%I #16 Dec 12 2021 19:59:21

%S 65,209,11009,38009,680609,2205209,3515609,4347209,10595009,12006209,

%T 31979009,89019209,169130009,244766009,247590209,258084209,325622009,

%U 357777209,377330609,441630209,496175609,640343009,1006475609

%N Composite numbers k such that both phi(k+12) = phi(k) + 12 and sigma(k+12) = sigma(k) + 12.

%C It is easy to show that if p, p+2, p+6 and p+8 are all prime (a prime quadruple as defined in A007530, which lists the values of p) with x=p(p+8), x+12=(p+2)(p+6), then x is in the sequence. I conjecture that all members of the sequence are of this form. - _Jud McCranie_, Oct 11 2000

%C Numbers so far are all congruent to 65 (mod 72). - _Ralf Stephan_, Jul 07 2003

%e k = 209 = 11*19, k + 12 = 221 = 13*17, phi(k + 12) = 192 = 180 + 12 = phi(k) + 12, also sigma(221) = 252 = sigma(209) + 12 = 240 + 12.

%e phi(65) + 12 = 60 = phi(65 + 12), sigma(65) + 12 = 96 = sigma(65 + 12), 65 is composite.

%o (PARI) isok(n) = !isprime(n) && (sigma(n+12) == sigma(n)+12) && (eulerphi(n+12)==eulerphi(n)+12); \\ _Michel Marcus_, Jul 14 2017

%Y Cf. A000010, A001838, A015917, A054902, A046133.

%K nonn

%O 1,1

%A _Labos Elemer_, Aug 17 2000

%E More terms from _Jud McCranie_, Oct 11 2000