Collatz extensions

Question

Because, in the basic Collatz algorithm, odds are always transformed into evens (via x=3x+1), and evens are transformed into evens with a probability of .5 (via x=x/2), there will be, on average, 2 divisions (x/4) for every multiplication (x*3). So probability predicts inevitable reduction.

Is the problem then proving there can be no collisions (i.e., internal loops) in test values before 1 is reached?

To try to understand, I've added a second test (only) after dividing an even in which, if the result is odd, I subtract 1 and divide again: i.e., x=(x-1)/2. Now probability predicts 6 divides for each multiplication, and the highest probabilistically descending multiplier is 2^6-1 (63). Here, instead of an average reduction of 3/4 (as per Collatz), it's only 63/64, and one would expect series to be a lot longer. Here's the pseudo-code:

while test>1
    if is_even(test)
        test = test/2
        if is_odd(test)
            // could test for 1 here
            test=(test-1)/2
        endif
    else
        test=test*63+1
    endif
endwhile

I've been running the above algo with huge integers for a few hours now and tested past 14 bits. So far the largest number of iterations has been 1,176,551 for the integer 10581 which reached a 2255-bit intermediary value.

Additional nested odd tests require larger multipliers and result in narrower probability spreads. E.g.:

0 tests (Collatz) : mult = 3,  reduces by 3/4
1 test (as here): mult=63, reduces by 63/64
2 tests: mult=16383 (2^14-1), reduces by 16383/16384
3 tests: mult=(2^30-1) ... 

I hypothesize that any above variation will eventually reduce any test value to 1 (though some could take longer than the universe will be around).

I assume this (these) extensions are equally un-provable conjectures, but might they shed light on Collatz by providing a broader context?

Answer 1

$$4k+3\to 12k+10\to 6k+5\to 18k+16\to 9k+8$$

The last number here is same parity as $k$ assuming it's odd we get:$$\to 54j+52$$ which will only divide by 2 twice if $j$ is even (aka if $k=4l+1$) otherwise it goes:$$\to 27j+26\to 81j+79$$ which only divides by 2 twice if $j=4m+1$ otherwise it goes $$\to 162m+161\to 486m+484\to 243m+242$$ which ... you get the drift. we can in theory continue to do this endlessly with conditions. That would mean in theory we can concoct numbers that go arbitrarily long without 2 divisions by 2 at once. We can also use this to show, the only way such a number is the lowest part of a cycle, is if 16k+12 or higher power of two multiplier links up to a number $4p+1$ going backwards.