Collatz 3n+1 problem

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Does it suffice to say that, using the tree/reverse of 3n+1 problem, if we can show that we can generate all positive integers starting from integer 1, including 1 itself, is equivalent to proving indirectly the Collatz Conjecture?




number-theory dynamical-systems collatz




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    What do you mean by "generate"?
    – Lord Shark the Unknown
    yesterday










  • yes, exactly, like say a generator: g(1) = x
    – busy Ang
    yesterday














up vote
0
down vote

favorite












Does it suffice to say that, using the tree/reverse of 3n+1 problem, if we can show that we can generate all positive integers starting from integer 1, including 1 itself, is equivalent to proving indirectly the Collatz Conjecture?




number-theory dynamical-systems collatz




share|cite|improve this question















  • 1




    What do you mean by "generate"?
    – Lord Shark the Unknown
    yesterday










  • yes, exactly, like say a generator: g(1) = x
    – busy Ang
    yesterday












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Does it suffice to say that, using the tree/reverse of 3n+1 problem, if we can show that we can generate all positive integers starting from integer 1, including 1 itself, is equivalent to proving indirectly the Collatz Conjecture?




number-theory dynamical-systems collatz




share|cite|improve this question











Does it suffice to say that, using the tree/reverse of 3n+1 problem, if we can show that we can generate all positive integers starting from integer 1, including 1 itself, is equivalent to proving indirectly the Collatz Conjecture?





number-theory dynamical-systems collatz





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked yesterday









busy Ang

183




183







  • 1




    What do you mean by "generate"?
    – Lord Shark the Unknown
    yesterday










  • yes, exactly, like say a generator: g(1) = x
    – busy Ang
    yesterday












  • 1




    What do you mean by "generate"?
    – Lord Shark the Unknown
    yesterday










  • yes, exactly, like say a generator: g(1) = x
    – busy Ang
    yesterday







1




1




What do you mean by "generate"?
– Lord Shark the Unknown
yesterday




What do you mean by "generate"?
– Lord Shark the Unknown
yesterday












yes, exactly, like say a generator: g(1) = x
– busy Ang
yesterday




yes, exactly, like say a generator: g(1) = x
– busy Ang
yesterday










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










We can indeed easily rephrase the Collatz conjecture to be about trying to generate every number starting from $1$ by applying a couple operations, but you have to be careful what you mean by "generate."



Let $f(x)=2x$ and $g(x)=x-1over 3$. It's natural to try to walk from $1$ to $n$ via $f$ and $g$ to show that $n$ isn't a counterexample to Collatz, but this doesn't always go smoothly: e.g. $g$ lets us map $13$ to $4$, but in a Collatz sequence we'll never go from $4$ to $13$ since $4$ is even (we would go instead from $4$ to $2$). So starting from $1$ and applying $f$ and $g$ repeatedly to get $n$ does not necessarily show that $n$ isn't a counterexample to Collatz, since when we reverse this walk we might get a sequence which doesn't follow the rules.



Specifically, we can never apply $g$ if the result will be even. The following is true:




Suppose there is a sequence $a_1,..., a_k$ with $a_1=1$, $a_k=n$, for each $i$ we have $a_i+1=f(a_i)$ or $a_i+1=g(a_i)$, and whenever $a_i+1$ is even we have $a_i+1=f(a_i)$. Then $n$ is not a counterexample to Collatz.




However, merely talking about "generating" numbers without this crucial restriction on when we're allowed to apply a given operation gives a very misleading idea.






share|cite|improve this answer





















  • yes I definitely agree, and certainly there will be restrictions by properly defining how the generator works in spitting out the set of positive integers, but my main concern is simply, if we have such generator that follows the premise of the conjecture, say there is g(k) = g(1) --> x:x∈ℕ, is that the same as proving the conjecture? To make it precise, we say that on particular operation on g(1) it maps to 3 that in turns 3 maps to 10 then finally 10 maps to 5, all as part of operations that defines the rule for g(1).
    – busy Ang
    yesterday










  • in addition, such generator will map those corresponding integers based on a parameter, k≥1. so for every unique k, g(1) generates corresponding integers.
    – busy Ang
    yesterday










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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










We can indeed easily rephrase the Collatz conjecture to be about trying to generate every number starting from $1$ by applying a couple operations, but you have to be careful what you mean by "generate."



Let $f(x)=2x$ and $g(x)=x-1over 3$. It's natural to try to walk from $1$ to $n$ via $f$ and $g$ to show that $n$ isn't a counterexample to Collatz, but this doesn't always go smoothly: e.g. $g$ lets us map $13$ to $4$, but in a Collatz sequence we'll never go from $4$ to $13$ since $4$ is even (we would go instead from $4$ to $2$). So starting from $1$ and applying $f$ and $g$ repeatedly to get $n$ does not necessarily show that $n$ isn't a counterexample to Collatz, since when we reverse this walk we might get a sequence which doesn't follow the rules.



Specifically, we can never apply $g$ if the result will be even. The following is true:




Suppose there is a sequence $a_1,..., a_k$ with $a_1=1$, $a_k=n$, for each $i$ we have $a_i+1=f(a_i)$ or $a_i+1=g(a_i)$, and whenever $a_i+1$ is even we have $a_i+1=f(a_i)$. Then $n$ is not a counterexample to Collatz.




However, merely talking about "generating" numbers without this crucial restriction on when we're allowed to apply a given operation gives a very misleading idea.






share|cite|improve this answer





















  • yes I definitely agree, and certainly there will be restrictions by properly defining how the generator works in spitting out the set of positive integers, but my main concern is simply, if we have such generator that follows the premise of the conjecture, say there is g(k) = g(1) --> x:x∈ℕ, is that the same as proving the conjecture? To make it precise, we say that on particular operation on g(1) it maps to 3 that in turns 3 maps to 10 then finally 10 maps to 5, all as part of operations that defines the rule for g(1).
    – busy Ang
    yesterday










  • in addition, such generator will map those corresponding integers based on a parameter, k≥1. so for every unique k, g(1) generates corresponding integers.
    – busy Ang
    yesterday














up vote
1
down vote



accepted










We can indeed easily rephrase the Collatz conjecture to be about trying to generate every number starting from $1$ by applying a couple operations, but you have to be careful what you mean by "generate."



Let $f(x)=2x$ and $g(x)=x-1over 3$. It's natural to try to walk from $1$ to $n$ via $f$ and $g$ to show that $n$ isn't a counterexample to Collatz, but this doesn't always go smoothly: e.g. $g$ lets us map $13$ to $4$, but in a Collatz sequence we'll never go from $4$ to $13$ since $4$ is even (we would go instead from $4$ to $2$). So starting from $1$ and applying $f$ and $g$ repeatedly to get $n$ does not necessarily show that $n$ isn't a counterexample to Collatz, since when we reverse this walk we might get a sequence which doesn't follow the rules.



Specifically, we can never apply $g$ if the result will be even. The following is true:




Suppose there is a sequence $a_1,..., a_k$ with $a_1=1$, $a_k=n$, for each $i$ we have $a_i+1=f(a_i)$ or $a_i+1=g(a_i)$, and whenever $a_i+1$ is even we have $a_i+1=f(a_i)$. Then $n$ is not a counterexample to Collatz.




However, merely talking about "generating" numbers without this crucial restriction on when we're allowed to apply a given operation gives a very misleading idea.






share|cite|improve this answer





















  • yes I definitely agree, and certainly there will be restrictions by properly defining how the generator works in spitting out the set of positive integers, but my main concern is simply, if we have such generator that follows the premise of the conjecture, say there is g(k) = g(1) --> x:x∈ℕ, is that the same as proving the conjecture? To make it precise, we say that on particular operation on g(1) it maps to 3 that in turns 3 maps to 10 then finally 10 maps to 5, all as part of operations that defines the rule for g(1).
    – busy Ang
    yesterday










  • in addition, such generator will map those corresponding integers based on a parameter, k≥1. so for every unique k, g(1) generates corresponding integers.
    – busy Ang
    yesterday












up vote
1
down vote



accepted







up vote
1
down vote



accepted






We can indeed easily rephrase the Collatz conjecture to be about trying to generate every number starting from $1$ by applying a couple operations, but you have to be careful what you mean by "generate."



Let $f(x)=2x$ and $g(x)=x-1over 3$. It's natural to try to walk from $1$ to $n$ via $f$ and $g$ to show that $n$ isn't a counterexample to Collatz, but this doesn't always go smoothly: e.g. $g$ lets us map $13$ to $4$, but in a Collatz sequence we'll never go from $4$ to $13$ since $4$ is even (we would go instead from $4$ to $2$). So starting from $1$ and applying $f$ and $g$ repeatedly to get $n$ does not necessarily show that $n$ isn't a counterexample to Collatz, since when we reverse this walk we might get a sequence which doesn't follow the rules.



Specifically, we can never apply $g$ if the result will be even. The following is true:




Suppose there is a sequence $a_1,..., a_k$ with $a_1=1$, $a_k=n$, for each $i$ we have $a_i+1=f(a_i)$ or $a_i+1=g(a_i)$, and whenever $a_i+1$ is even we have $a_i+1=f(a_i)$. Then $n$ is not a counterexample to Collatz.




However, merely talking about "generating" numbers without this crucial restriction on when we're allowed to apply a given operation gives a very misleading idea.






share|cite|improve this answer













We can indeed easily rephrase the Collatz conjecture to be about trying to generate every number starting from $1$ by applying a couple operations, but you have to be careful what you mean by "generate."



Let $f(x)=2x$ and $g(x)=x-1over 3$. It's natural to try to walk from $1$ to $n$ via $f$ and $g$ to show that $n$ isn't a counterexample to Collatz, but this doesn't always go smoothly: e.g. $g$ lets us map $13$ to $4$, but in a Collatz sequence we'll never go from $4$ to $13$ since $4$ is even (we would go instead from $4$ to $2$). So starting from $1$ and applying $f$ and $g$ repeatedly to get $n$ does not necessarily show that $n$ isn't a counterexample to Collatz, since when we reverse this walk we might get a sequence which doesn't follow the rules.



Specifically, we can never apply $g$ if the result will be even. The following is true:




Suppose there is a sequence $a_1,..., a_k$ with $a_1=1$, $a_k=n$, for each $i$ we have $a_i+1=f(a_i)$ or $a_i+1=g(a_i)$, and whenever $a_i+1$ is even we have $a_i+1=f(a_i)$. Then $n$ is not a counterexample to Collatz.




However, merely talking about "generating" numbers without this crucial restriction on when we're allowed to apply a given operation gives a very misleading idea.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered yesterday









Noah Schweber

110k9138258




110k9138258











  • yes I definitely agree, and certainly there will be restrictions by properly defining how the generator works in spitting out the set of positive integers, but my main concern is simply, if we have such generator that follows the premise of the conjecture, say there is g(k) = g(1) --> x:x∈ℕ, is that the same as proving the conjecture? To make it precise, we say that on particular operation on g(1) it maps to 3 that in turns 3 maps to 10 then finally 10 maps to 5, all as part of operations that defines the rule for g(1).
    – busy Ang
    yesterday










  • in addition, such generator will map those corresponding integers based on a parameter, k≥1. so for every unique k, g(1) generates corresponding integers.
    – busy Ang
    yesterday
















  • yes I definitely agree, and certainly there will be restrictions by properly defining how the generator works in spitting out the set of positive integers, but my main concern is simply, if we have such generator that follows the premise of the conjecture, say there is g(k) = g(1) --> x:x∈ℕ, is that the same as proving the conjecture? To make it precise, we say that on particular operation on g(1) it maps to 3 that in turns 3 maps to 10 then finally 10 maps to 5, all as part of operations that defines the rule for g(1).
    – busy Ang
    yesterday










  • in addition, such generator will map those corresponding integers based on a parameter, k≥1. so for every unique k, g(1) generates corresponding integers.
    – busy Ang
    yesterday















yes I definitely agree, and certainly there will be restrictions by properly defining how the generator works in spitting out the set of positive integers, but my main concern is simply, if we have such generator that follows the premise of the conjecture, say there is g(k) = g(1) --> x:x∈ℕ, is that the same as proving the conjecture? To make it precise, we say that on particular operation on g(1) it maps to 3 that in turns 3 maps to 10 then finally 10 maps to 5, all as part of operations that defines the rule for g(1).
– busy Ang
yesterday




yes I definitely agree, and certainly there will be restrictions by properly defining how the generator works in spitting out the set of positive integers, but my main concern is simply, if we have such generator that follows the premise of the conjecture, say there is g(k) = g(1) --> x:x∈ℕ, is that the same as proving the conjecture? To make it precise, we say that on particular operation on g(1) it maps to 3 that in turns 3 maps to 10 then finally 10 maps to 5, all as part of operations that defines the rule for g(1).
– busy Ang
yesterday












in addition, such generator will map those corresponding integers based on a parameter, k≥1. so for every unique k, g(1) generates corresponding integers.
– busy Ang
yesterday




in addition, such generator will map those corresponding integers based on a parameter, k≥1. so for every unique k, g(1) generates corresponding integers.
– busy Ang
yesterday












 

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