*s*by suggesting that

*s gets trapped in a finite field*for a lack of a better phrase to describe this process.

In particular, I imagine it gets trapped into a portion of the ((n+1)/2)

^{th}row of the exponentiation table of Z_{n}where n is some odd integer.The ((n+1)/2)

^{th}row of the exponentiation table of Z

_{n}, where n is some odd integer, is the 2

^{nd}row of the exponentiation table of Z

_{n}in reverse order, which means that the last few elements of it are consecutive powers of 2 in descending order. Furthermore, dividing any even element in this row by 2 produces the next element in the row and there are elements k,l in that row not necessarily distinct such that 3*k+1 = (l*((n+1)/2) mod n).

Most of this I haven't proven yet. So here are a few examples to (hopefully) better illustrate what I mean:

**Example:**Let n = 35. It can be verified that row 18 is equivalent to row 2 in reverse order, and so it ends in consecutive powers of 2 in descending order. Furthermore, dividing any even element in this row by 2 produces the next element in the row and there are elements k = 22, l = 29 in that row not necessarily distinct such that 3*22+1 (mod 35) = 29*18 (mod 35)

It may sound a bit counterintuitive that n is an odd integer because if n is even then it is divisible by 2 and so it would seem that it would fall right into the algorithm behind the Collatz conjecture.

However, this does not appear to be the case. I conjecture this to be the key to solving this problem.

**Example:**the exponentiation tables for n = 6 and n = 10 are given below:

Neither the (n/2)

^{th}nor the ((n/2)+1)

^{th}row ever reach consecutive powers of 2. In both of these cases n is a product of 2 and an odd integer, and in both cases (n/2)

^{i}= (n/2) mod n and ((n/2)+1)

^{i}= ((n/2)+1) mod n for all i < n.

In general I claim the following:

**Claim:**If n is an even integer then for all i such that 1 < i < n

(n/2)

^{i}= (n/2) mod n if n = 2m for some odd integer m

(n/2)

^{i}= 0 mod n if n = 2^{r}m for some odd integer m and some integer r**Claim:**If n is an even then for all i such that 0 < i < n

(n/2)

^{i}* 2^{i}= 0 mod n
In contrast:

**Claim:**If n is an odd integer and r is the order of 2 mod n then for all i such that 1 < i < r

((n+1)/2)

Below I attach a tool for finding all powers of k mod n for any integer n.

^{i}* 2^{i}= 1 mod nBelow I attach a tool for finding all powers of k mod n for any integer n.