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

1,1

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(n). 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.

Links

Jianing Song, Table of n, a(n) for n = 1..396 (all terms <= 20000)

Wikipedia, Multiplicative group of integers modulo n

Example

(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.

Prog

(PARI)
b(n) = my(i=0, k=eulerphi(n)); for(j=k+1, 3*k*log(log(k))+16, if(znstar(j)[2]==znstar(n)[2], i++)); i
isA319928(n) = if(n>2, b(n)==1, 0)

Crossrefs

Cf. A015126, A028476, A296233, A317993.

Sequence in context: A317534 A240068 A269424 * A025102 A188671 A166648

Adjacent sequences: A319925 A319926 A319927 * A319929 A319930 A319931

Keyword

nonn

Author

Jianing Song, Oct 03 2018

Status

approved