PROOF (?) COLLATZ CONJECTURE

[bill]

O_(i+1) = (3 * O_i + 1) / (2 ^n)
O_(i+1) = O_i * (3 + 1/O_i) / (2 ^n)

Let (3 + 1/O_i) = R_i. Then
O_(i+1) = O_i * (R_i) / (2 ^n)

For a sequence,

O_(i + k) = O_i * PROD{(R_i) / (2 ^n_i)

If CS loops, then O_(i + k) = O_i and

PROD{(R_i) / (2 ^n_i) = 1.0000000000000000000000 ad infinitum

3 < R_i < 4.

The numerator of each term in PROD is a non-integer between 3 and 4.
The denominator is a power of 2.

If PROD = 1, then its numerator must be 0 mod 2

Can a product be 0 mod 2 when none of the terms are 0 mod 2?

Bill J

[mensa...@aol.com]

This part is easy. If a number is 0 mod 2, it contains at least
one factor of two. If none of the terms are 0 mod 2, none of the
terms contain a factor of 2, so the product cannot contain a
factor of 2.

>
> Bill J