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Conjecture #1:

$a_(x,y)=(1/3)(2^{(2y+7)-2x)}(5(4^{x})-2)$

Generates positive even integers with a total stopping time S where $S=2y+13$, y is the set of all natural numbers and x is the set of all natural numbers such that x is less than or equal to y+4

Example: If y=1 we get $a_(x,1)=(1/3)(2^{9-2x})(5(4^{x})-2)$. This equation will now generate positive even integers with a TST of 15 only for x less than or equal to 5. So, it will generate 5 positive even integers:

768,832,848,852,853

If y=2 we get $a_(x,1)=(1/3)(2^{11-2x})(5(4^{x})-2)$. This equation will now generate positive even integers with a TST of 17 only for x less than or equal to 6. So, it will generate 6 positive even integers:

3072,3328,3392,3408,3412,3413

I conjecture that this continues for all x and y.

Conjecture #2:

$a_(x,y) =(1/3)(2^{(2y+8)-2x})(5(4^{x})-2)$

Generates positive even integers with a total stopping time S where $S=2y+14$, y is the set of all natural numbers and x is the set of all natural numbers such that x is less than or equal to y+4.

Example: If y=1 we get $a_(x,1)=(1/3)(2^{10-2x})(5(4^{x})-2)$. This equation will now generate positive even integers with a TST of 16 only for x less than or equal to 5. So, it will generate 5 positive even integers:

1536,1664,1696,1704,1706

If y=2 we get $a_(x,1)=(1/3)(2^{12-2x})(5(4^{x})-2)$. This equation will now generate positive even integers with a TST of 18 only for x less than or equal to 6. So, it will generate 6 positive even integers:

6144,6656,6784,6816,6824,6826

I conjecture that this continues for all x and y.

I actually have a generalized form but I want to know your opinion on this.

Does this look profound and has this already been discovered?

Tylersamuels643
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    Try putting your exponents in curly brackets & dollars around your formulea. This might make it readable. – Donald Splutterwit Apr 01 '20 at 01:00
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    You might look in MSE at https://math.stackexchange.com/questions/470782/longest-known-sequence-of-identical-consecutive-collatz-sequence-lengths/1251031#1251031 and https://math.stackexchange.com/questions/1243841/what-causes-long-sequences-of-consecutive-collatz-paths-to-share-the-same-leng/1253988#1253988 . I've given some observation and analysis of an equal-length problem. However, it seems a bit different from your question, but ideas of the approach might be helpful. – Gottfried Helms Apr 01 '20 at 07:18

1 Answers1

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All your numbers are multiple of $2^k$ and the well known numbers of the branch $\{3,13,53,213,853,3413,...\}$ of the Collatz tree.

The numbers on this branch are $a_{n+1}=4\cdot a_n+1$ with $a_0=3$ or $\frac{10\cdot 4^n-1}{3}$, and have all the same ("odd") stopping time (since they are on the same branch).

Multiplying by $2^k$ is the same as looking at the direct next branch of these numbers in the tree, which means they all have the same stopping time too (well, it depends on what you are looking at when determining the stopping time: Odd only, odd/even,...looking at "odd/even" stopping time is just a bit trickier, the exponent of 2 of the current/next branch plays a role in that case)

Your formulation is a bit strange (you go backward, limiting the results), but you could generalize to any branch. You could also try to go further than the direct next branch.

Collag3n
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  • Thanks for your comment. I actually have many more equations like this that generalize the branches. I just posted these for now. – Tylersamuels643 Apr 01 '20 at 11:34