Originally posted by Dejection
No, a loop is not required to disprove your conjecture. After a number goes up to some sort of peak, it could well turn back down quickly. But what's to say that it won't bounce back up, then down, then up, however going up overall?

Because that sequence would be infinitely long, and if it were infinitely long, there would be an infinite amount of numbers that would not converge to 1, and that would be easy to find.

*edit* and it wouldn't be like the prime numbers, where they are difficult to identify once they become sufficiently large... the next larger number in the sequence is AT MOST 3n+1, and smaller if you have to divide by 2 in between.

Originally posted by Dejection
No, a loop is not required to disprove your conjecture. After a number goes up to some sort of peak, it could well turn back down quickly. But what's to say that it won't bounce back up, then down, then up, however going up overall?

Perhaps not, but intuition tells me any exceptions to the theorem would eventually loop back onto itself, although it is very possible it may rise and fall many times before it does so.



I was surprised however, to find something out about a sort of analysis of moving from one odd number to the next odd number (as the even numbers behave in a very straightforward manner).

(Adding 1 here can be safely ignored at any decently high value.)

1 out of 2 times, I'll end up multiplying by 3 and dividing by only 2.
1 out of 4 times, I'll end up multiplying by 3 and dividing by 4.
1 out of 8 times, I'll end up multiplying by 3 and dividing by 8.
etc
etc

If I was to choose an odd number by random, and compare the next odd number in series to it, the expected (weighted average) value would equal 3 * sum(i=1 to INF, 4^-i), which works out to equal exactly 1, meaning on average the value stays the same for one to the next.

Since so many numbers seem to trend downwards in the long run, that would suggest some might trend upwards in the long run to balance out.

However, that is far from a proof either way. I still suspect all numbers trend towards one.

Originally posted by geepamoogle
The tricky part of that is showing that eventually all numbers greater than 1 will eventually reach a lower number.

You give me an arbitrary number of times an odd number leads to a higher odd number in 2 iterations consecutively, and I can find you one which will do it that many times in a row, which could lead to a small percentage of cases where th ...[text shortened]... n number loops into itself and would therefore repeat in a manner similar to 1, 4, 2, 1, 4, 2...

1 -> 4 -> 2 -> 1 -> 4
3 -> 10 -> 5 -> 16 -> 8 -> 4
6 -> 3
7 -> 22 -> 11 -> 34 -> 17 -> 52 -> 26 -> 13 -> 40 -> 20 -> 10
9 -> 28 -> 14 -> 7
12 -> 6
15 -> 46 -> 23 -> 70 -> 35 -> 106 -> 53 -> 160 -> 80 -> 40
18 -> 9
19 -> 58 -> 29 -> 88 -> 44 -> 22
21 -> 64 -> 32 -> 16
24 -> 12
25 -> 76 -> 38 -> 19

and in binary

1 -> 100 -> 10 -> 1
11 -> 1010 -> 101 -> 10000 -> 100 -> 100
110 -> 11
111 -> 10110 -> 1011 -> 100010 -> 10001 -> 110100 -> 11010 -> 1101 -> 101000 -> 10100 -> 1010
1001 -> 11100 -> 1110 -> 111
1100 -> 110
1111 -> 101110 -> 10111 -> 1000110 -> 100011 -> 1101010 -> 110101 -> 10100000 -> 1010000 -> 101000
10010 -> 1001
10011 -> 111010 -> 11101 -> 1011000 -> 101100 -> 10110
10101 -> 1000000 -> 100000 -> 1000
11000 -> 1100
11001 -> 1001100 -> 100110 -> 10011