[FONT=q_serif]So I saw the collatz conjecture gave it a try and ended up somewhere else . What is the name of these sequences patterns?[/FONT]
[FONT=q_serif]Top Odd number x 3 +1 to get the bottom Even number.[/FONT]
[FONT=q_serif]Or add 6 to the last even number. You will find a continuous pattern.[/FONT]
[FONT=q_serif]E 1. 3. 5. 7. 9. 11. 13. 15[/FONT]
[FONT=q_serif]O 4 10 16 22 28 34 40 46[/FONT]
[FONT=q_serif]Add 5 to the Bottom Even number. Turning it to an Odd number, divide by 3 to get the next sequences top Odd number. It goes on.[/FONT]
[FONT=q_serif]The last digit of the bottom Even number, as a continuous pattern of 4 0 6 8 2.[/FONT]
[FONT=q_serif]What could all this be. Thanks. [/FONT]
As I understand it, you started with the top row, consisting of all odd numbers (for some reason labeled E?); then you applied the formula y = 3x + 1 to get the corresponding even number in the bottom row (for some similarly odd reason labeled O). Then you're saying that if you apply the formula x = (y + 5)/3, you get the odd number in the next column.
I don't know what kind of answer you are looking for ("what could it be?"), but it can be easily explained using algebra.
Number the columns 1, 2, 3, ..., and call that index n:
Code:
[FONT=courier new]index n: 1 2 3 4 5
odd x: [/FONT][FONT=courier new]1 3 5 7 9
[/FONT][FONT=courier new]even y: [/FONT][FONT=courier new]4 10 16 22 28[/FONT]
The odd row is x = 2n - 1; for example, under n=1, we get 2(1) - 1 = 3.
The even row is y = 3x + 1, which applied to 2n - 1 gives y = 3(2n - 1) + 1 = 6n - 2. For example, under n=1, we have 6(1) - 2 = 4.
That tells us that each number in that row is 6 more than the one before, as you pointed out. We can call 6 the slope (rate of change).
Now you want to get the odd number x in the next column, n+1. That means we're starting with y = 6n - 2, and want to get x = 2(n+1) - 1. Do you see that? To find this formula, we can solve for n in terms of y: n = (y + 2)/6; and then put that into the formula for x: 2((y + 2)/6 + 1) - 1. Doing a little manipulation (nothing up my sleeve), this becomes x = 2(y + 2)/6 + 2 - 1 = (y + 2)/3 + 1 = (y + 2)/3 + 3/3 = (y + 2 + 3)/3 = (y + 5)/3. And that is just what you found by experiment.
Algebra can work directly with patterns, so we don't have to try something out over and over, but can just find the new pattern itself.
As for the pattern of last digits, the even row is, as I said, y = 6n - 2; the fifth number to the right is found by replacing n with n+5, and we get 6(n + 5) - 2 = 6n + 28. This is an increase of 30, so the ones digit is untouched, and we get the same ones digit every five places. That's the repetition you found. This whole thing could be explained better in terms of "modular arithmetic", but that would probably be more than you want to know ...
