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.

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

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^(e-3)) or (Z/mZ)* = C_2 X C_(2*3^(e-1)).

If k = 2^(e-3) > 1 and p = 2*k+1 is prime, then (Z/(3*p)Z)* = (Z/(2^e)Z)*; if k = 3^(e-1) > 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