

A328415


Numbers k such that (Z/mZ)* = C_2 X C_(2k) has exactly one solution, where (Z/mZ)* is the multiplicative group of integers modulo m.


1



4, 16, 27, 32, 64, 256, 512, 1024, 2048, 2187, 4096, 6561, 8192, 16384, 59049, 65536, 131072, 177147, 262144, 524288, 531441, 1048576, 1594323, 2097152, 4194304, 4782969, 8388608, 14348907, 16777216, 33554432, 67108864, 134217728, 268435456, 387420489, 536870912, 1073741824
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OFFSET

1,1


COMMENTS

Numbers k being powers of 2 or 3 such that 2*k+1 is not prime.
Proof. If m is a solution to (Z/mZ)* = C_2 X C_(2k) such that m is odd, then 2*m is also a solution, and vice versa. So if there is only one solution to (Z/mZ)* = C_2 X C_(2k), m must be a multiple of 4. If 8 divides m and m has odd prime factors, or if m has at least two distinct odd prime factors, then A046072(m) >= 3, a contradiction. So m = 2^e, e >= 3 or m = 4*p^e, p odd prime and e >= 1. If m = 4*p^e and p >= 5, then (Z/(3*p^e)Z)* = (Z/mZ)*. So we have m = 2^e, e >= 3 or m = 4*3^e, e >= 1, then (Z/mZ)* = C_2 X C_(2*2^(e3)) or (Z/mZ)* = C_2 X C_(2*3^(e1)).
If k = 2^(e3) > 1 and p = 2*k+1 is prime, then (Z/(3*p)Z)* = (Z/(2^e)Z)*; if k = 3^(e1) > 1 and p = 2*k+1 is prime, then (Z/(3*p)Z)* = (Z/(4*3^e)Z)*; on the other hand, if k is a power of 2 or a power of 3 such that 2*k+1 is not prime, then (Z/mZ)* = C_2 X C_(2k) indeed has only one solution.


LINKS

Table of n, a(n) for n=1..36.
Wikipedia, Multiplicative group of integers modulo n


EXAMPLE

The only solution to (Z/mZ)* = C_2 X C_54 is m = 324, so 54/2 = 27 is a term.


PROG

(PARI) select(i>!isprime(2*i+1), upto(10^9)) \\ See A006899 for the function upto(n)


CROSSREFS

Cf. A328412.
Sequence in context: A134330 A046346 A340852 * A097764 A227993 A072873
Adjacent sequences: A328412 A328413 A328414 * A328416 A328417 A328418


KEYWORD

nonn


AUTHOR

Jianing Song, Oct 14 2019


STATUS

approved



