A280988 Least k such that phi(k*n) is a perfect square, or 0 if no such k exists.
1, 1, 4, 2, 1, 2, 9, 1, 7, 1, 41, 1, 21, 9, 4, 2, 1, 6, 3, 2, 3, 41, 89, 2, 5, 14, 4, 13, 113, 2, 143, 1, 25, 1, 9, 3, 1, 2, 7, 1, 11, 3, 49, 25, 7, 89, 1151, 1, 43, 5, 4, 7, 553, 2, 15, 9, 1, 113, 233, 1, 77, 122, 1, 2, 21, 25, 299, 2, 356, 9, 281, 6, 3, 1, 11, 1, 61, 6, 313
Offset: 1
Keywords
Examples
a(11) = 41 because phi(k*11) is not a perfect square for 0 < k < 41 and phi(41*11) = 20^2.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
- Paul Pollack and Carl Pomerance, Square values of Euler's function, Bulletin of the London Mathematical Society, Vol. 46, No. 2 (2014), pp. 403-414, alternative link.
Programs
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Maple
f:= proc(n) local k; for k from 1 do if issqr(numtheory:-phi(k*n)) then return k fi od end proc: map(f, [$1..100]); # Robert Israel, Jan 12 2017 -
Mathematica
a[n_] := Module[{k = 1}, While[!IntegerQ[Sqrt[EulerPhi[k*n]]], k++]; k]; Array[a, 80] (* Amiram Eldar, Jul 13 2019 *) -
PARI
a(n) = {my(k = 1); while (!issquare(eulerphi(k*n)), k++); k; }
Comments