login
Least k such that phi(k*n) = phi(k*n+1), or 0 if no such k exists.
3

%I #18 Jul 27 2016 21:42:26

%S 1,52,1,26,3

%N Least k such that phi(k*n) = phi(k*n+1), or 0 if no such k exists.

%C From _Michael De Vlieger_, Jul 27 2016: (Start)

%C Terms 6 through 80, with 0 signifying no such k <= 10^6: {0, 375, 13, 55, 0, 45, 0, 8, 180121, 1, 2054, 15, 0, 116, 0, 125, 482, 2145, 0, 39, 4, 495, 144098, 76, 0, 105, 1027, 15, 37556, 75, 0, 495, 58, 25, 0, 4, 0, 7425, 241, 11, 2294, 178425, 0, 8475, 0, 5, 2, 5415, 0, 9, 72049, 65, 38, 976, 0, 944, 1702, 915, 761, 15, 0, 161175, 18778, 715, 0, 165, 0, 8, 1426, 13, 29, 20451, 0, 416, 0}.

%C n such that a(n) = 0 in above data: {6, 10, 12, 18, 20, 24, 30, 36, 40, 42, 48, 50, 54, 60, 66, 70, 72, 78, 80}, i.e., multiples of 2 and 3, and 2 and 5. (End)

%C If a(n) = 0, then a(m*n) = 0 for all m > 0. If c is a divisor of a(n), then a(c*n) = a(n)/c. Conjecture: If A275337(n) = 0, then a(n) = 0. - _Chai Wah Wu_, Jul 27 2016

%F a(A001274(n)) = 1.

%e a(2) = 52 because phi(52*2) = phi(52*2+1).

%t Table[k = 1; While[EulerPhi[k n] != EulerPhi[k n + 1], k++]; k, {n, 5}] (* _Michael De Vlieger_, Jul 27 2016 *)

%o (PARI) a(n) = {my(k = 1); while (eulerphi(k*n) != eulerphi(k*n+1), k++); k; }

%Y Cf. A000010, A001274, A275337.

%K nonn,more

%O 1,2

%A _Altug Alkan_, Jul 27 2016