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A320045 Smallest k such that (Z/kZ)* is isomorphic to (Z/nZ)*.

%I A320045 #16 Oct 08 2018 05:56:41
%S A320045 1,1,3,3,5,3,7,8,7,5,11,8,13,7,15,15,17,7,19,15,21,11,23,24,25,13,19,
%T A320045 21,29,15,31,32,33,17,35,21,37,19,35,40,41,21,43,33,35,23,47,40,43,25,
%U A320045 51,35,53,19,55,56,57,29,59,40,61,31,63,51,65,33,67,51,69,35
%N A320045 Smallest k such that (Z/kZ)* is isomorphic to (Z/nZ)*.
%C A320045 a(n) = n iff n is a term in A296233.
%C A320045 Most terms are odd. Among the first 10000 terms there are 8837 odd ones. The even terms are divisible by 4 because (Z/kZ)* is isomorphic to (Z/(2k)Z)* for odd k.
%C A320045 A015126(n) <= a(n) <= n.
%H A320045 Jianing Song, <a href="/A320045/b320045.txt">Table of n, a(n) for n = 1..10000</a>
%H A320045 Wikipedia, <a href="http://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n">Multiplicative group of integers modulo n</a>
%e A320045 The solutions to (Z/kZ)* = C_6 are k = 7, 9, 14 and 18, so a(7) = a(9) = a(14) = a(18) = 7.
%e A320045 The solutions to (Z/kZ)* = C_2 X C_20 are k = 55, 75, 100, 110 and 150, so a(55) = a(75) = a(100) = a(110) = a(150) = 55.
%e A320045 The solutions to (Z/kZ)* = C_2 X C_12 are k = 35, 39, 45, 52, 70, 78 and 90, so a(35) = a(39) = a(45) = a(52) = a(70) = a(78) = a(90) = 35.
%o A320045 (PARI) a(n) = my(i=eulerphi(n)); while(znstar(i)[2]!=znstar(n)[2], i++); i
%Y A320045 Cf. A015126, A296233, A320046.
%K A320045 nonn
%O A320045 1,3
%A A320045 _Jianing Song_, Oct 04 2018