How to initialize this process to make it produce a sequence with specified period?











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Let $S=(1,1,a)$. Then construct a new, infinite sequence $X$ using the following process:




  • First, $X_0=S_0=1$

  • Then add $S_0$ to each element of $S$, with letters being converted to their position in the alphabet and back after addition. The sequence $S$ is now $(2,2,b)$

  • Repeat with $S_1$. $X_1=S_1=2$.

  • Then add $2$ ($S_1$) to each element of $S$. $S$ is now $(4,4,d)$

  • Repeat. $X_2=S_2=d$, after addition, $S=(7,7,g)$

  • This time, $X_3=S_0=7$ and so on


All operations on numbers are modulo 10. All operations on letters are modulo 26 (e.g. $y+3=b$, $9+1=0$).



Is there a sequence which when this process is applied to, generates an $X$ which repeats itself? Until now, I've found $(1,1,a,s)$ which generates an $X$ of $(1,2,d,y,1,2,h,g,7,4,...)$










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    I've added a couple of tags and made an attempt to improve the title -- please correct anything that misrepresents your intentions.
    – r.e.s.
    Nov 25 at 20:15















up vote
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Let $S=(1,1,a)$. Then construct a new, infinite sequence $X$ using the following process:




  • First, $X_0=S_0=1$

  • Then add $S_0$ to each element of $S$, with letters being converted to their position in the alphabet and back after addition. The sequence $S$ is now $(2,2,b)$

  • Repeat with $S_1$. $X_1=S_1=2$.

  • Then add $2$ ($S_1$) to each element of $S$. $S$ is now $(4,4,d)$

  • Repeat. $X_2=S_2=d$, after addition, $S=(7,7,g)$

  • This time, $X_3=S_0=7$ and so on


All operations on numbers are modulo 10. All operations on letters are modulo 26 (e.g. $y+3=b$, $9+1=0$).



Is there a sequence which when this process is applied to, generates an $X$ which repeats itself? Until now, I've found $(1,1,a,s)$ which generates an $X$ of $(1,2,d,y,1,2,h,g,7,4,...)$










share|cite|improve this question




















  • 1




    I've added a couple of tags and made an attempt to improve the title -- please correct anything that misrepresents your intentions.
    – r.e.s.
    Nov 25 at 20:15













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $S=(1,1,a)$. Then construct a new, infinite sequence $X$ using the following process:




  • First, $X_0=S_0=1$

  • Then add $S_0$ to each element of $S$, with letters being converted to their position in the alphabet and back after addition. The sequence $S$ is now $(2,2,b)$

  • Repeat with $S_1$. $X_1=S_1=2$.

  • Then add $2$ ($S_1$) to each element of $S$. $S$ is now $(4,4,d)$

  • Repeat. $X_2=S_2=d$, after addition, $S=(7,7,g)$

  • This time, $X_3=S_0=7$ and so on


All operations on numbers are modulo 10. All operations on letters are modulo 26 (e.g. $y+3=b$, $9+1=0$).



Is there a sequence which when this process is applied to, generates an $X$ which repeats itself? Until now, I've found $(1,1,a,s)$ which generates an $X$ of $(1,2,d,y,1,2,h,g,7,4,...)$










share|cite|improve this question















Let $S=(1,1,a)$. Then construct a new, infinite sequence $X$ using the following process:




  • First, $X_0=S_0=1$

  • Then add $S_0$ to each element of $S$, with letters being converted to their position in the alphabet and back after addition. The sequence $S$ is now $(2,2,b)$

  • Repeat with $S_1$. $X_1=S_1=2$.

  • Then add $2$ ($S_1$) to each element of $S$. $S$ is now $(4,4,d)$

  • Repeat. $X_2=S_2=d$, after addition, $S=(7,7,g)$

  • This time, $X_3=S_0=7$ and so on


All operations on numbers are modulo 10. All operations on letters are modulo 26 (e.g. $y+3=b$, $9+1=0$).



Is there a sequence which when this process is applied to, generates an $X$ which repeats itself? Until now, I've found $(1,1,a,s)$ which generates an $X$ of $(1,2,d,y,1,2,h,g,7,4,...)$







sequences-and-series recurrence-relations periodic-functions






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edited Nov 25 at 20:13









r.e.s.

7,56411952




7,56411952










asked Nov 20 at 22:30









FireCubez

1084




1084








  • 1




    I've added a couple of tags and made an attempt to improve the title -- please correct anything that misrepresents your intentions.
    – r.e.s.
    Nov 25 at 20:15














  • 1




    I've added a couple of tags and made an attempt to improve the title -- please correct anything that misrepresents your intentions.
    – r.e.s.
    Nov 25 at 20:15








1




1




I've added a couple of tags and made an attempt to improve the title -- please correct anything that misrepresents your intentions.
– r.e.s.
Nov 25 at 20:15




I've added a couple of tags and made an attempt to improve the title -- please correct anything that misrepresents your intentions.
– r.e.s.
Nov 25 at 20:15










2 Answers
2






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Is there a sequence which when this process is applied to, generates an X which repeats itself?





$X$ is always eventually periodic



Any finite starting sequence $S$ will generate an $X$ that's eventually periodic.



Your $S=tt 11a$ generates $X$ composed of the following period of length $117$ repeated infinitely: $tt 12d74r50n36j12v50v12t36v36z74j74d12j36b36l36f74v98h36x50z50d36p74p36n50f50p50j98j50x36d50l12z98p12h12r12l50b12f98b74n$



Your $S=tt 11as$ generates $X$ composed of the following period of length $96$ repeated infinitely: $tt 12dy12hg74fc50ra74te12te74bu98ns74dy50jk98te98hg98lo12ns50xm50lo50pw74lo36te50vi98pw36jk12fc12pw$



Eventually periodic means that the periodic part may be preceded by finitely many elements. E.g., $S=tt 11b$ produces $X=tt color{blue}{1}(2e86w48e00i86e62q00g62u48a62i24w00s86y6)^infty$



Explanation



Your procedure can be seen as iterating the following rewrite rule, starting with $i=0$ and some initial tuple $S=(S_0,...,S_{l-1})$:



$$leftlangle i,(S_0,...,S_{l-1})rightrangle to leftlangle(i+1)bmod l, ((S_0+S_i)bmod b_0,...,(S_{l-1}+S_i)bmod b_{l-1})rightrangle
$$

where $(b_0,...,b_{k-1})$ is a given tuple of bases specifying the base in which arithmetic is to be done in each position of the $S$ tuple, and $X_i=S_i$ in each iteration. Iterating this rule generates an infinite sequence of $leftlangle i,Srightrangle$ values,
$$leftlangle i^0,S^0rightrangle to leftlangle i^1,S^1rightrangle toldotstoleftlangle i^k,S^krightrangletoldots,$$



in which $i$ takes values only in ${0,...,l-1}$, and the $j$th element of the $S$ tuple takes values only in ${0,...,b_{j}-1 }$; thus, there can be no more than $$ltimes b_0timesldotstimes b_{l-1} $$ distinct values of the pair $leftlangle i,Srightrangle$. Consequently, some pairs must repeat in the infinite sequence; that is, there must exist $0le j_1<j_2$ such that
$$leftlangle i^{j_1},S^{j_1}rightrangle = leftlangle i^{j_2},S^{j_2}rightrangle,
$$

which implies that the sequence of pairs (and hence the $X$ sequence) is eventually periodic.



Example



The following shows the initial $S=tt 11b$ generating $X=color{blue}{1}tt (2e86w48e00i86e62q00g62u48a62i24w00s86y6)^infty$:



   S = [1 , 1,  1]
base = [10, 10, 26]

count i S X[count]
----- -- --------- --------
0 0 [1, 1, 1] 1

1 1 [2, 2, 2] 2
2 2 [4, 4, 4] 4
3 0 [8, 8, 8] 8
4 1 [6, 6, 16] 6
5 2 [2, 2, 22] 22
6 0 [4, 4, 18] 4
7 1 [8, 8, 22] 8
8 2 [6, 6, 4] 4
9 0 [0, 0, 8] 0
10 1 [0, 0, 8] 0
11 2 [0, 0, 8] 8
12 0 [8, 8, 16] 8
13 1 [6, 6, 24] 6
14 2 [2, 2, 4] 4
15 0 [6, 6, 8] 6
16 1 [2, 2, 14] 2
17 2 [4, 4, 16] 16
18 0 [0, 0, 6] 0
19 1 [0, 0, 6] 0
20 2 [0, 0, 6] 6
21 0 [6, 6, 12] 6
22 1 [2, 2, 18] 2
23 2 [4, 4, 20] 20
24 0 [4, 4, 14] 4
25 1 [8, 8, 18] 8
26 2 [6, 6, 0] 0
27 0 [6, 6, 0] 6
28 1 [2, 2, 6] 2
29 2 [4, 4, 8] 8
30 0 [2, 2, 16] 2
31 1 [4, 4, 18] 4
32 2 [8, 8, 22] 22
33 0 [0, 0, 18] 0
34 1 [0, 0, 18] 0
35 2 [0, 0, 18] 18
36 0 [8, 8, 10] 8
37 1 [6, 6, 18] 6
38 2 [2, 2, 24] 24
39 0 [6, 6, 22] 6

40 1 [2, 2, 2] 2
41 2 [4, 4, 4] 4
...


Finding an $S$ that generates a given $X$



Let $S=(S_0,S_1,ldots,S_{l-1})^infty$ (i.e. the infinite sequence obtained by repeating $(S_0,S_1,ldots,S_{l-1})$), and let $b=(b_0,b_1,ldots,b_{l-1})^infty$ be the sequence of bases in which arithmetic is to be performed in the corresponding positions of $S$ and $X$. Now your procedure generating $X$ from a given $S$ can be conveniently written as follows:
$$begin{align}X_0&=S_0quadbmod b_0\
X_1&=S_1+X_0quadbmod b_1\
X_2&=S_2+X_1+X_0quadbmod b_2\
X_3&=S_3+X_2+X_1+X_0quadbmod b_3\
&ldots\
X_k&=S_k+X_{k-1}+X_{k-2}+ldots+X_0quadbmod b_k\
&ldots\
end{align}$$

Inverting this we can produce an $S$ that will generate a given eventually periodic $X$, as follows:
$$begin{align}S_0&=X_0quadbmod b_0\
S_1&=X_1-X_0quadbmod b_1\
S_2&=X_2-X_1-X_0quadbmod b_2\
S_3&=X_3-X_2-X_1-X_0quadbmod b_3\
&ldots\
S_k&=X_k-X_{k-1}-X_{k-2}-ldots-X_0quadbmod b_k\
&ldots\
end{align}$$



NB: The period of $S$ may be very much longer than the period of the $X$ that it generates. For example, $X=tt (3d)^infty$ is generated by $S$ with the following period:
$tt 3a7u1o5i9c3w7q1k5e9y3s7m1g5a9u3o7i1c5w9q3k7e1y5s9m3g7a1u5o9i3c7w1q5k9e3y7s1m5g9a3u7o1i5c9w3q7k1e5y9s3m7g1a5u9o3i7c1w5q9k3e7y1s5m9g$






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  • Could you elaborate on why this is?
    – FireCubez
    Nov 21 at 20:05










  • @FireCubez - I've added some explanation.
    – r.e.s.
    Nov 22 at 18:36










  • @FireCubez - From the answer you posted, it seems you're interested in knowing how to find an $S$ that will generate a given $X$, so I've added an explanation of that as well.
    – r.e.s.
    Nov 25 at 20:04


















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0
down vote













After working on it for a bit (yes, this is what I do in my free time), I found that the starting sequence just needs to sum up to a multiple of both $10$ and $26$. So for example, a sequence consisting of $130$ nines would work and would generate $(9,8,6,2,4,8,6,2,4,...)$.



When only dealing with letters, you can have the sequence sum to a multiple of only $26$. So for example, the sequence $(s,m,u,i,y,d,f,f,u)$ produces $(s,e,q,u,e,n,c,e,x,s,e,q,u,e,n,c,e,x,...)$. Note the $x$, it's a padder to get the shift it back into position so that the next time the original sequence is restored.






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    2 Answers
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    up vote
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    down vote



    accepted











    Is there a sequence which when this process is applied to, generates an X which repeats itself?





    $X$ is always eventually periodic



    Any finite starting sequence $S$ will generate an $X$ that's eventually periodic.



    Your $S=tt 11a$ generates $X$ composed of the following period of length $117$ repeated infinitely: $tt 12d74r50n36j12v50v12t36v36z74j74d12j36b36l36f74v98h36x50z50d36p74p36n50f50p50j98j50x36d50l12z98p12h12r12l50b12f98b74n$



    Your $S=tt 11as$ generates $X$ composed of the following period of length $96$ repeated infinitely: $tt 12dy12hg74fc50ra74te12te74bu98ns74dy50jk98te98hg98lo12ns50xm50lo50pw74lo36te50vi98pw36jk12fc12pw$



    Eventually periodic means that the periodic part may be preceded by finitely many elements. E.g., $S=tt 11b$ produces $X=tt color{blue}{1}(2e86w48e00i86e62q00g62u48a62i24w00s86y6)^infty$



    Explanation



    Your procedure can be seen as iterating the following rewrite rule, starting with $i=0$ and some initial tuple $S=(S_0,...,S_{l-1})$:



    $$leftlangle i,(S_0,...,S_{l-1})rightrangle to leftlangle(i+1)bmod l, ((S_0+S_i)bmod b_0,...,(S_{l-1}+S_i)bmod b_{l-1})rightrangle
    $$

    where $(b_0,...,b_{k-1})$ is a given tuple of bases specifying the base in which arithmetic is to be done in each position of the $S$ tuple, and $X_i=S_i$ in each iteration. Iterating this rule generates an infinite sequence of $leftlangle i,Srightrangle$ values,
    $$leftlangle i^0,S^0rightrangle to leftlangle i^1,S^1rightrangle toldotstoleftlangle i^k,S^krightrangletoldots,$$



    in which $i$ takes values only in ${0,...,l-1}$, and the $j$th element of the $S$ tuple takes values only in ${0,...,b_{j}-1 }$; thus, there can be no more than $$ltimes b_0timesldotstimes b_{l-1} $$ distinct values of the pair $leftlangle i,Srightrangle$. Consequently, some pairs must repeat in the infinite sequence; that is, there must exist $0le j_1<j_2$ such that
    $$leftlangle i^{j_1},S^{j_1}rightrangle = leftlangle i^{j_2},S^{j_2}rightrangle,
    $$

    which implies that the sequence of pairs (and hence the $X$ sequence) is eventually periodic.



    Example



    The following shows the initial $S=tt 11b$ generating $X=color{blue}{1}tt (2e86w48e00i86e62q00g62u48a62i24w00s86y6)^infty$:



       S = [1 , 1,  1]
    base = [10, 10, 26]

    count i S X[count]
    ----- -- --------- --------
    0 0 [1, 1, 1] 1

    1 1 [2, 2, 2] 2
    2 2 [4, 4, 4] 4
    3 0 [8, 8, 8] 8
    4 1 [6, 6, 16] 6
    5 2 [2, 2, 22] 22
    6 0 [4, 4, 18] 4
    7 1 [8, 8, 22] 8
    8 2 [6, 6, 4] 4
    9 0 [0, 0, 8] 0
    10 1 [0, 0, 8] 0
    11 2 [0, 0, 8] 8
    12 0 [8, 8, 16] 8
    13 1 [6, 6, 24] 6
    14 2 [2, 2, 4] 4
    15 0 [6, 6, 8] 6
    16 1 [2, 2, 14] 2
    17 2 [4, 4, 16] 16
    18 0 [0, 0, 6] 0
    19 1 [0, 0, 6] 0
    20 2 [0, 0, 6] 6
    21 0 [6, 6, 12] 6
    22 1 [2, 2, 18] 2
    23 2 [4, 4, 20] 20
    24 0 [4, 4, 14] 4
    25 1 [8, 8, 18] 8
    26 2 [6, 6, 0] 0
    27 0 [6, 6, 0] 6
    28 1 [2, 2, 6] 2
    29 2 [4, 4, 8] 8
    30 0 [2, 2, 16] 2
    31 1 [4, 4, 18] 4
    32 2 [8, 8, 22] 22
    33 0 [0, 0, 18] 0
    34 1 [0, 0, 18] 0
    35 2 [0, 0, 18] 18
    36 0 [8, 8, 10] 8
    37 1 [6, 6, 18] 6
    38 2 [2, 2, 24] 24
    39 0 [6, 6, 22] 6

    40 1 [2, 2, 2] 2
    41 2 [4, 4, 4] 4
    ...


    Finding an $S$ that generates a given $X$



    Let $S=(S_0,S_1,ldots,S_{l-1})^infty$ (i.e. the infinite sequence obtained by repeating $(S_0,S_1,ldots,S_{l-1})$), and let $b=(b_0,b_1,ldots,b_{l-1})^infty$ be the sequence of bases in which arithmetic is to be performed in the corresponding positions of $S$ and $X$. Now your procedure generating $X$ from a given $S$ can be conveniently written as follows:
    $$begin{align}X_0&=S_0quadbmod b_0\
    X_1&=S_1+X_0quadbmod b_1\
    X_2&=S_2+X_1+X_0quadbmod b_2\
    X_3&=S_3+X_2+X_1+X_0quadbmod b_3\
    &ldots\
    X_k&=S_k+X_{k-1}+X_{k-2}+ldots+X_0quadbmod b_k\
    &ldots\
    end{align}$$

    Inverting this we can produce an $S$ that will generate a given eventually periodic $X$, as follows:
    $$begin{align}S_0&=X_0quadbmod b_0\
    S_1&=X_1-X_0quadbmod b_1\
    S_2&=X_2-X_1-X_0quadbmod b_2\
    S_3&=X_3-X_2-X_1-X_0quadbmod b_3\
    &ldots\
    S_k&=X_k-X_{k-1}-X_{k-2}-ldots-X_0quadbmod b_k\
    &ldots\
    end{align}$$



    NB: The period of $S$ may be very much longer than the period of the $X$ that it generates. For example, $X=tt (3d)^infty$ is generated by $S$ with the following period:
    $tt 3a7u1o5i9c3w7q1k5e9y3s7m1g5a9u3o7i1c5w9q3k7e1y5s9m3g7a1u5o9i3c7w1q5k9e3y7s1m5g9a3u7o1i5c9w3q7k1e5y9s3m7g1a5u9o3i7c1w5q9k3e7y1s5m9g$






    share|cite|improve this answer























    • Could you elaborate on why this is?
      – FireCubez
      Nov 21 at 20:05










    • @FireCubez - I've added some explanation.
      – r.e.s.
      Nov 22 at 18:36










    • @FireCubez - From the answer you posted, it seems you're interested in knowing how to find an $S$ that will generate a given $X$, so I've added an explanation of that as well.
      – r.e.s.
      Nov 25 at 20:04















    up vote
    1
    down vote



    accepted











    Is there a sequence which when this process is applied to, generates an X which repeats itself?





    $X$ is always eventually periodic



    Any finite starting sequence $S$ will generate an $X$ that's eventually periodic.



    Your $S=tt 11a$ generates $X$ composed of the following period of length $117$ repeated infinitely: $tt 12d74r50n36j12v50v12t36v36z74j74d12j36b36l36f74v98h36x50z50d36p74p36n50f50p50j98j50x36d50l12z98p12h12r12l50b12f98b74n$



    Your $S=tt 11as$ generates $X$ composed of the following period of length $96$ repeated infinitely: $tt 12dy12hg74fc50ra74te12te74bu98ns74dy50jk98te98hg98lo12ns50xm50lo50pw74lo36te50vi98pw36jk12fc12pw$



    Eventually periodic means that the periodic part may be preceded by finitely many elements. E.g., $S=tt 11b$ produces $X=tt color{blue}{1}(2e86w48e00i86e62q00g62u48a62i24w00s86y6)^infty$



    Explanation



    Your procedure can be seen as iterating the following rewrite rule, starting with $i=0$ and some initial tuple $S=(S_0,...,S_{l-1})$:



    $$leftlangle i,(S_0,...,S_{l-1})rightrangle to leftlangle(i+1)bmod l, ((S_0+S_i)bmod b_0,...,(S_{l-1}+S_i)bmod b_{l-1})rightrangle
    $$

    where $(b_0,...,b_{k-1})$ is a given tuple of bases specifying the base in which arithmetic is to be done in each position of the $S$ tuple, and $X_i=S_i$ in each iteration. Iterating this rule generates an infinite sequence of $leftlangle i,Srightrangle$ values,
    $$leftlangle i^0,S^0rightrangle to leftlangle i^1,S^1rightrangle toldotstoleftlangle i^k,S^krightrangletoldots,$$



    in which $i$ takes values only in ${0,...,l-1}$, and the $j$th element of the $S$ tuple takes values only in ${0,...,b_{j}-1 }$; thus, there can be no more than $$ltimes b_0timesldotstimes b_{l-1} $$ distinct values of the pair $leftlangle i,Srightrangle$. Consequently, some pairs must repeat in the infinite sequence; that is, there must exist $0le j_1<j_2$ such that
    $$leftlangle i^{j_1},S^{j_1}rightrangle = leftlangle i^{j_2},S^{j_2}rightrangle,
    $$

    which implies that the sequence of pairs (and hence the $X$ sequence) is eventually periodic.



    Example



    The following shows the initial $S=tt 11b$ generating $X=color{blue}{1}tt (2e86w48e00i86e62q00g62u48a62i24w00s86y6)^infty$:



       S = [1 , 1,  1]
    base = [10, 10, 26]

    count i S X[count]
    ----- -- --------- --------
    0 0 [1, 1, 1] 1

    1 1 [2, 2, 2] 2
    2 2 [4, 4, 4] 4
    3 0 [8, 8, 8] 8
    4 1 [6, 6, 16] 6
    5 2 [2, 2, 22] 22
    6 0 [4, 4, 18] 4
    7 1 [8, 8, 22] 8
    8 2 [6, 6, 4] 4
    9 0 [0, 0, 8] 0
    10 1 [0, 0, 8] 0
    11 2 [0, 0, 8] 8
    12 0 [8, 8, 16] 8
    13 1 [6, 6, 24] 6
    14 2 [2, 2, 4] 4
    15 0 [6, 6, 8] 6
    16 1 [2, 2, 14] 2
    17 2 [4, 4, 16] 16
    18 0 [0, 0, 6] 0
    19 1 [0, 0, 6] 0
    20 2 [0, 0, 6] 6
    21 0 [6, 6, 12] 6
    22 1 [2, 2, 18] 2
    23 2 [4, 4, 20] 20
    24 0 [4, 4, 14] 4
    25 1 [8, 8, 18] 8
    26 2 [6, 6, 0] 0
    27 0 [6, 6, 0] 6
    28 1 [2, 2, 6] 2
    29 2 [4, 4, 8] 8
    30 0 [2, 2, 16] 2
    31 1 [4, 4, 18] 4
    32 2 [8, 8, 22] 22
    33 0 [0, 0, 18] 0
    34 1 [0, 0, 18] 0
    35 2 [0, 0, 18] 18
    36 0 [8, 8, 10] 8
    37 1 [6, 6, 18] 6
    38 2 [2, 2, 24] 24
    39 0 [6, 6, 22] 6

    40 1 [2, 2, 2] 2
    41 2 [4, 4, 4] 4
    ...


    Finding an $S$ that generates a given $X$



    Let $S=(S_0,S_1,ldots,S_{l-1})^infty$ (i.e. the infinite sequence obtained by repeating $(S_0,S_1,ldots,S_{l-1})$), and let $b=(b_0,b_1,ldots,b_{l-1})^infty$ be the sequence of bases in which arithmetic is to be performed in the corresponding positions of $S$ and $X$. Now your procedure generating $X$ from a given $S$ can be conveniently written as follows:
    $$begin{align}X_0&=S_0quadbmod b_0\
    X_1&=S_1+X_0quadbmod b_1\
    X_2&=S_2+X_1+X_0quadbmod b_2\
    X_3&=S_3+X_2+X_1+X_0quadbmod b_3\
    &ldots\
    X_k&=S_k+X_{k-1}+X_{k-2}+ldots+X_0quadbmod b_k\
    &ldots\
    end{align}$$

    Inverting this we can produce an $S$ that will generate a given eventually periodic $X$, as follows:
    $$begin{align}S_0&=X_0quadbmod b_0\
    S_1&=X_1-X_0quadbmod b_1\
    S_2&=X_2-X_1-X_0quadbmod b_2\
    S_3&=X_3-X_2-X_1-X_0quadbmod b_3\
    &ldots\
    S_k&=X_k-X_{k-1}-X_{k-2}-ldots-X_0quadbmod b_k\
    &ldots\
    end{align}$$



    NB: The period of $S$ may be very much longer than the period of the $X$ that it generates. For example, $X=tt (3d)^infty$ is generated by $S$ with the following period:
    $tt 3a7u1o5i9c3w7q1k5e9y3s7m1g5a9u3o7i1c5w9q3k7e1y5s9m3g7a1u5o9i3c7w1q5k9e3y7s1m5g9a3u7o1i5c9w3q7k1e5y9s3m7g1a5u9o3i7c1w5q9k3e7y1s5m9g$






    share|cite|improve this answer























    • Could you elaborate on why this is?
      – FireCubez
      Nov 21 at 20:05










    • @FireCubez - I've added some explanation.
      – r.e.s.
      Nov 22 at 18:36










    • @FireCubez - From the answer you posted, it seems you're interested in knowing how to find an $S$ that will generate a given $X$, so I've added an explanation of that as well.
      – r.e.s.
      Nov 25 at 20:04













    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted







    Is there a sequence which when this process is applied to, generates an X which repeats itself?





    $X$ is always eventually periodic



    Any finite starting sequence $S$ will generate an $X$ that's eventually periodic.



    Your $S=tt 11a$ generates $X$ composed of the following period of length $117$ repeated infinitely: $tt 12d74r50n36j12v50v12t36v36z74j74d12j36b36l36f74v98h36x50z50d36p74p36n50f50p50j98j50x36d50l12z98p12h12r12l50b12f98b74n$



    Your $S=tt 11as$ generates $X$ composed of the following period of length $96$ repeated infinitely: $tt 12dy12hg74fc50ra74te12te74bu98ns74dy50jk98te98hg98lo12ns50xm50lo50pw74lo36te50vi98pw36jk12fc12pw$



    Eventually periodic means that the periodic part may be preceded by finitely many elements. E.g., $S=tt 11b$ produces $X=tt color{blue}{1}(2e86w48e00i86e62q00g62u48a62i24w00s86y6)^infty$



    Explanation



    Your procedure can be seen as iterating the following rewrite rule, starting with $i=0$ and some initial tuple $S=(S_0,...,S_{l-1})$:



    $$leftlangle i,(S_0,...,S_{l-1})rightrangle to leftlangle(i+1)bmod l, ((S_0+S_i)bmod b_0,...,(S_{l-1}+S_i)bmod b_{l-1})rightrangle
    $$

    where $(b_0,...,b_{k-1})$ is a given tuple of bases specifying the base in which arithmetic is to be done in each position of the $S$ tuple, and $X_i=S_i$ in each iteration. Iterating this rule generates an infinite sequence of $leftlangle i,Srightrangle$ values,
    $$leftlangle i^0,S^0rightrangle to leftlangle i^1,S^1rightrangle toldotstoleftlangle i^k,S^krightrangletoldots,$$



    in which $i$ takes values only in ${0,...,l-1}$, and the $j$th element of the $S$ tuple takes values only in ${0,...,b_{j}-1 }$; thus, there can be no more than $$ltimes b_0timesldotstimes b_{l-1} $$ distinct values of the pair $leftlangle i,Srightrangle$. Consequently, some pairs must repeat in the infinite sequence; that is, there must exist $0le j_1<j_2$ such that
    $$leftlangle i^{j_1},S^{j_1}rightrangle = leftlangle i^{j_2},S^{j_2}rightrangle,
    $$

    which implies that the sequence of pairs (and hence the $X$ sequence) is eventually periodic.



    Example



    The following shows the initial $S=tt 11b$ generating $X=color{blue}{1}tt (2e86w48e00i86e62q00g62u48a62i24w00s86y6)^infty$:



       S = [1 , 1,  1]
    base = [10, 10, 26]

    count i S X[count]
    ----- -- --------- --------
    0 0 [1, 1, 1] 1

    1 1 [2, 2, 2] 2
    2 2 [4, 4, 4] 4
    3 0 [8, 8, 8] 8
    4 1 [6, 6, 16] 6
    5 2 [2, 2, 22] 22
    6 0 [4, 4, 18] 4
    7 1 [8, 8, 22] 8
    8 2 [6, 6, 4] 4
    9 0 [0, 0, 8] 0
    10 1 [0, 0, 8] 0
    11 2 [0, 0, 8] 8
    12 0 [8, 8, 16] 8
    13 1 [6, 6, 24] 6
    14 2 [2, 2, 4] 4
    15 0 [6, 6, 8] 6
    16 1 [2, 2, 14] 2
    17 2 [4, 4, 16] 16
    18 0 [0, 0, 6] 0
    19 1 [0, 0, 6] 0
    20 2 [0, 0, 6] 6
    21 0 [6, 6, 12] 6
    22 1 [2, 2, 18] 2
    23 2 [4, 4, 20] 20
    24 0 [4, 4, 14] 4
    25 1 [8, 8, 18] 8
    26 2 [6, 6, 0] 0
    27 0 [6, 6, 0] 6
    28 1 [2, 2, 6] 2
    29 2 [4, 4, 8] 8
    30 0 [2, 2, 16] 2
    31 1 [4, 4, 18] 4
    32 2 [8, 8, 22] 22
    33 0 [0, 0, 18] 0
    34 1 [0, 0, 18] 0
    35 2 [0, 0, 18] 18
    36 0 [8, 8, 10] 8
    37 1 [6, 6, 18] 6
    38 2 [2, 2, 24] 24
    39 0 [6, 6, 22] 6

    40 1 [2, 2, 2] 2
    41 2 [4, 4, 4] 4
    ...


    Finding an $S$ that generates a given $X$



    Let $S=(S_0,S_1,ldots,S_{l-1})^infty$ (i.e. the infinite sequence obtained by repeating $(S_0,S_1,ldots,S_{l-1})$), and let $b=(b_0,b_1,ldots,b_{l-1})^infty$ be the sequence of bases in which arithmetic is to be performed in the corresponding positions of $S$ and $X$. Now your procedure generating $X$ from a given $S$ can be conveniently written as follows:
    $$begin{align}X_0&=S_0quadbmod b_0\
    X_1&=S_1+X_0quadbmod b_1\
    X_2&=S_2+X_1+X_0quadbmod b_2\
    X_3&=S_3+X_2+X_1+X_0quadbmod b_3\
    &ldots\
    X_k&=S_k+X_{k-1}+X_{k-2}+ldots+X_0quadbmod b_k\
    &ldots\
    end{align}$$

    Inverting this we can produce an $S$ that will generate a given eventually periodic $X$, as follows:
    $$begin{align}S_0&=X_0quadbmod b_0\
    S_1&=X_1-X_0quadbmod b_1\
    S_2&=X_2-X_1-X_0quadbmod b_2\
    S_3&=X_3-X_2-X_1-X_0quadbmod b_3\
    &ldots\
    S_k&=X_k-X_{k-1}-X_{k-2}-ldots-X_0quadbmod b_k\
    &ldots\
    end{align}$$



    NB: The period of $S$ may be very much longer than the period of the $X$ that it generates. For example, $X=tt (3d)^infty$ is generated by $S$ with the following period:
    $tt 3a7u1o5i9c3w7q1k5e9y3s7m1g5a9u3o7i1c5w9q3k7e1y5s9m3g7a1u5o9i3c7w1q5k9e3y7s1m5g9a3u7o1i5c9w3q7k1e5y9s3m7g1a5u9o3i7c1w5q9k3e7y1s5m9g$






    share|cite|improve this answer















    Is there a sequence which when this process is applied to, generates an X which repeats itself?





    $X$ is always eventually periodic



    Any finite starting sequence $S$ will generate an $X$ that's eventually periodic.



    Your $S=tt 11a$ generates $X$ composed of the following period of length $117$ repeated infinitely: $tt 12d74r50n36j12v50v12t36v36z74j74d12j36b36l36f74v98h36x50z50d36p74p36n50f50p50j98j50x36d50l12z98p12h12r12l50b12f98b74n$



    Your $S=tt 11as$ generates $X$ composed of the following period of length $96$ repeated infinitely: $tt 12dy12hg74fc50ra74te12te74bu98ns74dy50jk98te98hg98lo12ns50xm50lo50pw74lo36te50vi98pw36jk12fc12pw$



    Eventually periodic means that the periodic part may be preceded by finitely many elements. E.g., $S=tt 11b$ produces $X=tt color{blue}{1}(2e86w48e00i86e62q00g62u48a62i24w00s86y6)^infty$



    Explanation



    Your procedure can be seen as iterating the following rewrite rule, starting with $i=0$ and some initial tuple $S=(S_0,...,S_{l-1})$:



    $$leftlangle i,(S_0,...,S_{l-1})rightrangle to leftlangle(i+1)bmod l, ((S_0+S_i)bmod b_0,...,(S_{l-1}+S_i)bmod b_{l-1})rightrangle
    $$

    where $(b_0,...,b_{k-1})$ is a given tuple of bases specifying the base in which arithmetic is to be done in each position of the $S$ tuple, and $X_i=S_i$ in each iteration. Iterating this rule generates an infinite sequence of $leftlangle i,Srightrangle$ values,
    $$leftlangle i^0,S^0rightrangle to leftlangle i^1,S^1rightrangle toldotstoleftlangle i^k,S^krightrangletoldots,$$



    in which $i$ takes values only in ${0,...,l-1}$, and the $j$th element of the $S$ tuple takes values only in ${0,...,b_{j}-1 }$; thus, there can be no more than $$ltimes b_0timesldotstimes b_{l-1} $$ distinct values of the pair $leftlangle i,Srightrangle$. Consequently, some pairs must repeat in the infinite sequence; that is, there must exist $0le j_1<j_2$ such that
    $$leftlangle i^{j_1},S^{j_1}rightrangle = leftlangle i^{j_2},S^{j_2}rightrangle,
    $$

    which implies that the sequence of pairs (and hence the $X$ sequence) is eventually periodic.



    Example



    The following shows the initial $S=tt 11b$ generating $X=color{blue}{1}tt (2e86w48e00i86e62q00g62u48a62i24w00s86y6)^infty$:



       S = [1 , 1,  1]
    base = [10, 10, 26]

    count i S X[count]
    ----- -- --------- --------
    0 0 [1, 1, 1] 1

    1 1 [2, 2, 2] 2
    2 2 [4, 4, 4] 4
    3 0 [8, 8, 8] 8
    4 1 [6, 6, 16] 6
    5 2 [2, 2, 22] 22
    6 0 [4, 4, 18] 4
    7 1 [8, 8, 22] 8
    8 2 [6, 6, 4] 4
    9 0 [0, 0, 8] 0
    10 1 [0, 0, 8] 0
    11 2 [0, 0, 8] 8
    12 0 [8, 8, 16] 8
    13 1 [6, 6, 24] 6
    14 2 [2, 2, 4] 4
    15 0 [6, 6, 8] 6
    16 1 [2, 2, 14] 2
    17 2 [4, 4, 16] 16
    18 0 [0, 0, 6] 0
    19 1 [0, 0, 6] 0
    20 2 [0, 0, 6] 6
    21 0 [6, 6, 12] 6
    22 1 [2, 2, 18] 2
    23 2 [4, 4, 20] 20
    24 0 [4, 4, 14] 4
    25 1 [8, 8, 18] 8
    26 2 [6, 6, 0] 0
    27 0 [6, 6, 0] 6
    28 1 [2, 2, 6] 2
    29 2 [4, 4, 8] 8
    30 0 [2, 2, 16] 2
    31 1 [4, 4, 18] 4
    32 2 [8, 8, 22] 22
    33 0 [0, 0, 18] 0
    34 1 [0, 0, 18] 0
    35 2 [0, 0, 18] 18
    36 0 [8, 8, 10] 8
    37 1 [6, 6, 18] 6
    38 2 [2, 2, 24] 24
    39 0 [6, 6, 22] 6

    40 1 [2, 2, 2] 2
    41 2 [4, 4, 4] 4
    ...


    Finding an $S$ that generates a given $X$



    Let $S=(S_0,S_1,ldots,S_{l-1})^infty$ (i.e. the infinite sequence obtained by repeating $(S_0,S_1,ldots,S_{l-1})$), and let $b=(b_0,b_1,ldots,b_{l-1})^infty$ be the sequence of bases in which arithmetic is to be performed in the corresponding positions of $S$ and $X$. Now your procedure generating $X$ from a given $S$ can be conveniently written as follows:
    $$begin{align}X_0&=S_0quadbmod b_0\
    X_1&=S_1+X_0quadbmod b_1\
    X_2&=S_2+X_1+X_0quadbmod b_2\
    X_3&=S_3+X_2+X_1+X_0quadbmod b_3\
    &ldots\
    X_k&=S_k+X_{k-1}+X_{k-2}+ldots+X_0quadbmod b_k\
    &ldots\
    end{align}$$

    Inverting this we can produce an $S$ that will generate a given eventually periodic $X$, as follows:
    $$begin{align}S_0&=X_0quadbmod b_0\
    S_1&=X_1-X_0quadbmod b_1\
    S_2&=X_2-X_1-X_0quadbmod b_2\
    S_3&=X_3-X_2-X_1-X_0quadbmod b_3\
    &ldots\
    S_k&=X_k-X_{k-1}-X_{k-2}-ldots-X_0quadbmod b_k\
    &ldots\
    end{align}$$



    NB: The period of $S$ may be very much longer than the period of the $X$ that it generates. For example, $X=tt (3d)^infty$ is generated by $S$ with the following period:
    $tt 3a7u1o5i9c3w7q1k5e9y3s7m1g5a9u3o7i1c5w9q3k7e1y5s9m3g7a1u5o9i3c7w1q5k9e3y7s1m5g9a3u7o1i5c9w3q7k1e5y9s3m7g1a5u9o3i7c1w5q9k3e7y1s5m9g$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 25 at 20:20

























    answered Nov 21 at 18:04









    r.e.s.

    7,56411952




    7,56411952












    • Could you elaborate on why this is?
      – FireCubez
      Nov 21 at 20:05










    • @FireCubez - I've added some explanation.
      – r.e.s.
      Nov 22 at 18:36










    • @FireCubez - From the answer you posted, it seems you're interested in knowing how to find an $S$ that will generate a given $X$, so I've added an explanation of that as well.
      – r.e.s.
      Nov 25 at 20:04


















    • Could you elaborate on why this is?
      – FireCubez
      Nov 21 at 20:05










    • @FireCubez - I've added some explanation.
      – r.e.s.
      Nov 22 at 18:36










    • @FireCubez - From the answer you posted, it seems you're interested in knowing how to find an $S$ that will generate a given $X$, so I've added an explanation of that as well.
      – r.e.s.
      Nov 25 at 20:04
















    Could you elaborate on why this is?
    – FireCubez
    Nov 21 at 20:05




    Could you elaborate on why this is?
    – FireCubez
    Nov 21 at 20:05












    @FireCubez - I've added some explanation.
    – r.e.s.
    Nov 22 at 18:36




    @FireCubez - I've added some explanation.
    – r.e.s.
    Nov 22 at 18:36












    @FireCubez - From the answer you posted, it seems you're interested in knowing how to find an $S$ that will generate a given $X$, so I've added an explanation of that as well.
    – r.e.s.
    Nov 25 at 20:04




    @FireCubez - From the answer you posted, it seems you're interested in knowing how to find an $S$ that will generate a given $X$, so I've added an explanation of that as well.
    – r.e.s.
    Nov 25 at 20:04










    up vote
    0
    down vote













    After working on it for a bit (yes, this is what I do in my free time), I found that the starting sequence just needs to sum up to a multiple of both $10$ and $26$. So for example, a sequence consisting of $130$ nines would work and would generate $(9,8,6,2,4,8,6,2,4,...)$.



    When only dealing with letters, you can have the sequence sum to a multiple of only $26$. So for example, the sequence $(s,m,u,i,y,d,f,f,u)$ produces $(s,e,q,u,e,n,c,e,x,s,e,q,u,e,n,c,e,x,...)$. Note the $x$, it's a padder to get the shift it back into position so that the next time the original sequence is restored.






    share|cite|improve this answer



























      up vote
      0
      down vote













      After working on it for a bit (yes, this is what I do in my free time), I found that the starting sequence just needs to sum up to a multiple of both $10$ and $26$. So for example, a sequence consisting of $130$ nines would work and would generate $(9,8,6,2,4,8,6,2,4,...)$.



      When only dealing with letters, you can have the sequence sum to a multiple of only $26$. So for example, the sequence $(s,m,u,i,y,d,f,f,u)$ produces $(s,e,q,u,e,n,c,e,x,s,e,q,u,e,n,c,e,x,...)$. Note the $x$, it's a padder to get the shift it back into position so that the next time the original sequence is restored.






      share|cite|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        After working on it for a bit (yes, this is what I do in my free time), I found that the starting sequence just needs to sum up to a multiple of both $10$ and $26$. So for example, a sequence consisting of $130$ nines would work and would generate $(9,8,6,2,4,8,6,2,4,...)$.



        When only dealing with letters, you can have the sequence sum to a multiple of only $26$. So for example, the sequence $(s,m,u,i,y,d,f,f,u)$ produces $(s,e,q,u,e,n,c,e,x,s,e,q,u,e,n,c,e,x,...)$. Note the $x$, it's a padder to get the shift it back into position so that the next time the original sequence is restored.






        share|cite|improve this answer














        After working on it for a bit (yes, this is what I do in my free time), I found that the starting sequence just needs to sum up to a multiple of both $10$ and $26$. So for example, a sequence consisting of $130$ nines would work and would generate $(9,8,6,2,4,8,6,2,4,...)$.



        When only dealing with letters, you can have the sequence sum to a multiple of only $26$. So for example, the sequence $(s,m,u,i,y,d,f,f,u)$ produces $(s,e,q,u,e,n,c,e,x,s,e,q,u,e,n,c,e,x,...)$. Note the $x$, it's a padder to get the shift it back into position so that the next time the original sequence is restored.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 21 at 16:35

























        answered Nov 21 at 15:39









        FireCubez

        1084




        1084






























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