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%I #44 May 12 2024 18:11:25
%S 24,32,80,96,120,128,160,168,240,252,256,264,324,384,400,408,416,456,
%T 480,504,512,544,552,640,648,672,696,768,840,928,1040,1088,1128,1272,
%U 1280,1312,1320,1360,1408,1416,1504,1536,1632,1696,1704,1840,1848,1896,1920,1992
%N Numbers k such that there is no other m such that (Z/mZ)* is isomorphic to (Z/kZ)*, where (Z/kZ)* is the multiplicative group of integers modulo k.
%C Numbers such that A317993(k) = 1.
%C To find such k, it's sufficient to check for A015126(k) <= m <= A028476(k).
%C This is a subsequence of A296233. As a result, all members in this sequence should not satisfy any congruence mentioned there. Specially, all terms here are divisible by 4.
%C There are only 218 terms <= 10000 and 396 terms <= 20000.
%H Jianing Song, <a href="/A319928/b319928.txt">Table of n, a(n) for n = 1..396</a> (all terms <= 20000)
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n">Multiplicative group of integers modulo n</a>
%e (Z/24Z)* = C_2 X C_2 X C_2, and there is no other m such that (Z/mZ)* = C_2 X C_2 X C_2, so 24 is a term.
%e (Z/96Z)* = C_2 X C_2 X C_8, and there is no other m such that (Z/mZ)* = C_2 X C_2 X C_8, so 24 is a term.
%o (PARI) b(n) = my(i=0, search_max = A057635(eulerphi(n))); for(j=eulerphi(n)+1, search_max, if(znstar(j)[2]==znstar(n)[2], i++)); i \\ search_max is the largest k such that phi(k) = phi(n). See A057635 for its program
%o isA319928(n) = if(n>2, b(n)==1, 0)
%Y Cf. A015126, A028476, A057635, A296233, A317993.
%K nonn
%O 1,1
%A _Jianing Song_, Oct 03 2018