9690,9942,11214,19938,21702,23210,44498,86244,110504,132050,
%U 216092,756840,859434,1257788,1398270,2976222,3021378,6972594,13466918,20996012,24036584,25964952,30402458,32582658
%N Numbers n such that x = 2^n-2 satisfies phi(x)+2=phi(x+2).
%C Other solutions of this equation are in A001838.
%C Also, n such that 2^(n-1)-1 is prime. Proof: If x=2^n-2, phi(x)+2=phi(x+2) <==> phi(2^n-2)+2=phi(2^n) <==> phi(2(2^(n-1)-1)) + 2 = 2^n(1-1/2) <==> phi(2)*phi(2^(n-1)-1)+2=2^(n-1) <==> phi(2^(n-1)-1) = 2^(n-1)-2 if y=2^(n-1)-1. We have ph(y)=y-1 <==> y=2^(n-1)-1 is prime. Therefore a(n) = A000043(n)+1. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 19 2004
%H Ivan Panchenko, <a href="/A050475/b050475.txt">Table of n, a(n) for n = 1..47</a>
%e phi(2^18-2)+2=131072=phi(2^18), so 18 is in the sequence.
%t Flatten[Position[EulerPhi[2^# - 2] + 2 == EulerPhi[2^# ] & /@ Range[1, 250], True]] (* Vit Planocka *)
%Y Cf. A001838.
%K nonn
%O 1,1
%A _Jud McCranie_, Dec 24 1999
%E a(39)-a(44) from _Ivan Panchenko_, Apr 11 2018
|
