A282016 Least k > 0 such that phi(n) divides phi(n+k).
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, 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, 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
Offset: 1
Keywords
Examples
a(5) = 3 because phi(5) = 4 divides phi(5 + 3) = 4 and 3 is the least positive number with this property.
Links
- Altug Alkan, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
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 *) -
PARI
a(n) = my(k = 1); while (eulerphi(n+k) % eulerphi(n) != 0, k++); k;
Formula
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
a(n) = A069797(n) - n. - Altug Alkan, Feb 10 2017
Comments