I conjectured that the order of 2 mod n for such n is i + (i - 2) regardless of whether i is even or odd.

The recurrence relation for generating all such n up to a limit is :

2

^{i}+ (2^{(i-1)}- 3) if i is even
or 2

^{i}+ (2^{i-1}+ 3) if i is oddFor the fun of it, I wrote a simple script to generate the first few such integers up to some limit.

```
/* -------------------------------------------------
This content is released under the GNU License
http://www.gnu.org/copyleft/gpl.html
Author: Marina Ibrishimova
Version: 1.0
Purpose: Generate the 3sequence
---------------------------------------------------- */
function gen_seq(i, limit, a)
{
var j = i - 1;
var n = Math.pow(2,j);
var m = Math.pow(2,i);
if (i < limit)
{
if (i % 2 != 0)
{
n = n + 3;
}
else
{
n = n - 3;
}
a.push(m+n);
i = i + 1;
gen_seq(i, limit, a)
}
return a;
}
```

Here's the output for limit = 12:

15, 21, 51, 93, 195, 381, 771, 1533, 3075, 6141

Obviously, there is another way to generate this sequence than the one I use in my script. Each term in the sequence can also be generated by multiplying 3 and 2^i - 1 if i is odd or 3 and 2^i + 1 if i is even.

3*5 = 15 where 5 = 2^2 + 1

3*7 = 21 where 7 = 2^3 - 1

3*17 = 51 where 17 = 2^4 + 1

3*31 = 93 where 31 = 2^5 - 1

and so on.