A277261 Least k>0 such that phi(Fibonacci(n)) divides phi(Fibonacci(n+k)).

1, 1, 1, 1, 1, 1, 1, 4, 3, 5, 11, 2, 13, 5, 6, 16, 17, 6, 16, 10, 14, 11, 23, 20, 25, 13, 22, 14, 29, 10, 31, 32, 22, 17, 35, 36, 37, 38, 26, 20, 41, 42, 43, 44, 18, 46, 47, 48, 49, 50, 51, 26, 53, 54, 55, 56, 57, 58, 59, 60, 61, 31, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 37, 75, 76, 77, 78, 79, 80, 81, 41

Offset: 1

Author

Altug Alkan, Oct 07 2016

Keywords

• nonn

Comments

a(n) <= n, since Fibonacci(n) divides Fibonacci(2n) and phi(x) divides phi(y) if x divides y. - Robert Israel, Dec 01 2016

Examples

			a(7) = 1 because phi(Fibonacci(7)) = phi(Fibonacci(8)) = 12.
		

Crossrefs

Cf. A000010, A000045, A065449.

Programs

Maple

f:= proc(n) uses combinat, numtheory; local k, phin;
    phin:= phi(fibonacci(n));
    for k from 1 do if phi(fibonacci(n+k)) mod phin = 0 then return k fi od
end proc;
map(f, [$1..100]); # Robert Israel, Dec 01 2016

Mathematica

Table[k = 1; While[Mod[EulerPhi@ Fibonacci[n + k], EulerPhi@ Fibonacci@ n] != 0, k++]; k, {n, 82}] (* Michael De Vlieger, Nov 23 2016 *)

PARI

a(n) = {my(k=1); while (eulerphi(fibonacci(n+k)) % eulerphi(fibonacci(n)), k++); k;} \\ Michel Marcus, Nov 19 2016