A319928 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.

24, 32, 80, 96, 120, 128, 160, 168, 240, 252, 256, 264, 324, 384, 400, 408, 416, 456, 480, 504, 512, 544, 552, 640, 648, 672, 696, 768, 840, 928, 1040, 1088, 1128, 1272, 1280, 1312, 1320, 1360, 1408, 1416, 1504, 1536, 1632, 1696, 1704, 1840, 1848, 1896, 1920, 1992

Offset: 1

Author

Jianing Song, Oct 03 2018

Keywords

• nonn

Comments

Numbers such that A317993(k) = 1.

To find such k, it's sufficient to check for A015126(k) <= m <= A028476(k).

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.

There are only 218 terms <= 10000 and 396 terms <= 20000.

Examples

			(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.
(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.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..396 (all terms <= 20000)
• Wikipedia, Multiplicative group of integers modulo n

Crossrefs

Cf. A015126, A028476, A057635, A296233, A317993.

Programs

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
isA319928(n) = if(n>2, b(n)==1, 0)