A328414 -id:A328414 - 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

A328413 Numbers k such that (Z/mZ)* = C_2 X C_(2k) has solutions m, where (Z/mZ)* is the multiplicative group of integers modulo m.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 14, 15, 16, 18, 20, 21, 22, 23, 24, 26, 27, 29, 30, 32, 33, 35, 36, 39, 40, 41, 42, 44, 45, 46, 48, 50, 51, 53, 54, 55, 56, 58, 60, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 78, 81, 82, 83, 86, 87, 88, 89, 90, 95, 96, 98, 99, 102, 105, 106, 110, 111

Offset: 1

Jianing Song, Oct 14 2019

nonn

For n > 1, it is easy to see A114871(n)/2 is a term of this sequence. The smallest term here not of the form A114871(k)/2 is 24: 48 is not of the form (p-1)*p^k for any prime p, but (Z/mZ)* = C_2 X C_48 has solutions m = 119, 153, 238, 306.

			(Z/mZ)* = C_2 X C_2 has solutions m = 8, 12; (Z/mZ)* = C_2 X C_4 has solutions m = 15, 16, 20, 30; (Z/mZ)* = C_2 X C_6 has solutions m = 21, 28, 36, 42; (Z/mZ)* = C_2 X C_8 has solutions m = 32; (Z/mZ)* = C_2 X C_10 has solutions m = 33, 44, 66; (Z/mZ)* = C_2 X C_12 has solutions m = 35, 39, 45, 52, 70, 78, 90. So 1, 2, 3, 4, 5, 6 are all terms.

Cf. A328412. Complement of A328414.

Cf. also A114871.

isA328413(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(1)); if(k==N+1, return(0)))
for(n=1, 100, if(isA328413(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