^{i}where i is the largest power of 2 such that 2

^{i}< n. I called k a Collatz exit point and n a Collatz trap.

It turns out that the largest power i of 2 such that 2

^{i}< n for any odd integer n is involved in many other interesting theorems and I'm very much tempted to give it a special definition but I shall refrain from doing so until I prove more of the theorems associated with it.

**Theorem:**If n is an odd integer and i is the largest power of 2 such that 2

^{i}< n then

2

^{i+1}(mod n) = 2^{i}- r for some integer r such that r = n - 2^{i}^{ }**Proof:**If n is an odd integer then n can be represented as n = 2

^{i}+ r for some integer r such that 2

^{i-1}< r < 2

^{i}

Since n = 2

^{i}+ r then 2

^{i}= (n - r) and also since 2

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

^{i}(mod n) then 2

^{i+1}(mod n) = 2*(n - r) (mod n)

Clearly, since 2

^{i-1}< r < 2

^{i}then 2*(n - r) < 2*n so

2

^{i+1}(mod n) = 2*(n - r) (mod n)= 2*(n - r) - n = ( 2*2

^{i}) - n = ( 2*2

^{i})- (2

^{i}+ r) = 2

^{i}+2

^{i}- 2

^{i}- r = 2

^{i}- r .'.

Example: Let n = 35, then the largest power i of 2 mod 35 such that 2

^{i}< 35 is i = 5 since 2

^{5}= 32 and 32 is congruent to 32 (mod 35) so no modular arithmetic is needed to compute this power.

However, for all i > 6 then 2

^{i }> 35 and so modular arithmetic will be needed to compute such powers unless of course I use the method I described here and its improved version here, which generate all powers of 2 mod n without using modular arithmetic, multiplication, or exponentiation.

If I'm only interested in the value of 2

^{6 }mod 35 then by the theorem above 2

^{i+1 }mod n = 2

^{6 }mod 35 = 32 - (35 - 32) = 29

This is obviously insignificant but it is still interesting because computing 2

^{i+1 }will not exceed n using this formula.