For the past several years I have been chasing after a far grander idea, the idea that there is a simple formula for finding the order of 2 mod n. It seems like there is a quick formula indeed but it is not the same for all.

**Claim:**If n is an odd integer such that n = 2

^{i}+ 1 then the order of 2 mod n is 2*i.

Furthermore, given the set of all powers of 2 smaller than n, say S1 = [2, 4, 8, ..., 2

^{i}] then the set of the remaining powers of 2 mod n, is S2 = [2

^{i}- 1, 2

^{i}- 3, 2

^{i}- 7, ..., 1]

Proof: Later

**Example**: Let n = 33. Clearly 33 = 2

^{5 }+ 1 so the order of 2 mod n is 2*5 = 10. It can be verified that 2

^{10 }= 1 mod 33

**Note:**2

^{i}+ 1 can be either a prime or a composite integer but the formula stays the same. For all other n it seems that there are two different formulas depending on whether n is a prime or a composite.

Interestingly enough, if I construct a set of powers of 2 plus 1, namely the set of 2

^{i}+ 1 for i = 1,2,3 4, 5, 6... then obviously this set is [3, 5, 9, 17, 33, 65, …]Reminder: the order of 2 mod n is

*the smallest integer*k such that 2

^{k}= 1 mod n

Below is a slow calculator for finding the order of 2 mod n for any odd integer n:

So I boldly claim what no one (hopefully) has claimed before:

**Claim**: The set of the order of 2 mod n for each element of the set [3, 5, 9, 17, 33, 65,…]

is precisely the set of all even integers [2, 4, 6, 8, 10, 12, …]