cp's OEIS Frontend

A282016 Least k > 0 such that phi(n) divides phi(n+k).

%I A282016 #46 Feb 10 2017 20:13:22
%S A282016 1,1,1,1,3,1,2,2,4,2,11,1,8,4,1,1,15,1,8,4,5,3,23,6,8,2,10,7,29,2,30,
%T A282016 2,8,6,4,1,20,16,6,8,14,3,6,6,7,21,47,3,37,5,13,4,53,3,20,9,6,29,59,4,
%U A282016 16,15,10,4,32,9,67,12,20,2,71,1,18,2,7,15,16,6,78,5,28,6,83,6,43,12,26,12
%N A282016 Least k > 0 such that phi(n) divides phi(n+k).
%C A282016 See A171935 for least positive k such that phi(n) = phi(n+k), or 0 if no such k exists.
%C A282016 See also logarithmic scatterplot of this sequence. - _Altug Alkan_, Feb 09 2017
%H A282016 Altug Alkan, <a href="/A282016/b282016.txt">Table of n, a(n) for n = 1..10000</a>
%F A282016 a(n) << n^5 as a consequence of Xylouris' form of Linnik's theorem: phi(n) is at most n-1, and a(n) is at most the least prime which is 1 mod phi(n). - _Charles R Greathouse IV_, Feb 09 2017
%F A282016 a(n) = A069797(n) - n. - _Altug Alkan_, Feb 10 2017
%e A282016 a(5) = 3 because phi(5) = 4 divides phi(5 + 3) = 4 and 3 is the least positive number with this property.
%t A282016 f[n_] := Block[{k = 1}, While[ Mod[ EulerPhi[n + k], EulerPhi[ n]] > 0, k++]; k]; Array[f, 88] (* _Robert G. Wilson v_, Feb 09 2017 *)
%o A282016 (PARI) a(n) = my(k = 1); while (eulerphi(n+k) % eulerphi(n) != 0, k++); k;
%Y A282016 Cf. A000010, A069797, A171935.
%K A282016 nonn
%O A282016 1,5
%A A282016 _Altug Alkan_, Feb 09 2017