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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.
3
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. 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)
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, 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)
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Oct 03 2018
STATUS
approved