A328410 -id:A328410 - cp's OEIS frontend

A328411 Largest m such that (Z/mZ)* = C_2 X C_(2n), or 0 if no such m exists, where (Z/mZ)* is the multiplicative group of integers modulo m.

12, 30, 42, 32, 66, 90, 0, 102, 114, 150, 138, 0, 0, 174, 198, 128, 0, 270, 0, 246, 294, 230, 282, 306, 0, 318, 324, 0, 354, 450, 0, 256, 414, 0, 426, 438, 0, 0, 474, 374, 498, 522, 0, 534, 594, 470, 0, 582, 0, 750, 618, 0, 642, 810, 726, 678, 0, 590, 0, 738, 0, 0, 762, 512

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.

If (Z/mZ)* is isomorphic to C_2 X C_(2k) for some k, let x be any element in (Z/mZ)* such that the multiplicative order of x is 2k and that x != -1, then {-1, x} generates (Z/mZ)*. For example, (Z/16Z)* = {+-1, +-3, +-9, +-11}, (Z/32Z)* = {+-1, +-3, +-9, +-27, +-17, +-19, +-25, +-11}.

			The solutions to (Z/mZ)* = C_2 X C_12 are m = 35, 39, 45, 52, 70, 78 and 90, the largest of which is 90, so a(6) = 90.

Wikipedia, Multiplicative group of integers modulo n

Cf. A062373, A328410 (largest m).

Cf. also A049283, A057635.

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

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

A328411 Largest m such that (Z/mZ)* = C_2 X C_(2n), or 0 if no such m exists, where (Z/mZ)* is the multiplicative group of integers modulo m.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI