A328413 -id:A328413 - cp's OEIS frontend

A328412 Number of solutions to (Z/mZ)* = C_2 X C_(2n), where (Z/mZ)* is the multiplicative group of integers modulo m.

2, 4, 4, 1, 3, 7, 0, 4, 4, 5, 3, 0, 0, 3, 7, 1, 0, 7, 0, 3, 6, 2, 3, 4, 0, 3, 1, 0, 3, 11, 0, 1, 7, 0, 3, 3, 0, 0, 3, 2, 3, 8, 0, 3, 4, 2, 0, 3, 0, 6, 3, 0, 3, 5, 5, 3, 0, 2, 0, 4, 0, 0, 3, 1, 3, 4, 0, 3, 7, 4, 0, 4, 0, 3, 3, 0, 0, 12, 0, 0, 4, 2, 3, 0, 0, 3, 4, 2, 3, 9, 0, 0

Offset: 1

Jianing Song, Oct 14 2019

nonn

It is sufficient to check all numbers in the range [A049283(4n), A057635(4n)] for m if 4n is a totient number.
Conjecture: every number occurs in this sequence. That is to say, A328416(n) > 0 for every n.
Conjecture: this sequence is unbounded. That is to say, A328417 and A328418 are infinite.

			See the a-file for the solutions to (Z/mZ)* = C_2 X C_(2n) for n <= 5000.

Jianing Song, Table of n, a(n) for n = 1..10000
Jianing Song, Solutions to (Z/mZ)* = C_2 X C_(2n), n <= 5000
Wikipedia, Multiplicative group of integers modulo n

Cf. A328413 (numbers k such that a(k) > 0), A328414 (indices of 0), A328415 (indices of 1).
Cf. A328416 (smallest k such that a(k) = n).
Cf. A328417, A328418 (records in this sequence).
Cf. also A049823, A057635.

a(n) = my(i=0, r=4*n, N=floor(exp(Euler)*r*log(log(r^2))+2.5*r/log(log(r^2)))); for(k=r+1, N, if(eulerphi(k)==r && lcm(znstar(k)[2])==r/2, i++)); i

A328414 Numbers k such that (Z/mZ)* = C_2 X C_(2k) has no solutions m, where (Z/mZ)* is the multiplicative group of integers modulo m.

Original entry on oeis.org

7, 12, 13, 17, 19, 25, 28, 31, 34, 37, 38, 43, 47, 49, 52, 57, 59, 61, 62, 67, 71, 73, 76, 77, 79, 80, 84, 85, 91, 92, 93, 94, 97, 100, 101, 103, 104, 107, 108, 109, 112, 117, 118, 121, 122, 124, 127, 129, 133, 137, 139, 142, 143, 144, 148, 149, 151, 152, 157, 160, 161, 163, 164

Offset: 1

Jianing Song, Oct 14 2019

nonn

Indices of 0 in A328410, A328411 and A328412.

By definition, if there is no such m that psi(m) = 2k, psi = A002322, then m is a term of this sequence.

			12 is a term: if there exists m such that (Z/mZ)* = C_2 X C_24 = C_2 X C_8 X C_3, then m must have a factor q such that q is an odd prime power and phi(q) = 8 or phi(q) = 24, phi = A000010, which is impossible.
80 is a term: if there exists m such that (Z/mZ)* = C_2 X C_80 = C_2 X C_16 X C_5, then m must have a factor q such that q is an odd prime power and phi(q) = 80 or phi(q) = 16, which is impossible.

Wikipedia, Multiplicative group of integers modulo n

Cf. A328410, A328411, A328412. Complement of A328413.

Cf. also A000010, A002322, A005277, A079695.

isA328414(n) = my(r=4*n, N=floor(exp(Euler)*r*log(log(r^2))+2.5*r/log(log(r^2)))); for(k=r+1, N+1, if(eulerphi(k)==r && lcm(znstar(k)[2])==r/2, return(0)); if(k==N+1, return(1)))
for(n=1, 200, if(isA328414(n), print1(n, ", ")))

cp's OEIS Frontend

A328412 Number of solutions to (Z/mZ)* = C_2 X C_(2n), where (Z/mZ)* is the multiplicative group of integers modulo m.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI