Is it possible to find such a discrete non-cyclic mathematical function?












1












$begingroup$


Is it possible to find such a discrete non-cyclic mathematical function,with the following features?




Let $$f: mathbb{Z^{+}} rightarrow left{0,1,2 right}$$




And for any $ninmathbb{Z^{+}},left{ f(3n-2),f(3n-1),f(3n)right}$ must be equal $left{0,1,2 right}$, or $left{ 0,2,1right},left{1,2,0 right},left{1,0,2 right} left{2,1,0 right},left{ 2,0,1right}$



Discrete non-cyclic means, for example:



$$color{purple}{left{f(1),f(2),f(3),f(4),f(5),f(6),f(7),f(8),...f(n) right}:=}color{red}{left{color{red}{0,1,2,2,1,0,1,0,2;} color{green}{0,1,2,2,1,0,1,0,2;} color{blue}{0,1,2,2,1,0,1,0,2;}... right}}$$



Thus, we have discrete cyclic sequence. Because,



$f(9)=f(1), f(10)=f(2), f(11)=f(3),...,f(16)=f(8).$ In other words,
$f(n)=f(n-8)$, where $n≥9, ninmathbb{Z^{+}}$ and cycle length is equal to $8.$



I hope I'm asking the question clearly.



I am looking for a mathematical function that takes discrete non-cyclic values. I'm not saying to MSE "Find such a function." What I want to know is whether such a special function can exist mathematically.



Thank you very much.






sequences-and-series functions discrete-mathematics soft-question research







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes, such a function "exists mathematically", and we can easily describe such a function. For instance, take $$ f(x) = begin{cases} 1 & x text{ is a perfect square}\ 0 & text{otherwise} end{cases} $$
    $endgroup$
    – Omnomnomnom
    Dec 20 '18 at 20:54








  • 1




    $begingroup$
    The typical term for what you're calling "cyclic" is periodic.
    $endgroup$
    – Omnomnomnom
    Dec 20 '18 at 21:00










  • $begingroup$
    Your question is completely unclear: why do you introduce the sets $D(f)$ and $E(f)$ but never mention then again? What is $n$ in the "definition" of $D(f)$. An example is an inadequate replacement for a definition, so we have no idea what you mean by "discrete non-cyclic" particularly as your example makes no sense: how do $f$ and $n$ on the left-hand side of the symbol ${:}{=}$ relate to the numbers on the right-hand side?
    $endgroup$
    – Rob Arthan
    Dec 20 '18 at 21:01












  • $begingroup$
    Concretely, the sequence I'm describing is $$ {1,overbrace{0,0}^2,1,overbrace{0,dots,0}^4,1,overbrace{0,dots,0}^6,1,dots} $$
    $endgroup$
    – Omnomnomnom
    Dec 20 '18 at 21:03








  • 1




    $begingroup$
    I suppose $D(f)$ means the domain of $f$. But this is non-standard notation. And in any case, the domain of $f$ is $Bbb Z^+$, not ${1,2,3,4,5,ldots,n}$. And $E(f)$ means the range of $f$? (I can't think what $E$ stands for.) But if so, why not just write $f:Bbb Z^+to{0,1,2}$?
    $endgroup$
    – TonyK
    Dec 20 '18 at 21:07
















1












$begingroup$


Is it possible to find such a discrete non-cyclic mathematical function,with the following features?




Let $$f: mathbb{Z^{+}} rightarrow left{0,1,2 right}$$




And for any $ninmathbb{Z^{+}},left{ f(3n-2),f(3n-1),f(3n)right}$ must be equal $left{0,1,2 right}$, or $left{ 0,2,1right},left{1,2,0 right},left{1,0,2 right} left{2,1,0 right},left{ 2,0,1right}$



Discrete non-cyclic means, for example:



$$color{purple}{left{f(1),f(2),f(3),f(4),f(5),f(6),f(7),f(8),...f(n) right}:=}color{red}{left{color{red}{0,1,2,2,1,0,1,0,2;} color{green}{0,1,2,2,1,0,1,0,2;} color{blue}{0,1,2,2,1,0,1,0,2;}... right}}$$



Thus, we have discrete cyclic sequence. Because,



$f(9)=f(1), f(10)=f(2), f(11)=f(3),...,f(16)=f(8).$ In other words,
$f(n)=f(n-8)$, where $n≥9, ninmathbb{Z^{+}}$ and cycle length is equal to $8.$



I hope I'm asking the question clearly.



I am looking for a mathematical function that takes discrete non-cyclic values. I'm not saying to MSE "Find such a function." What I want to know is whether such a special function can exist mathematically.



Thank you very much.






sequences-and-series functions discrete-mathematics soft-question research







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Yes, such a function "exists mathematically", and we can easily describe such a function. For instance, take $$ f(x) = begin{cases} 1 & x text{ is a perfect square}\ 0 & text{otherwise} end{cases} $$
    $endgroup$
    – Omnomnomnom
    Dec 20 '18 at 20:54








  • 1




    $begingroup$
    The typical term for what you're calling "cyclic" is periodic.
    $endgroup$
    – Omnomnomnom
    Dec 20 '18 at 21:00










  • $begingroup$
    Your question is completely unclear: why do you introduce the sets $D(f)$ and $E(f)$ but never mention then again? What is $n$ in the "definition" of $D(f)$. An example is an inadequate replacement for a definition, so we have no idea what you mean by "discrete non-cyclic" particularly as your example makes no sense: how do $f$ and $n$ on the left-hand side of the symbol ${:}{=}$ relate to the numbers on the right-hand side?
    $endgroup$
    – Rob Arthan
    Dec 20 '18 at 21:01












  • $begingroup$
    Concretely, the sequence I'm describing is $$ {1,overbrace{0,0}^2,1,overbrace{0,dots,0}^4,1,overbrace{0,dots,0}^6,1,dots} $$
    $endgroup$
    – Omnomnomnom
    Dec 20 '18 at 21:03








  • 1




    $begingroup$
    I suppose $D(f)$ means the domain of $f$. But this is non-standard notation. And in any case, the domain of $f$ is $Bbb Z^+$, not ${1,2,3,4,5,ldots,n}$. And $E(f)$ means the range of $f$? (I can't think what $E$ stands for.) But if so, why not just write $f:Bbb Z^+to{0,1,2}$?
    $endgroup$
    – TonyK
    Dec 20 '18 at 21:07














1












1








1





$begingroup$


Is it possible to find such a discrete non-cyclic mathematical function,with the following features?




Let $$f: mathbb{Z^{+}} rightarrow left{0,1,2 right}$$




And for any $ninmathbb{Z^{+}},left{ f(3n-2),f(3n-1),f(3n)right}$ must be equal $left{0,1,2 right}$, or $left{ 0,2,1right},left{1,2,0 right},left{1,0,2 right} left{2,1,0 right},left{ 2,0,1right}$



Discrete non-cyclic means, for example:



$$color{purple}{left{f(1),f(2),f(3),f(4),f(5),f(6),f(7),f(8),...f(n) right}:=}color{red}{left{color{red}{0,1,2,2,1,0,1,0,2;} color{green}{0,1,2,2,1,0,1,0,2;} color{blue}{0,1,2,2,1,0,1,0,2;}... right}}$$



Thus, we have discrete cyclic sequence. Because,



$f(9)=f(1), f(10)=f(2), f(11)=f(3),...,f(16)=f(8).$ In other words,
$f(n)=f(n-8)$, where $n≥9, ninmathbb{Z^{+}}$ and cycle length is equal to $8.$



I hope I'm asking the question clearly.



I am looking for a mathematical function that takes discrete non-cyclic values. I'm not saying to MSE "Find such a function." What I want to know is whether such a special function can exist mathematically.



Thank you very much.






sequences-and-series functions discrete-mathematics soft-question research







share|cite|improve this question











$endgroup$




Is it possible to find such a discrete non-cyclic mathematical function,with the following features?




Let $$f: mathbb{Z^{+}} rightarrow left{0,1,2 right}$$




And for any $ninmathbb{Z^{+}},left{ f(3n-2),f(3n-1),f(3n)right}$ must be equal $left{0,1,2 right}$, or $left{ 0,2,1right},left{1,2,0 right},left{1,0,2 right} left{2,1,0 right},left{ 2,0,1right}$



Discrete non-cyclic means, for example:



$$color{purple}{left{f(1),f(2),f(3),f(4),f(5),f(6),f(7),f(8),...f(n) right}:=}color{red}{left{color{red}{0,1,2,2,1,0,1,0,2;} color{green}{0,1,2,2,1,0,1,0,2;} color{blue}{0,1,2,2,1,0,1,0,2;}... right}}$$



Thus, we have discrete cyclic sequence. Because,



$f(9)=f(1), f(10)=f(2), f(11)=f(3),...,f(16)=f(8).$ In other words,
$f(n)=f(n-8)$, where $n≥9, ninmathbb{Z^{+}}$ and cycle length is equal to $8.$



I hope I'm asking the question clearly.



I am looking for a mathematical function that takes discrete non-cyclic values. I'm not saying to MSE "Find such a function." What I want to know is whether such a special function can exist mathematically.



Thank you very much.






sequences-and-series functions discrete-mathematics soft-question research




sequences-and-series functions discrete-mathematics soft-question research






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 21:44







Student

















asked Dec 20 '18 at 20:51









StudentStudent

6301418




6301418








  • 1




    $begingroup$
    Yes, such a function "exists mathematically", and we can easily describe such a function. For instance, take $$ f(x) = begin{cases} 1 & x text{ is a perfect square}\ 0 & text{otherwise} end{cases} $$
    $endgroup$
    – Omnomnomnom
    Dec 20 '18 at 20:54








  • 1




    $begingroup$
    The typical term for what you're calling "cyclic" is periodic.
    $endgroup$
    – Omnomnomnom
    Dec 20 '18 at 21:00










  • $begingroup$
    Your question is completely unclear: why do you introduce the sets $D(f)$ and $E(f)$ but never mention then again? What is $n$ in the "definition" of $D(f)$. An example is an inadequate replacement for a definition, so we have no idea what you mean by "discrete non-cyclic" particularly as your example makes no sense: how do $f$ and $n$ on the left-hand side of the symbol ${:}{=}$ relate to the numbers on the right-hand side?
    $endgroup$
    – Rob Arthan
    Dec 20 '18 at 21:01












  • $begingroup$
    Concretely, the sequence I'm describing is $$ {1,overbrace{0,0}^2,1,overbrace{0,dots,0}^4,1,overbrace{0,dots,0}^6,1,dots} $$
    $endgroup$
    – Omnomnomnom
    Dec 20 '18 at 21:03








  • 1




    $begingroup$
    I suppose $D(f)$ means the domain of $f$. But this is non-standard notation. And in any case, the domain of $f$ is $Bbb Z^+$, not ${1,2,3,4,5,ldots,n}$. And $E(f)$ means the range of $f$? (I can't think what $E$ stands for.) But if so, why not just write $f:Bbb Z^+to{0,1,2}$?
    $endgroup$
    – TonyK
    Dec 20 '18 at 21:07














  • 1




    $begingroup$
    Yes, such a function "exists mathematically", and we can easily describe such a function. For instance, take $$ f(x) = begin{cases} 1 & x text{ is a perfect square}\ 0 & text{otherwise} end{cases} $$
    $endgroup$
    – Omnomnomnom
    Dec 20 '18 at 20:54








  • 1




    $begingroup$
    The typical term for what you're calling "cyclic" is periodic.
    $endgroup$
    – Omnomnomnom
    Dec 20 '18 at 21:00










  • $begingroup$
    Your question is completely unclear: why do you introduce the sets $D(f)$ and $E(f)$ but never mention then again? What is $n$ in the "definition" of $D(f)$. An example is an inadequate replacement for a definition, so we have no idea what you mean by "discrete non-cyclic" particularly as your example makes no sense: how do $f$ and $n$ on the left-hand side of the symbol ${:}{=}$ relate to the numbers on the right-hand side?
    $endgroup$
    – Rob Arthan
    Dec 20 '18 at 21:01












  • $begingroup$
    Concretely, the sequence I'm describing is $$ {1,overbrace{0,0}^2,1,overbrace{0,dots,0}^4,1,overbrace{0,dots,0}^6,1,dots} $$
    $endgroup$
    – Omnomnomnom
    Dec 20 '18 at 21:03








  • 1




    $begingroup$
    I suppose $D(f)$ means the domain of $f$. But this is non-standard notation. And in any case, the domain of $f$ is $Bbb Z^+$, not ${1,2,3,4,5,ldots,n}$. And $E(f)$ means the range of $f$? (I can't think what $E$ stands for.) But if so, why not just write $f:Bbb Z^+to{0,1,2}$?
    $endgroup$
    – TonyK
    Dec 20 '18 at 21:07








1




1




$begingroup$
Yes, such a function "exists mathematically", and we can easily describe such a function. For instance, take $$ f(x) = begin{cases} 1 & x text{ is a perfect square}\ 0 & text{otherwise} end{cases} $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 20:54






$begingroup$
Yes, such a function "exists mathematically", and we can easily describe such a function. For instance, take $$ f(x) = begin{cases} 1 & x text{ is a perfect square}\ 0 & text{otherwise} end{cases} $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 20:54






1




1




$begingroup$
The typical term for what you're calling "cyclic" is periodic.
$endgroup$
– Omnomnomnom
Dec 20 '18 at 21:00




$begingroup$
The typical term for what you're calling "cyclic" is periodic.
$endgroup$
– Omnomnomnom
Dec 20 '18 at 21:00












$begingroup$
Your question is completely unclear: why do you introduce the sets $D(f)$ and $E(f)$ but never mention then again? What is $n$ in the "definition" of $D(f)$. An example is an inadequate replacement for a definition, so we have no idea what you mean by "discrete non-cyclic" particularly as your example makes no sense: how do $f$ and $n$ on the left-hand side of the symbol ${:}{=}$ relate to the numbers on the right-hand side?
$endgroup$
– Rob Arthan
Dec 20 '18 at 21:01






$begingroup$
Your question is completely unclear: why do you introduce the sets $D(f)$ and $E(f)$ but never mention then again? What is $n$ in the "definition" of $D(f)$. An example is an inadequate replacement for a definition, so we have no idea what you mean by "discrete non-cyclic" particularly as your example makes no sense: how do $f$ and $n$ on the left-hand side of the symbol ${:}{=}$ relate to the numbers on the right-hand side?
$endgroup$
– Rob Arthan
Dec 20 '18 at 21:01














$begingroup$
Concretely, the sequence I'm describing is $$ {1,overbrace{0,0}^2,1,overbrace{0,dots,0}^4,1,overbrace{0,dots,0}^6,1,dots} $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 21:03






$begingroup$
Concretely, the sequence I'm describing is $$ {1,overbrace{0,0}^2,1,overbrace{0,dots,0}^4,1,overbrace{0,dots,0}^6,1,dots} $$
$endgroup$
– Omnomnomnom
Dec 20 '18 at 21:03






1




1




$begingroup$
I suppose $D(f)$ means the domain of $f$. But this is non-standard notation. And in any case, the domain of $f$ is $Bbb Z^+$, not ${1,2,3,4,5,ldots,n}$. And $E(f)$ means the range of $f$? (I can't think what $E$ stands for.) But if so, why not just write $f:Bbb Z^+to{0,1,2}$?
$endgroup$
– TonyK
Dec 20 '18 at 21:07




$begingroup$
I suppose $D(f)$ means the domain of $f$. But this is non-standard notation. And in any case, the domain of $f$ is $Bbb Z^+$, not ${1,2,3,4,5,ldots,n}$. And $E(f)$ means the range of $f$? (I can't think what $E$ stands for.) But if so, why not just write $f:Bbb Z^+to{0,1,2}$?
$endgroup$
– TonyK
Dec 20 '18 at 21:07










1 Answer
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$begingroup$

Not only are such sequences easy to define (as shown in the comments to your post), but there are also some very cool ones that find applications elsewhere in mathematics. In particular, the Thue-Morse sequence is aperiodic, and even has a ternary version.






share|cite|improve this answer











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    active

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    $begingroup$

    Not only are such sequences easy to define (as shown in the comments to your post), but there are also some very cool ones that find applications elsewhere in mathematics. In particular, the Thue-Morse sequence is aperiodic, and even has a ternary version.






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      Not only are such sequences easy to define (as shown in the comments to your post), but there are also some very cool ones that find applications elsewhere in mathematics. In particular, the Thue-Morse sequence is aperiodic, and even has a ternary version.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Not only are such sequences easy to define (as shown in the comments to your post), but there are also some very cool ones that find applications elsewhere in mathematics. In particular, the Thue-Morse sequence is aperiodic, and even has a ternary version.






        share|cite|improve this answer











        $endgroup$



        Not only are such sequences easy to define (as shown in the comments to your post), but there are also some very cool ones that find applications elsewhere in mathematics. In particular, the Thue-Morse sequence is aperiodic, and even has a ternary version.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 20 '18 at 21:01









        Frpzzd

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        answered Dec 20 '18 at 21:01









        DeficientMathDudeDeficientMathDude

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