Timeline for sum of all subgroup elements
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Oct 9, 2020 at 21:43 | comment | added | user6690 | and question with * : what is the limit when N -> inf = number_of _Ns_where(number of even > number of odd)/number_of _Ns_where(number of even <= number of odd) is it 1? ...574983 15362 15466 104 0.861205 | |
| Oct 9, 2020 at 17:44 | comment | added | user6690 | I have checked many Ns, I have hypothesis : any difference between odd and even is possible for some N, how it can be proved? | |
| Oct 9, 2020 at 17:35 | comment | added | user6690 | I would say , the most interesting number for me is difference between odd and even :number which is "free" from order of group, for small N it has some connections with quadratic class number ,later they diverge... | |
| Oct 9, 2020 at 17:20 | comment | added | user6690 | thank you, but it is trivial case - odd == even. N = 35, no solution. | |
| Oct 9, 2020 at 17:18 | comment | added | Richard Stanley | @user6690 The equation $2^j\equiv -1$ (mod 35) does not have a solution, so $n=35$ is not a counterexample. | |
| Oct 9, 2020 at 17:05 | comment | added | Richard Stanley | If the equation $2^j\equiv -1$ (mod $n$) has a solution, then $k=\frac 12 \mathrm{ord}_n(2)$, where $\mathrm{ord}_n(2)$ denotes the order of 2 modulo $n$. This is because the powers of 2 come in pairs $c,n-c$. | |
| Oct 9, 2020 at 16:16 | review | Close votes | |||
| Oct 24, 2020 at 3:06 | |||||
| Oct 9, 2020 at 12:52 | comment | added | user6690 | if $N$ is prime, factorization of $N-1$ is not “slow” problem.Then just check division 2 ** factors - 1 by $N$ | |
| Oct 9, 2020 at 6:45 | comment | added | Gerry Myerson | I think even if $N$ is prime you need to know the order of $2$ modulo $N$, and there's no known fast way to compute that. | |
| Oct 9, 2020 at 4:31 | review | First posts | |||
| Oct 9, 2020 at 15:54 | |||||
| Oct 9, 2020 at 4:29 | history | asked | user6690 | CC BY-SA 4.0 |