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edited Oct 26 2011 at 7:17 |
[deleted question]Implication for m-cycles in Collatz-type problems.
[deleted][deleted] Background Consider Collatz-type problems of the form $an + 1$, where $a > 2$ is a positive, odd integer (e.g., $3n + 1$, $5n +1 $, $7n + 1$, etc.). For convenience, automatically divide by two. The $3n + 1$ problem has a single known positive cycle $(1, 2)$. This cycle has period $2$, and the maximum value this cycle obtains differs from its minimum value by a power of two; that is $2 = 2 \cdot 1$. The $7n + 1$ problem has a positive cycle $(1, 4, 2)$. This cycle has period $3$, and the maximum value $4 = 2^2 \cdot 1$ is the minimum value multiplied by some power of two. The $5n + 1$ problem has three positive cycles: $(1, 3, 8, 4, 2)$, $(13, 33, 83, 208, 104, 52, 26)$, and $(17, 43, 108, 54, 27, 68, 34)$. The first cycle has period $5$, and its maximum $8 = 2^3 \cdot 1$. The other two cycles have period $7$, and one of these $7$-cycles has a maximum value $208 = 2^4 \cdot 13$. Question Is it known whether the existence of a positive $m$-cycle in such a Collatz-type problem implies the existence of a positive $m$-cycle such that its maximum $M$ and its minimum $\mu$ are related by $M = 2^k \mu$, for some $k \in \mathbb{N}$? Edit I used $m$-cycle to denote a cycle of period $m$ when it has already been standardized that $m$-cycle means a cycle with $m$ local minima. Either affirmative answer would serve just as well, however. Question 2 Does the existence of a nontrivial (positive) cycle in the $3n+1$ problem imply the existence of a nontrivial (positive) cycle with maximum $M$ and minimum $m$ such that $M = 2^k m$, for some $k \in \mathbb{N}$? Note: If you prefer to work with the odd-only transformation, the condition would be that the maximum element transforms into the minimum element.
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edited Oct 26 2011 at 7:12 |
Implication for m-cycles in Collatz-type problems.[deleted question]
Background Consider Collatz-type problems of the form $an + 1$, where $a > 2$ is a positive, odd integer (e.g., $3n + 1$, $5n +1 $, $7n + 1$, etc.). For convenience, automatically divide by two. The $3n + 1$ problem has a single known positive cycle $(1, 2)$. This cycle has period $2$, and the maximum value this cycle obtains differs from its minimum value by a power of two; that is $2 = 2 \cdot 1$. The $7n + 1$ problem has a positive cycle $(1, 4, 2)$. This cycle has period $3$, and the maximum value $4 = 2^2 \cdot 1$ is the minimum value multiplied by some power of two. The $5n + 1$ problem has three positive cycles: $(1, 3, 8, 4, 2)$, $(13, 33, 83, 208, 104, 52, 26)$, and $(17, 43, 108, 54, 27, 68, 34)$. The first cycle has period $5$, and its maximum $8 = 2^3 \cdot 1$. The other two cycles have period $7$, and one of these $7$-cycles has a maximum value $208 = 2^4 \cdot 13$. Question Is it known whether the existence of a positive $m$-cycle in such a Collatz-type problem implies the existence of a positive $m$-cycle such that its maximum $M$ and its minimum $\mu$ are related by $M = 2^k \mu$, for some $k \in \mathbb{N}$? Edit I used $m$-cycle to denote a cycle of period $m$ when it has already been standardized that $m$-cycle means a cycle with $m$ local minima. Either affirmative answer would serve just as well, however. Question 2 Does the existence of a nontrivial (positive) cycle in the $3n+1$ problem imply the existence of a nontrivial (positive) cycle with maximum $M$ and minimum $m$ such that $M = 2^k m$, for some $k \in \mathbb{N}$? Note: If you prefer to work with the odd-only transformation, the condition would be that the maximum element transforms into the minimum element. [deleted][deleted]
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edited Oct 17 2011 at 11:33 |
Background
Consider Collatz-type problems of the form $an + 1$, where $a > 2$ is a positive, odd integer (e.g., $3n + 1$, $5n +1 $, $7n + 1$, etc.). For convenience, automatically divide by two.
The $3n + 1$ problem has a single known positive cycle $(1, 2)$. This cycle has period $2$, and the maximum value this cycle obtains differs from its minimum value by a power of two; that is $2 = 2 \cdot 1$.
The $7n + 1$ problem has a positive cycle $(1, 4, 2)$. This cycle has period $3$, and the maximum value $4 = 2^2 \cdot 1$ is the minimum value multiplied by some power of two.
The $5n + 1$ problem has three positive cycles: $(1, 3, 8, 4, 2)$, $(13, 33, 83, 208, 104, 52, 26)$, and $(17, 43, 108, 54, 27, 68, 34)$. The first cycle has period $5$, and its maximum $8 = 2^3 \cdot 1$. The other two cycles have period $7$, and one of these $7$-cycles has a maximum value $208 = 2^4 \cdot 13$.
Question
Is it known whether the existence of a positive $m$-cycle in such a Collatz-type problem implies the existence of a positive $m$-cycle such that its maximum $M$ and its minimum $\mu$ are related by $M = 2^k \mu$, for some $k \in \mathbb{N}$?
Edit
I used $m$-cycle to denote a cycle of period $m$ when it has already been standardized that $m$-cycle means a cycle with $m$ local minima. Either affirmative answer would serve just as well, however.
Question 2
Does the existence of a nontrivial (positive) cycle in the $3n+1$ problem imply the existence of a nontrivial (positive) cycle with maximum $M$ and minimum $m$ such that $M = 2^k m$, for some $k \in \mathbb{N}$?
Note: If you prefer to work with only the odd-only transformation, the condition would be that the maximum element transforms into the minimum element.
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edited Oct 17 2011 at 10:59 |
Background
Consider Collatz-type problems of the form $an + 1$, where $a > 2$ is a positive, odd integer (e.g., $3n + 1$, $5n +1 $, $7n + 1$, etc.). For convenience, automatically divide by two.
The $3n + 1$ problem has a single known positive cycle $(1, 2)$. This cycle has period $2$, and the maximum value this cycle obtains differs from its minimum value by a power of two; that is $2 = 2 \cdot 1$.
The $7n + 1$ problem has a positive cycle $(1, 4, 2)$. This cycle has period $3$, and the maximum value $4 = 2^2 \cdot 1$ is the minimum value multiplied by some power of two.
The $5n + 1$ problem has three positive cycles: $(1, 3, 8, 4, 2)$, $(13, 33, 83, 208, 104, 52, 26)$, and $(17, 43, 108, 54, 27, 68, 34)$. The first cycle has period $5$, and its maximum $8 = 2^3 \cdot 1$. The other two cycles have period $7$, and one of these $7$-cycles has a maximum value $208 = 2^4 \cdot 13$.
Question
Is it known whether the existence of a positive $m$-cycle in such a Collatz-type problem implies the existence of a positive $m$-cycle such that its maximum $M$ and its minimum $m$ \mu$ are related by $M = 2^k m$, \mu$, for some $k \in \mathbb{N}$?
Edit
I used $m$-cycle to denote a cycle of period $m$ when it has already been standardized that $m$-cycle means a cycle with $m$ local minima. Either affirmative answer would serve just as well, however.
Question 2
Does the existence of a nontrivial (positive) cycle in the $3n+1$ problem imply the existence of a nontrivial (positive) cycle with maximum $M$ and minimum $m$ such that $M = 2^k m$, for some $k \in \mathbb{N}$?
Note: If you prefer to work with only the odd-only transformation, the condition would be that the maximum element transforms into the minimum element.
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edited Oct 17 2011 at 10:47 |
Background
Consider Collatz-type problems of the form $an + 1$, where $a > 2$ is a positive, odd integer (e.g., $3n + 1$, $5n +1 $, $7n + 1$, etc.). For convenience, automatically divide by two.
The $3n + 1$ problem has a single known positive cycle $(1, 2)$. This cycle has period $2$, and the maximum value this cycle obtains differs from its minimum value by a power of two; that is $2 = 2 \cdot 1$.
The $7n + 1$ problem has a positive cycle $(1, 24, 4)$2)$. This cycle has period $3$, and the maximum value $4 = 2^2 \cdot 1$ is the minimum value multiplied by some power of two.
The $5n + 1$ problem has three positive cycles: $(1, 3, 8, 4, 2)$, $(13, 33, 83, 208, 104, 52, 26)$, and $(17, 43, 108, 54, 27, 68, 34)$. The first cycle has period $5$, and its maximum $8 = 2^3 \cdot 1$. The other two cycles have period $7$, and one of these $7$-cycles has a maximum value $208 = 2^4 \cdot 13$.
Question
Is it known whether the existence of a positive $m$-cycle in such a Collatz-type problem implies the existence of a positive $m$-cycle such that its maximum $M$ and its minimum $m$ are related by $M = 2^k m$, for some $k \in \mathbb{N}$?
Edit
I used $m$-cycle to denote a cycle of period $m$ when it has already been standardized that $m$-cycle means a cycle with $m$ local minima. Either affirmative answer would serve just as well, however.
Question 2
Does the existence of a nontrivial (positive) cycle in the $3n+1$ problem imply the existence of a nontrivial (positive) cycle with maximum $M$ and minimum $m$ such that $M = 2^k m$, for some $k \in \mathbb{N}$?
Note: If you prefer to work with only the odd-only transformation, the condition would be that the maximum element transforms into the minimum element.
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edited Oct 16 2011 at 5:11 |
Background
Consider Collatz-type problems of the form $an + 1$, where $a > 2$ is a positive, odd integer (e.g., $3n + 1$, $5n +1 $, $7n + 1$, etc.). For convenience, automatically divide by two.
The $3n + 1$ problem has a single known positive cycle $(1, 2)$. This cycle has period $2$, and the maximum value this cycle obtains differs from its minimum value by a power of two; that is $2 = 2 \cdot 1$.
The $7n + 1$ problem has a positive cycle $(1, 2, 4)$. This cycle has period $3$, and the maximum value $4 = 2^2 \cdot 1$ is the minimum value multiplied by some power of two.
The $5n + 1$ problem has three positive cycles: $(1, 3, 8, 4, 2)$, $(13, 33, 83, 208, 104, 52, 26)$, and $(17, 43, 108, 54, 27, 68, 34)$. The first cycle has period $5$, and its maximum $8 = 2^3 \cdot 1$. The other two cycles have period $7$, and one of these $7$-cycles has a maximum value $208 = 2^4 \cdot 13$.
Question
Is it known whether the existence of a positive $m$-cycle in such a Collatz-type problem implies the existence of a positive $m$-cycle such that its maximum $M$ and its minimum $m$ are related by $M = 2^k m$, for some $k \in \mathbb{N}$?
Edit
I used $m$-cycle to denote a cycle of period $m$ when it has already been standardized that $m$-cycle means a cycle with $m$ local minima. Either affirmative answer would serve just as well, however.
Question 2
Does the existence of a nontrivial (positive) cycle in the $3n+1$ problem imply the existence of a nontrivial (positive) cycle with maximum $M$ and minimum $m$ such that $M = 2^k m$, for all some $k \in \mathbb{N}$?
Note: If you prefer to work with only the odd-only transformation, the condition would be that the maximum element transforms into the minimum element.
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edited Oct 16 2011 at 4:25 |
Background
Consider Collatz-type problems of the form $an + 1$, where $a > 2$ is a positive, odd integer (e.g., $3n + 1$, $5n +1 $, $7n + 1$, etc.). For convenience, automatically divide by two.
The $3n + 1$ problem has a single known positive cycle $(1, 2)$. This cycle has period $2$, and the maximum value this cycle obtains differs from its minimum value by a power of two; that is $2 = 2 \cdot 1$.
The $7n + 1$ problem has a positive cycle $(1, 2, 4)$. This cycle has period $3$, and the maximum value $4 = 2^2 \cdot 1$ is the minimum value multiplied by some power of two.
The $5n + 1$ problem has three positive cycles: $(1, 3, 8, 4, 2)$, $(13, 33, 83, 208, 104, 52, 26)$, and $(17, 43, 108, 54, 27, 68, 34)$. The first cycle has period $5$, and its maximum $8 = 2^3 \cdot 1$. The other two cycles have period $7$, and one of these $7$-cycles has a maximum value $208 = 2^4 \cdot 13$.
Question
Is it known whether the existence of a positive $m$-cycle in such a Collatz-type problem implies the existence of a positive $m$-cycle such that its maximum $M$ and its minimum $m$ are related by $M = 2^k m$, for some $k \in \mathbb{N}$?
Edit
I used $m$-cycle to denote a cycle of period $m$ when it has already been standardized that $m$-cycle means a cycle with $m$ local minima. Either answer would serve just as well, however.
Question 2
Does the existence of a nontrivial (positive) cycle in the $3n+1$ problem imply the existence of a nontrivial (positive) cycle with maximum $M$ and minimum $m$ such that $M = 2^k m$, for all $k \in \mathbb{N}$?
Note: If you prefer to work with only the odd-only transformation, the condition would be that the maximum element transforms into the minimum element.
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edited Oct 14 2011 at 13:01 |
Implication for m-cycles in Collatz type Collatz-type problems.
Background
Consider Collatz-type problems of the form $an + 1$, where $a > 2$ is a positive, odd integer (e.g., $3n + 1$, $5n +1 $, $7n + 1$, etc.). For convenience, automatically divide by two.
The $3n + 1$ problem has a single known positive cycle $(1, 2)$. This cycle has period two, $2$, and the maximum value this cycle obtains differs from its minimum value by a power of two; that is $2 = 2 \cdot 1$.
The $7n + 1$ problem has a positive cycle $(1, 2, 4)$. This cycle has period three, $3$, and the maximum value $4 = 2^2 \cdot 1$ is the minimum value times multiplied by some power of two.
The $5n + 1$ problem has three positive cycles: $(1, 3, 8, 4, 2)$, $(13, 33, 83, 208, 104, 52, 26)$, and $(17, 43, 108, 54, 27, 68, 34)$. The first cycle has period $5$, and its maximum $8 = 2^3 \cdot 1$. The other two cycles have period $7$, and one of these $7$-cycles has a maximum value $208 = 2^4 \cdot 13$.
Question
Is it known whether the existence of a positive $m$-cycle in such a Collatz-type problem implies the existence of a positive $m$-cycle such that its maximum $M$ and its minimum $m$ are related by $M = 2^k m$, for some $k \in \mathbb{N}$?
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edited Oct 14 2011 at 12:32 |
Background
Consider Collatz-type problems of the form $an + 1$, where $a > 2$ is a positive, odd integer (e.g., $3n + 1$, $5n +1 $, $7n + 1$, etc.). For convenience, automatically divide by two.
The $3n + 1$ problem has a single known positive cycle $(1, 2)$. This cycle has period two, and the maximum value this cycle obtains differs from its minimum value by a power of two; that is $4 2 = 2 \cdot 1$.
The $7n + 1$ problem has a positive cycle $(1, 2, 4)$. This cycle has period three, and the maximum value $4 = 2^2 \cdot 1$ is the minimum value times some power of two.
The $5n + 1$ problem has three positive cycles: $(1, 3, 8, 4, 2)$, $(13, 33, 83, 208, 104, 52, 26)$, and $(17, 43, 108, 54, 27, 68, 34)$. The first cycle has period $5$, and its maximum $8 = 2^3 \cdot 1$. The other two cycles have period $7$, and one of these $7$-cycles has a maximum value $208 = 2^4 \cdot 13$.
Question
Is it known whether the existence of a positive $m$-cycle in such a Collatz-type problem implies the existence of a positive $m$-cycle such that its maximum $M$ and its minimum $m$ are related by $M = 2^k m$, for some $k \in \mathbb{N}$?
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asked Oct 14 2011 at 12:15 |
Implication for m-cycles in Collatz type problems.
Background
Consider Collatz-type problems of the form $an + 1$, where $a > 2$ is a positive, odd integer (e.g., $3n + 1$, $5n +1 $, $7n + 1$, etc.). For convenience, automatically divide by two.
The $3n + 1$ problem has a single known positive cycle $(1, 2)$. This cycle has period two, and the maximum value this cycle obtains differs from its minimum value by a power of two; that is $4 = 2 \cdot 1$.
The $7n + 1$ problem has a positive cycle $(1, 2, 4)$. This cycle has period three, and the maximum value $4 = 2^2 \cdot 1$ is the minimum value times some power of two.
The $5n + 1$ problem has three positive cycles: $(1, 3, 8, 4, 2)$, $(13, 33, 83, 208, 104, 52, 26)$, and $(17, 43, 108, 54, 27, 68, 34)$. The first cycle has period $5$, and its maximum $8 = 2^3 \cdot 1$. The other two cycles have period $7$, and one of these $7$-cycles has a maximum value $208 = 2^4 \cdot 13$
Question
Is it known whether the existence of a positive $m$-cycle in such a Collatz-type problem implies the existence of a positive $m$-cycle such that its maximum $M$ and its minimum $m$ are related by $M = 2^k m$, for some $k \in \mathbb{N}$?
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