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A277261 Least k>0 such that phi(Fibonacci(n)) divides phi(Fibonacci(n+k)). 1

%I

%S 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,

%T 14,29,10,31,32,22,17,35,36,37,38,26,20,41,42,43,44,18,46,47,48,49,50,

%U 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

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

%C a(n) <= n, since Fibonacci(n) divides Fibonacci(2n) and phi(x) divides phi(y) if x divides y. - _Robert Israel_, Dec 01 2016

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

%p f:= proc(n) uses combinat, numtheory; local k, phin;

%p phin:= phi(fibonacci(n));

%p for k from 1 do if phi(fibonacci(n+k)) mod phin = 0 then return k fi od

%p end proc;

%p map(f, [$1..100]); # _Robert Israel_, Dec 01 2016

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

%o (PARI) a(n) = {my(k=1); while (eulerphi(fibonacci(n+k)) % eulerphi(fibonacci(n)), k++); k;} \\ _Michel Marcus_, Nov 19 2016

%Y Cf. A000010, A000045, A065449.

%K nonn

%O 1,8

%A _Altug Alkan_, Oct 07 2016

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Last modified October 14 00:10 EDT 2019. Contains 327990 sequences. (Running on oeis4.)