Kolakoski sequence collapse
$begingroup$
Look on wikipedia for more information on the Kolakoski sequence if you're unfamiliar with it. The Kolakoski sequence is supposed to be a fractal because you can get the same sequence by taking the length of each "run" in the sequence. I was in sage math, messing around with collapsing the Kolakoski sequence to a single digit by taking the length of the runs in the sequences. ex.
$[1,2,2,1,1,2,1,2,2,1]$
$[1,2,2,1,1,2,1]$
$[1,2,2,1,1]$
$[1,2,2]$
$[1,2]$
$[1,1]$
$[2]$
Now, sometimes this works out perfectly normal, but not always. It's quite frequent that by doing this you'll actually end up with sequences containing runs of lengths of 3 or greater. Here's an example of a length which does not collapse properly.
$[1,2,2,1,1,2,1,2,2]$
$[1,2,2,1,1,2]$
$[1,2,2,1]$
$[1,2,1]$
$[1,1,1]$
$[3?]$
So, I wrote a program in python to find which length sequences would actually collapse properly. The first ones go as follows: $1,2,3,5,7,10,$ $11,15,17,23,25,$ $34,37,50,55,75,82$. All of this brings up lots of questions for me. First of all, does this mean that the Kolaski sequence isn't really a fractal since it doesn't really collapse properly as has been shown? Is there some algorithm that can calculate the sequence which consists of all the lengths that do properly collapse?
sequences-and-series
asked Jul 13 '18 at 17:14
Cadyn BombaciCadyn Bombaci
162
$endgroup$
add a comment |
$begingroup$
Look on wikipedia for more information on the Kolakoski sequence if you're unfamiliar with it. The Kolakoski sequence is supposed to be a fractal because you can get the same sequence by taking the length of each "run" in the sequence. I was in sage math, messing around with collapsing the Kolakoski sequence to a single digit by taking the length of the runs in the sequences. ex.
$[1,2,2,1,1,2,1,2,2,1]$
$[1,2,2,1,1,2,1]$
$[1,2,2,1,1]$
$[1,2,2]$
$[1,2]$
$[1,1]$
$[2]$
Now, sometimes this works out perfectly normal, but not always. It's quite frequent that by doing this you'll actually end up with sequences containing runs of lengths of 3 or greater. Here's an example of a length which does not collapse properly.
$[1,2,2,1,1,2,1,2,2]$
$[1,2,2,1,1,2]$
$[1,2,2,1]$
$[1,2,1]$
$[1,1,1]$
$[3?]$
So, I wrote a program in python to find which length sequences would actually collapse properly. The first ones go as follows: $1,2,3,5,7,10,$ $11,15,17,23,25,$ $34,37,50,55,75,82$. All of this brings up lots of questions for me. First of all, does this mean that the Kolaski sequence isn't really a fractal since it doesn't really collapse properly as has been shown? Is there some algorithm that can calculate the sequence which consists of all the lengths that do properly collapse?
sequences-and-series
asked Jul 13 '18 at 17:14
Cadyn BombaciCadyn Bombaci
162
$endgroup$
1
$begingroup$
The Kolakoski sequence is an infinite integer sequence. You are only looking at initial finite truncations of it. A better way to look at it is to reverse your procedure. Start with [1,2] and go upwards.
$endgroup$
– Somos
Jul 13 '18 at 17:19
2
$begingroup$
Nice first question. Welcome to Math Stack Exchange!
$endgroup$
– qwr
Jul 13 '18 at 17:31
$begingroup$
Yes, nice question. Note that your first example also goes through a $[1,1]$ step. The finish would generally be cleaner without the initial $1$ (which is also Kolakoski).
$endgroup$
– Joffan
Jul 13 '18 at 18:04
$begingroup$
Hi, nice question. But what do you mean by "properly collapse"? Do you just mean that it ends in $1$ or $2$? Or do you mean that the whole procedure only gives $1$s and $2$s?
$endgroup$
– 6005
Jan 11 at 20:02
add a comment |
$begingroup$
Look on wikipedia for more information on the Kolakoski sequence if you're unfamiliar with it. The Kolakoski sequence is supposed to be a fractal because you can get the same sequence by taking the length of each "run" in the sequence. I was in sage math, messing around with collapsing the Kolakoski sequence to a single digit by taking the length of the runs in the sequences. ex.
$[1,2,2,1,1,2,1,2,2,1]$
$[1,2,2,1,1,2,1]$
$[1,2,2,1,1]$
$[1,2,2]$
$[1,2]$
$[1,1]$
$[2]$
Now, sometimes this works out perfectly normal, but not always. It's quite frequent that by doing this you'll actually end up with sequences containing runs of lengths of 3 or greater. Here's an example of a length which does not collapse properly.
$[1,2,2,1,1,2,1,2,2]$
$[1,2,2,1,1,2]$
$[1,2,2,1]$
$[1,2,1]$
$[1,1,1]$
$[3?]$
So, I wrote a program in python to find which length sequences would actually collapse properly. The first ones go as follows: $1,2,3,5,7,10,$ $11,15,17,23,25,$ $34,37,50,55,75,82$. All of this brings up lots of questions for me. First of all, does this mean that the Kolaski sequence isn't really a fractal since it doesn't really collapse properly as has been shown? Is there some algorithm that can calculate the sequence which consists of all the lengths that do properly collapse?
sequences-and-series
asked Jul 13 '18 at 17:14
Cadyn BombaciCadyn Bombaci
162
$endgroup$
Look on wikipedia for more information on the Kolakoski sequence if you're unfamiliar with it. The Kolakoski sequence is supposed to be a fractal because you can get the same sequence by taking the length of each "run" in the sequence. I was in sage math, messing around with collapsing the Kolakoski sequence to a single digit by taking the length of the runs in the sequences. ex.
$[1,2,2,1,1,2,1,2,2,1]$
$[1,2,2,1,1,2,1]$
$[1,2,2,1,1]$
$[1,2,2]$
$[1,2]$
$[1,1]$
$[2]$
Now, sometimes this works out perfectly normal, but not always. It's quite frequent that by doing this you'll actually end up with sequences containing runs of lengths of 3 or greater. Here's an example of a length which does not collapse properly.
$[1,2,2,1,1,2,1,2,2]$
$[1,2,2,1,1,2]$
$[1,2,2,1]$
$[1,2,1]$
$[1,1,1]$
$[3?]$
So, I wrote a program in python to find which length sequences would actually collapse properly. The first ones go as follows: $1,2,3,5,7,10,$ $11,15,17,23,25,$ $34,37,50,55,75,82$. All of this brings up lots of questions for me. First of all, does this mean that the Kolaski sequence isn't really a fractal since it doesn't really collapse properly as has been shown? Is there some algorithm that can calculate the sequence which consists of all the lengths that do properly collapse?
sequences-and-series
sequences-and-series
asked Jul 13 '18 at 17:14
Cadyn BombaciCadyn Bombaci
162
asked Jul 13 '18 at 17:14
Cadyn BombaciCadyn Bombaci
162
edited Jul 13 '18 at 18:08
Cadyn Bombaci
asked Jul 13 '18 at 17:14
Cadyn BombaciCadyn Bombaci
162
asked Jul 13 '18 at 17:14
Cadyn BombaciCadyn Bombaci
162
asked Jul 13 '18 at 17:14
Cadyn BombaciCadyn Bombaci
162
162
1
$begingroup$
The Kolakoski sequence is an infinite integer sequence. You are only looking at initial finite truncations of it. A better way to look at it is to reverse your procedure. Start with [1,2] and go upwards.
$endgroup$
– Somos
Jul 13 '18 at 17:19
2
$begingroup$
Nice first question. Welcome to Math Stack Exchange!
$endgroup$
– qwr
Jul 13 '18 at 17:31
$begingroup$
Yes, nice question. Note that your first example also goes through a $[1,1]$ step. The finish would generally be cleaner without the initial $1$ (which is also Kolakoski).
$endgroup$
– Joffan
Jul 13 '18 at 18:04
$begingroup$
Hi, nice question. But what do you mean by "properly collapse"? Do you just mean that it ends in $1$ or $2$? Or do you mean that the whole procedure only gives $1$s and $2$s?
$endgroup$
– 6005
Jan 11 at 20:02
add a comment |
1
$begingroup$
The Kolakoski sequence is an infinite integer sequence. You are only looking at initial finite truncations of it. A better way to look at it is to reverse your procedure. Start with [1,2] and go upwards.
$endgroup$
– Somos
Jul 13 '18 at 17:19
2
$begingroup$
Nice first question. Welcome to Math Stack Exchange!
$endgroup$
– qwr
Jul 13 '18 at 17:31
$begingroup$
Yes, nice question. Note that your first example also goes through a $[1,1]$ step. The finish would generally be cleaner without the initial $1$ (which is also Kolakoski).
$endgroup$
– Joffan
Jul 13 '18 at 18:04
$begingroup$
Hi, nice question. But what do you mean by "properly collapse"? Do you just mean that it ends in $1$ or $2$? Or do you mean that the whole procedure only gives $1$s and $2$s?
$endgroup$
– 6005
Jan 11 at 20:02
1
1
$begingroup$
The Kolakoski sequence is an infinite integer sequence. You are only looking at initial finite truncations of it. A better way to look at it is to reverse your procedure. Start with [1,2] and go upwards.
$endgroup$
– Somos
Jul 13 '18 at 17:19
$begingroup$
The Kolakoski sequence is an infinite integer sequence. You are only looking at initial finite truncations of it. A better way to look at it is to reverse your procedure. Start with [1,2] and go upwards.
$endgroup$
– Somos
Jul 13 '18 at 17:19
2
2
$begingroup$
Nice first question. Welcome to Math Stack Exchange!
$endgroup$
– qwr
Jul 13 '18 at 17:31
$begingroup$
Nice first question. Welcome to Math Stack Exchange!
$endgroup$
– qwr
Jul 13 '18 at 17:31
$begingroup$
Yes, nice question. Note that your first example also goes through a $[1,1]$ step. The finish would generally be cleaner without the initial $1$ (which is also Kolakoski).
$endgroup$
– Joffan
Jul 13 '18 at 18:04
$begingroup$
Yes, nice question. Note that your first example also goes through a $[1,1]$ step. The finish would generally be cleaner without the initial $1$ (which is also Kolakoski).
$endgroup$
– Joffan
Jul 13 '18 at 18:04
$begingroup$
Hi, nice question. But what do you mean by "properly collapse"? Do you just mean that it ends in $1$ or $2$? Or do you mean that the whole procedure only gives $1$s and $2$s?
$endgroup$
– 6005
Jan 11 at 20:02
$begingroup$
Hi, nice question. But what do you mean by "properly collapse"? Do you just mean that it ends in $1$ or $2$? Or do you mean that the whole procedure only gives $1$s and $2$s?
$endgroup$
– 6005
Jan 11 at 20:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $$S(p)=K_1+K_2+...K_p$$
the length will properly collapse if
$$n=S(S(S...(S(2)))))).$$
for example, if $n=S(S(2))$, it will collapse after three encoding operations.
answered Jan 11 at 19:36
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2849897%2fkolakoski-sequence-collapse%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $$S(p)=K_1+K_2+...K_p$$
the length will properly collapse if
$$n=S(S(S...(S(2)))))).$$
for example, if $n=S(S(2))$, it will collapse after three encoding operations.
answered Jan 11 at 19:36
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
$begingroup$
Let $$S(p)=K_1+K_2+...K_p$$
the length will properly collapse if
$$n=S(S(S...(S(2)))))).$$
for example, if $n=S(S(2))$, it will collapse after three encoding operations.
answered Jan 11 at 19:36
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
add a comment |
$begingroup$
Let $$S(p)=K_1+K_2+...K_p$$
the length will properly collapse if
$$n=S(S(S...(S(2)))))).$$
for example, if $n=S(S(2))$, it will collapse after three encoding operations.
answered Jan 11 at 19:36
hamam_Abdallahhamam_Abdallah
38.1k21634
$endgroup$
Let $$S(p)=K_1+K_2+...K_p$$
the length will properly collapse if
$$n=S(S(S...(S(2)))))).$$
for example, if $n=S(S(2))$, it will collapse after three encoding operations.
answered Jan 11 at 19:36
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Jan 11 at 19:36
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Jan 11 at 19:36
hamam_Abdallahhamam_Abdallah
38.1k21634
answered Jan 11 at 19:36
hamam_Abdallahhamam_Abdallah
38.1k21634
38.1k21634
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2849897%2fkolakoski-sequence-collapse%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
The Kolakoski sequence is an infinite integer sequence. You are only looking at initial finite truncations of it. A better way to look at it is to reverse your procedure. Start with [1,2] and go upwards.
$endgroup$
– Somos
Jul 13 '18 at 17:19
2
$begingroup$
Nice first question. Welcome to Math Stack Exchange!
$endgroup$
– qwr
Jul 13 '18 at 17:31
$begingroup$
Yes, nice question. Note that your first example also goes through a $[1,1]$ step. The finish would generally be cleaner without the initial $1$ (which is also Kolakoski).
$endgroup$
– Joffan
Jul 13 '18 at 18:04
$begingroup$
Hi, nice question. But what do you mean by "properly collapse"? Do you just mean that it ends in $1$ or $2$? Or do you mean that the whole procedure only gives $1$s and $2$s?
$endgroup$
– 6005
Jan 11 at 20:02