Finding whether $dfrac2^k - (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 … + 2^t_0+t_1+…+t_n-1)3^n$ can describe all integers

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I was working on the Collatz conjecture(while not expecting to get anywhere) and I suspect that if it is possible to show that
$$dfrac2^k - (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
can represent all integers for some $k, n$ and $t_i$s, it is possible to prove the Collatz conjecture. If $t_0 geq 1$ and $t_i > 1$ for $i>0$, $iin mathbbZ^+$



While I am aware that there is no easy way to actually prove this, I was hoping for a recommended method of approaching this problem. If requested, I will put the derivation below. However, even if the derivation is incorrect, I am still interested in whether there is a way to prove or disprove that the above can represent all integers. I have only tried one approach so far and I will put it below(even if I was unsuccessful). If this has not been a simplification, I will appreciate if that was pointed out as well.



Proof by induction



Given
$$z = dfrac2^k - (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
must satisfy $z in mathbbZ^+$. At $z=1$, as
$$1 = dfrac2^4-(2cdot3 + 2^0)3^2 = dfrac99 = 1$$
I here assumed that at $z=a$,
$$a = dfrac2^k - (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
Here, if we were to look at $z=a+1$,
$$a+1 = dfrac2^k + 3^n- (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
However, I was not able to think about anything from this point.




number-theory collatz




share|cite|improve this question



















  • This ansatz is well known, and there have been several attempts to prove that indeed all numbers can be represented by this structure - but with no success so far. Only with restrictions on the sequence of $t_0,t_1,...$ some progress could be made, for instance if all $t_k$ except the last would equal $1$ this leads to the "1-cycle"-problem and there has been some success with analyses on this (proof of nonexistence etc).
    – Gottfried Helms
    Jul 24 at 12:25










  • Ah, got it. Thank you very much! @GottfriedHelms !
    – Isamu Isozaki
    Jul 25 at 21:04










  • I'll try another approach!
    – Isamu Isozaki
    Jul 25 at 21:05










  • Yes, this looks equivalent to some previously-discovered representations of the Collatz problem, which are summarised fairly exhaustively by Lagarias' well-known paper. I think this equation is more revealing once you have divided the numerator throughout by $3^n$ and you can also express $t_0..$ as individual terms of a strictly increasing sequence of powers of $2$. My personal opinion is not to give up on this approach, rather I prefer to better understand what forms of such sequences are possible, and how they can be put into bijection with the integers.
    – Robert Frost
    Jul 26 at 11:37






  • 1




    @RobertFrost Oh, that is quite interesting! I have not yet read that famous paper and I do not understand the second portion completely yet I will read it and I will give it another try! I have tried another approach yet while I think I made a bit of progress, I was actually thinking about finding another method like the one here. Thank you!
    – Isamu Isozaki
    Jul 28 at 8:36














up vote
2
down vote

favorite












I was working on the Collatz conjecture(while not expecting to get anywhere) and I suspect that if it is possible to show that
$$dfrac2^k - (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
can represent all integers for some $k, n$ and $t_i$s, it is possible to prove the Collatz conjecture. If $t_0 geq 1$ and $t_i > 1$ for $i>0$, $iin mathbbZ^+$



While I am aware that there is no easy way to actually prove this, I was hoping for a recommended method of approaching this problem. If requested, I will put the derivation below. However, even if the derivation is incorrect, I am still interested in whether there is a way to prove or disprove that the above can represent all integers. I have only tried one approach so far and I will put it below(even if I was unsuccessful). If this has not been a simplification, I will appreciate if that was pointed out as well.



Proof by induction



Given
$$z = dfrac2^k - (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
must satisfy $z in mathbbZ^+$. At $z=1$, as
$$1 = dfrac2^4-(2cdot3 + 2^0)3^2 = dfrac99 = 1$$
I here assumed that at $z=a$,
$$a = dfrac2^k - (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
Here, if we were to look at $z=a+1$,
$$a+1 = dfrac2^k + 3^n- (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
However, I was not able to think about anything from this point.




number-theory collatz




share|cite|improve this question



















  • This ansatz is well known, and there have been several attempts to prove that indeed all numbers can be represented by this structure - but with no success so far. Only with restrictions on the sequence of $t_0,t_1,...$ some progress could be made, for instance if all $t_k$ except the last would equal $1$ this leads to the "1-cycle"-problem and there has been some success with analyses on this (proof of nonexistence etc).
    – Gottfried Helms
    Jul 24 at 12:25










  • Ah, got it. Thank you very much! @GottfriedHelms !
    – Isamu Isozaki
    Jul 25 at 21:04










  • I'll try another approach!
    – Isamu Isozaki
    Jul 25 at 21:05










  • Yes, this looks equivalent to some previously-discovered representations of the Collatz problem, which are summarised fairly exhaustively by Lagarias' well-known paper. I think this equation is more revealing once you have divided the numerator throughout by $3^n$ and you can also express $t_0..$ as individual terms of a strictly increasing sequence of powers of $2$. My personal opinion is not to give up on this approach, rather I prefer to better understand what forms of such sequences are possible, and how they can be put into bijection with the integers.
    – Robert Frost
    Jul 26 at 11:37






  • 1




    @RobertFrost Oh, that is quite interesting! I have not yet read that famous paper and I do not understand the second portion completely yet I will read it and I will give it another try! I have tried another approach yet while I think I made a bit of progress, I was actually thinking about finding another method like the one here. Thank you!
    – Isamu Isozaki
    Jul 28 at 8:36












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I was working on the Collatz conjecture(while not expecting to get anywhere) and I suspect that if it is possible to show that
$$dfrac2^k - (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
can represent all integers for some $k, n$ and $t_i$s, it is possible to prove the Collatz conjecture. If $t_0 geq 1$ and $t_i > 1$ for $i>0$, $iin mathbbZ^+$



While I am aware that there is no easy way to actually prove this, I was hoping for a recommended method of approaching this problem. If requested, I will put the derivation below. However, even if the derivation is incorrect, I am still interested in whether there is a way to prove or disprove that the above can represent all integers. I have only tried one approach so far and I will put it below(even if I was unsuccessful). If this has not been a simplification, I will appreciate if that was pointed out as well.



Proof by induction



Given
$$z = dfrac2^k - (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
must satisfy $z in mathbbZ^+$. At $z=1$, as
$$1 = dfrac2^4-(2cdot3 + 2^0)3^2 = dfrac99 = 1$$
I here assumed that at $z=a$,
$$a = dfrac2^k - (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
Here, if we were to look at $z=a+1$,
$$a+1 = dfrac2^k + 3^n- (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
However, I was not able to think about anything from this point.




number-theory collatz




share|cite|improve this question











I was working on the Collatz conjecture(while not expecting to get anywhere) and I suspect that if it is possible to show that
$$dfrac2^k - (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
can represent all integers for some $k, n$ and $t_i$s, it is possible to prove the Collatz conjecture. If $t_0 geq 1$ and $t_i > 1$ for $i>0$, $iin mathbbZ^+$



While I am aware that there is no easy way to actually prove this, I was hoping for a recommended method of approaching this problem. If requested, I will put the derivation below. However, even if the derivation is incorrect, I am still interested in whether there is a way to prove or disprove that the above can represent all integers. I have only tried one approach so far and I will put it below(even if I was unsuccessful). If this has not been a simplification, I will appreciate if that was pointed out as well.



Proof by induction



Given
$$z = dfrac2^k - (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
must satisfy $z in mathbbZ^+$. At $z=1$, as
$$1 = dfrac2^4-(2cdot3 + 2^0)3^2 = dfrac99 = 1$$
I here assumed that at $z=a$,
$$a = dfrac2^k - (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
Here, if we were to look at $z=a+1$,
$$a+1 = dfrac2^k + 3^n- (2cdot3^n-1 + 2^t_03^n-2 + 2^t_0+t_13^n-3 .... + 2^t_0+t_1+...+t_n-1)3^n$$
However, I was not able to think about anything from this point.





number-theory collatz





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 24 at 2:00









Isamu Isozaki

839




839











  • This ansatz is well known, and there have been several attempts to prove that indeed all numbers can be represented by this structure - but with no success so far. Only with restrictions on the sequence of $t_0,t_1,...$ some progress could be made, for instance if all $t_k$ except the last would equal $1$ this leads to the "1-cycle"-problem and there has been some success with analyses on this (proof of nonexistence etc).
    – Gottfried Helms
    Jul 24 at 12:25










  • Ah, got it. Thank you very much! @GottfriedHelms !
    – Isamu Isozaki
    Jul 25 at 21:04










  • I'll try another approach!
    – Isamu Isozaki
    Jul 25 at 21:05










  • Yes, this looks equivalent to some previously-discovered representations of the Collatz problem, which are summarised fairly exhaustively by Lagarias' well-known paper. I think this equation is more revealing once you have divided the numerator throughout by $3^n$ and you can also express $t_0..$ as individual terms of a strictly increasing sequence of powers of $2$. My personal opinion is not to give up on this approach, rather I prefer to better understand what forms of such sequences are possible, and how they can be put into bijection with the integers.
    – Robert Frost
    Jul 26 at 11:37






  • 1




    @RobertFrost Oh, that is quite interesting! I have not yet read that famous paper and I do not understand the second portion completely yet I will read it and I will give it another try! I have tried another approach yet while I think I made a bit of progress, I was actually thinking about finding another method like the one here. Thank you!
    – Isamu Isozaki
    Jul 28 at 8:36
















  • This ansatz is well known, and there have been several attempts to prove that indeed all numbers can be represented by this structure - but with no success so far. Only with restrictions on the sequence of $t_0,t_1,...$ some progress could be made, for instance if all $t_k$ except the last would equal $1$ this leads to the "1-cycle"-problem and there has been some success with analyses on this (proof of nonexistence etc).
    – Gottfried Helms
    Jul 24 at 12:25










  • Ah, got it. Thank you very much! @GottfriedHelms !
    – Isamu Isozaki
    Jul 25 at 21:04










  • I'll try another approach!
    – Isamu Isozaki
    Jul 25 at 21:05










  • Yes, this looks equivalent to some previously-discovered representations of the Collatz problem, which are summarised fairly exhaustively by Lagarias' well-known paper. I think this equation is more revealing once you have divided the numerator throughout by $3^n$ and you can also express $t_0..$ as individual terms of a strictly increasing sequence of powers of $2$. My personal opinion is not to give up on this approach, rather I prefer to better understand what forms of such sequences are possible, and how they can be put into bijection with the integers.
    – Robert Frost
    Jul 26 at 11:37






  • 1




    @RobertFrost Oh, that is quite interesting! I have not yet read that famous paper and I do not understand the second portion completely yet I will read it and I will give it another try! I have tried another approach yet while I think I made a bit of progress, I was actually thinking about finding another method like the one here. Thank you!
    – Isamu Isozaki
    Jul 28 at 8:36















This ansatz is well known, and there have been several attempts to prove that indeed all numbers can be represented by this structure - but with no success so far. Only with restrictions on the sequence of $t_0,t_1,...$ some progress could be made, for instance if all $t_k$ except the last would equal $1$ this leads to the "1-cycle"-problem and there has been some success with analyses on this (proof of nonexistence etc).
– Gottfried Helms
Jul 24 at 12:25




This ansatz is well known, and there have been several attempts to prove that indeed all numbers can be represented by this structure - but with no success so far. Only with restrictions on the sequence of $t_0,t_1,...$ some progress could be made, for instance if all $t_k$ except the last would equal $1$ this leads to the "1-cycle"-problem and there has been some success with analyses on this (proof of nonexistence etc).
– Gottfried Helms
Jul 24 at 12:25












Ah, got it. Thank you very much! @GottfriedHelms !
– Isamu Isozaki
Jul 25 at 21:04




Ah, got it. Thank you very much! @GottfriedHelms !
– Isamu Isozaki
Jul 25 at 21:04












I'll try another approach!
– Isamu Isozaki
Jul 25 at 21:05




I'll try another approach!
– Isamu Isozaki
Jul 25 at 21:05












Yes, this looks equivalent to some previously-discovered representations of the Collatz problem, which are summarised fairly exhaustively by Lagarias' well-known paper. I think this equation is more revealing once you have divided the numerator throughout by $3^n$ and you can also express $t_0..$ as individual terms of a strictly increasing sequence of powers of $2$. My personal opinion is not to give up on this approach, rather I prefer to better understand what forms of such sequences are possible, and how they can be put into bijection with the integers.
– Robert Frost
Jul 26 at 11:37




Yes, this looks equivalent to some previously-discovered representations of the Collatz problem, which are summarised fairly exhaustively by Lagarias' well-known paper. I think this equation is more revealing once you have divided the numerator throughout by $3^n$ and you can also express $t_0..$ as individual terms of a strictly increasing sequence of powers of $2$. My personal opinion is not to give up on this approach, rather I prefer to better understand what forms of such sequences are possible, and how they can be put into bijection with the integers.
– Robert Frost
Jul 26 at 11:37




1




1




@RobertFrost Oh, that is quite interesting! I have not yet read that famous paper and I do not understand the second portion completely yet I will read it and I will give it another try! I have tried another approach yet while I think I made a bit of progress, I was actually thinking about finding another method like the one here. Thank you!
– Isamu Isozaki
Jul 28 at 8:36




@RobertFrost Oh, that is quite interesting! I have not yet read that famous paper and I do not understand the second portion completely yet I will read it and I will give it another try! I have tried another approach yet while I think I made a bit of progress, I was actually thinking about finding another method like the one here. Thank you!
– Isamu Isozaki
Jul 28 at 8:36















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