Help finding the formula for this sequence {$23$, $114$, $187$, $473$, $2792$, $5624$, $19640$, $75884$,...












0












$begingroup$


Excuse the large title (The 'good title' page said not to be afraid to make it too long)



${23,114,187,473,2792,5624,19640,75884,187211,479797,1452795,5102858,14872865,72392867,146262888}$



I'm trying to figure out the nth term for these numbers. To help:





  • I'm generating these numbers on a python program:



    import math
    n=23
    digit=0
    while(True):
    pi=n*math.sin(math.radians(((90*n)-180)/(n)))*math.sin(math.radians(180/n))
    strpi=str(pi)
    count1=2
    while(count1<len(strpi)-1):
    if(strpi[count1]==str(math.pi)[count1]):
    if(digit<count1):
    print(n,"nn",pi,"n")
    digit=count1
    else:
    break
    count1+=1
    n+=1



  • For the non-programmers the program uses this formula:



    $ π=n cdot sin(frac{90n-180}{n}) cdot sin(frac{180}{n})$



    What this does is calculate $pi$ more accurately the higher the value of $n$. The program
    recorded the value of $n$ each time the next decimal value of $pi$ was found (increasing $n$ by $1$ starting from $0$ each time). So if $n$ is $23$ the
    output is $3.1$ and if it is $114$ the value is is $3.14$. Note that these values have
    decimal places after them that are not digits of $pi$. And this is in degrees, not radians.




  • This formula is derived from this one (Done with the sine rule):



    $ A=nr^2 cdot sin(frac{90n-180}{n}) cdot sin(frac{180}{n})$



    $A=$ The area of a proper polygon, $r=$ length from centre to corner, $n=$ number of sides




I substituted $A$ with $πr^2$ and simplified




  • It took me a while to realise that the difference between the first two values is larger than the difference between the second and third ($91$ then $73$)


  • If my program gets a new value I'll be adding it in. And if you want me to create an algorithm to find different data relating to this to help you solve it, or have an improved program do not hesitate to ask/suggest.


  • Technically speaking, the first element could be labelled as $12$ which gives the first digit as $3$.


  • Below are the pi values given when n is increased



enter image description here




  • Here is a GIF showing the increase of numbers as $n$ increases
    enter image description here










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    In radians and after simplification, you compute $frac n2sin frac{2pi}{n}$. It's half the perimeter of a polygon with $n/2$ sides.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 15 '18 at 15:47












  • $begingroup$
    This is in degrees. I'll add this in
    $endgroup$
    – yolo
    Dec 15 '18 at 15:48










  • $begingroup$
    Maybe I'm wrong, but I highly doubt you're going to find a formula for this.
    $endgroup$
    – Matt Samuel
    Dec 15 '18 at 15:58










  • $begingroup$
    Of course it's in degrees, that's why I translated this to radians. Degrees are useless when you want to use some calculus, and with calculus you get pretty easily that $frac n2 sin frac{2pi}{n}topi$ as $ntoinfty$, albeit very slowly. That's basically the method used by Archimedes to get his approximation $frac{22}7$. You could find an asymptotic approximation for your integer sequence, but like @MattSamuel I don't think there is a simple closed form.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 15 '18 at 15:58












  • $begingroup$
    A pi on both sides of the equation? $π=frac{n}{2} × sin(frac{2π}{n})$
    $endgroup$
    – yolo
    Dec 15 '18 at 16:02


















0












$begingroup$


Excuse the large title (The 'good title' page said not to be afraid to make it too long)



${23,114,187,473,2792,5624,19640,75884,187211,479797,1452795,5102858,14872865,72392867,146262888}$



I'm trying to figure out the nth term for these numbers. To help:





  • I'm generating these numbers on a python program:



    import math
    n=23
    digit=0
    while(True):
    pi=n*math.sin(math.radians(((90*n)-180)/(n)))*math.sin(math.radians(180/n))
    strpi=str(pi)
    count1=2
    while(count1<len(strpi)-1):
    if(strpi[count1]==str(math.pi)[count1]):
    if(digit<count1):
    print(n,"nn",pi,"n")
    digit=count1
    else:
    break
    count1+=1
    n+=1



  • For the non-programmers the program uses this formula:



    $ π=n cdot sin(frac{90n-180}{n}) cdot sin(frac{180}{n})$



    What this does is calculate $pi$ more accurately the higher the value of $n$. The program
    recorded the value of $n$ each time the next decimal value of $pi$ was found (increasing $n$ by $1$ starting from $0$ each time). So if $n$ is $23$ the
    output is $3.1$ and if it is $114$ the value is is $3.14$. Note that these values have
    decimal places after them that are not digits of $pi$. And this is in degrees, not radians.




  • This formula is derived from this one (Done with the sine rule):



    $ A=nr^2 cdot sin(frac{90n-180}{n}) cdot sin(frac{180}{n})$



    $A=$ The area of a proper polygon, $r=$ length from centre to corner, $n=$ number of sides




I substituted $A$ with $πr^2$ and simplified




  • It took me a while to realise that the difference between the first two values is larger than the difference between the second and third ($91$ then $73$)


  • If my program gets a new value I'll be adding it in. And if you want me to create an algorithm to find different data relating to this to help you solve it, or have an improved program do not hesitate to ask/suggest.


  • Technically speaking, the first element could be labelled as $12$ which gives the first digit as $3$.


  • Below are the pi values given when n is increased



enter image description here




  • Here is a GIF showing the increase of numbers as $n$ increases
    enter image description here










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    In radians and after simplification, you compute $frac n2sin frac{2pi}{n}$. It's half the perimeter of a polygon with $n/2$ sides.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 15 '18 at 15:47












  • $begingroup$
    This is in degrees. I'll add this in
    $endgroup$
    – yolo
    Dec 15 '18 at 15:48










  • $begingroup$
    Maybe I'm wrong, but I highly doubt you're going to find a formula for this.
    $endgroup$
    – Matt Samuel
    Dec 15 '18 at 15:58










  • $begingroup$
    Of course it's in degrees, that's why I translated this to radians. Degrees are useless when you want to use some calculus, and with calculus you get pretty easily that $frac n2 sin frac{2pi}{n}topi$ as $ntoinfty$, albeit very slowly. That's basically the method used by Archimedes to get his approximation $frac{22}7$. You could find an asymptotic approximation for your integer sequence, but like @MattSamuel I don't think there is a simple closed form.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 15 '18 at 15:58












  • $begingroup$
    A pi on both sides of the equation? $π=frac{n}{2} × sin(frac{2π}{n})$
    $endgroup$
    – yolo
    Dec 15 '18 at 16:02
















0












0








0





$begingroup$


Excuse the large title (The 'good title' page said not to be afraid to make it too long)



${23,114,187,473,2792,5624,19640,75884,187211,479797,1452795,5102858,14872865,72392867,146262888}$



I'm trying to figure out the nth term for these numbers. To help:





  • I'm generating these numbers on a python program:



    import math
    n=23
    digit=0
    while(True):
    pi=n*math.sin(math.radians(((90*n)-180)/(n)))*math.sin(math.radians(180/n))
    strpi=str(pi)
    count1=2
    while(count1<len(strpi)-1):
    if(strpi[count1]==str(math.pi)[count1]):
    if(digit<count1):
    print(n,"nn",pi,"n")
    digit=count1
    else:
    break
    count1+=1
    n+=1



  • For the non-programmers the program uses this formula:



    $ π=n cdot sin(frac{90n-180}{n}) cdot sin(frac{180}{n})$



    What this does is calculate $pi$ more accurately the higher the value of $n$. The program
    recorded the value of $n$ each time the next decimal value of $pi$ was found (increasing $n$ by $1$ starting from $0$ each time). So if $n$ is $23$ the
    output is $3.1$ and if it is $114$ the value is is $3.14$. Note that these values have
    decimal places after them that are not digits of $pi$. And this is in degrees, not radians.




  • This formula is derived from this one (Done with the sine rule):



    $ A=nr^2 cdot sin(frac{90n-180}{n}) cdot sin(frac{180}{n})$



    $A=$ The area of a proper polygon, $r=$ length from centre to corner, $n=$ number of sides




I substituted $A$ with $πr^2$ and simplified




  • It took me a while to realise that the difference between the first two values is larger than the difference between the second and third ($91$ then $73$)


  • If my program gets a new value I'll be adding it in. And if you want me to create an algorithm to find different data relating to this to help you solve it, or have an improved program do not hesitate to ask/suggest.


  • Technically speaking, the first element could be labelled as $12$ which gives the first digit as $3$.


  • Below are the pi values given when n is increased



enter image description here




  • Here is a GIF showing the increase of numbers as $n$ increases
    enter image description here










share|cite|improve this question











$endgroup$




Excuse the large title (The 'good title' page said not to be afraid to make it too long)



${23,114,187,473,2792,5624,19640,75884,187211,479797,1452795,5102858,14872865,72392867,146262888}$



I'm trying to figure out the nth term for these numbers. To help:





  • I'm generating these numbers on a python program:



    import math
    n=23
    digit=0
    while(True):
    pi=n*math.sin(math.radians(((90*n)-180)/(n)))*math.sin(math.radians(180/n))
    strpi=str(pi)
    count1=2
    while(count1<len(strpi)-1):
    if(strpi[count1]==str(math.pi)[count1]):
    if(digit<count1):
    print(n,"nn",pi,"n")
    digit=count1
    else:
    break
    count1+=1
    n+=1



  • For the non-programmers the program uses this formula:



    $ π=n cdot sin(frac{90n-180}{n}) cdot sin(frac{180}{n})$



    What this does is calculate $pi$ more accurately the higher the value of $n$. The program
    recorded the value of $n$ each time the next decimal value of $pi$ was found (increasing $n$ by $1$ starting from $0$ each time). So if $n$ is $23$ the
    output is $3.1$ and if it is $114$ the value is is $3.14$. Note that these values have
    decimal places after them that are not digits of $pi$. And this is in degrees, not radians.




  • This formula is derived from this one (Done with the sine rule):



    $ A=nr^2 cdot sin(frac{90n-180}{n}) cdot sin(frac{180}{n})$



    $A=$ The area of a proper polygon, $r=$ length from centre to corner, $n=$ number of sides




I substituted $A$ with $πr^2$ and simplified




  • It took me a while to realise that the difference between the first two values is larger than the difference between the second and third ($91$ then $73$)


  • If my program gets a new value I'll be adding it in. And if you want me to create an algorithm to find different data relating to this to help you solve it, or have an improved program do not hesitate to ask/suggest.


  • Technically speaking, the first element could be labelled as $12$ which gives the first digit as $3$.


  • Below are the pi values given when n is increased



enter image description here




  • Here is a GIF showing the increase of numbers as $n$ increases
    enter image description here







sequences-and-series pi






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 19:43







yolo

















asked Dec 15 '18 at 15:36









yoloyolo

1919




1919








  • 3




    $begingroup$
    In radians and after simplification, you compute $frac n2sin frac{2pi}{n}$. It's half the perimeter of a polygon with $n/2$ sides.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 15 '18 at 15:47












  • $begingroup$
    This is in degrees. I'll add this in
    $endgroup$
    – yolo
    Dec 15 '18 at 15:48










  • $begingroup$
    Maybe I'm wrong, but I highly doubt you're going to find a formula for this.
    $endgroup$
    – Matt Samuel
    Dec 15 '18 at 15:58










  • $begingroup$
    Of course it's in degrees, that's why I translated this to radians. Degrees are useless when you want to use some calculus, and with calculus you get pretty easily that $frac n2 sin frac{2pi}{n}topi$ as $ntoinfty$, albeit very slowly. That's basically the method used by Archimedes to get his approximation $frac{22}7$. You could find an asymptotic approximation for your integer sequence, but like @MattSamuel I don't think there is a simple closed form.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 15 '18 at 15:58












  • $begingroup$
    A pi on both sides of the equation? $π=frac{n}{2} × sin(frac{2π}{n})$
    $endgroup$
    – yolo
    Dec 15 '18 at 16:02
















  • 3




    $begingroup$
    In radians and after simplification, you compute $frac n2sin frac{2pi}{n}$. It's half the perimeter of a polygon with $n/2$ sides.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 15 '18 at 15:47












  • $begingroup$
    This is in degrees. I'll add this in
    $endgroup$
    – yolo
    Dec 15 '18 at 15:48










  • $begingroup$
    Maybe I'm wrong, but I highly doubt you're going to find a formula for this.
    $endgroup$
    – Matt Samuel
    Dec 15 '18 at 15:58










  • $begingroup$
    Of course it's in degrees, that's why I translated this to radians. Degrees are useless when you want to use some calculus, and with calculus you get pretty easily that $frac n2 sin frac{2pi}{n}topi$ as $ntoinfty$, albeit very slowly. That's basically the method used by Archimedes to get his approximation $frac{22}7$. You could find an asymptotic approximation for your integer sequence, but like @MattSamuel I don't think there is a simple closed form.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 15 '18 at 15:58












  • $begingroup$
    A pi on both sides of the equation? $π=frac{n}{2} × sin(frac{2π}{n})$
    $endgroup$
    – yolo
    Dec 15 '18 at 16:02










3




3




$begingroup$
In radians and after simplification, you compute $frac n2sin frac{2pi}{n}$. It's half the perimeter of a polygon with $n/2$ sides.
$endgroup$
– Jean-Claude Arbaut
Dec 15 '18 at 15:47






$begingroup$
In radians and after simplification, you compute $frac n2sin frac{2pi}{n}$. It's half the perimeter of a polygon with $n/2$ sides.
$endgroup$
– Jean-Claude Arbaut
Dec 15 '18 at 15:47














$begingroup$
This is in degrees. I'll add this in
$endgroup$
– yolo
Dec 15 '18 at 15:48




$begingroup$
This is in degrees. I'll add this in
$endgroup$
– yolo
Dec 15 '18 at 15:48












$begingroup$
Maybe I'm wrong, but I highly doubt you're going to find a formula for this.
$endgroup$
– Matt Samuel
Dec 15 '18 at 15:58




$begingroup$
Maybe I'm wrong, but I highly doubt you're going to find a formula for this.
$endgroup$
– Matt Samuel
Dec 15 '18 at 15:58












$begingroup$
Of course it's in degrees, that's why I translated this to radians. Degrees are useless when you want to use some calculus, and with calculus you get pretty easily that $frac n2 sin frac{2pi}{n}topi$ as $ntoinfty$, albeit very slowly. That's basically the method used by Archimedes to get his approximation $frac{22}7$. You could find an asymptotic approximation for your integer sequence, but like @MattSamuel I don't think there is a simple closed form.
$endgroup$
– Jean-Claude Arbaut
Dec 15 '18 at 15:58






$begingroup$
Of course it's in degrees, that's why I translated this to radians. Degrees are useless when you want to use some calculus, and with calculus you get pretty easily that $frac n2 sin frac{2pi}{n}topi$ as $ntoinfty$, albeit very slowly. That's basically the method used by Archimedes to get his approximation $frac{22}7$. You could find an asymptotic approximation for your integer sequence, but like @MattSamuel I don't think there is a simple closed form.
$endgroup$
– Jean-Claude Arbaut
Dec 15 '18 at 15:58














$begingroup$
A pi on both sides of the equation? $π=frac{n}{2} × sin(frac{2π}{n})$
$endgroup$
– yolo
Dec 15 '18 at 16:02






$begingroup$
A pi on both sides of the equation? $π=frac{n}{2} × sin(frac{2π}{n})$
$endgroup$
– yolo
Dec 15 '18 at 16:02












2 Answers
2






active

oldest

votes


















0












$begingroup$

As Jean-Claude Arbaut says, you are approximating $pi$ with $frac n2sin frac {2pi}n$. For $n$ large the argument of $sin$ is small and we can use the Taylor series. You are then computing $pi-frac n{12}left(frac {2pi}nright)^3+O(frac 1{n^4})$ The alternating series theorem says that the error of truncation is of the same sign as the first ignored term and smaller than it. You are looking for the times when the error gets smaller than the digits of $pi$ starting somewhere. For your first example, you need the error to be less than $0.04159265ldots$. The small gap between $114$ and $187$ comes because that digit of $pi$ is $1$. I am sure there is no easy way to compute the exact terms because they depend on the remainder after some number of digits, which can vary over a wide range. We can ask when $frac n{12}left(frac {2pi}nright)^3 lt 10^{-k}$, which comes when $ngt left(frac{12cdot 10^k}{8pi^3}right)^{1/3}$ and we would expect terms to be roughly multiplied by $sqrt {10} approx 3.162$



Added: I plotted $n$, the number of correct decimals, vs $log_{10}a(n)$ where $a(n)$ is the $n^{th}$ term from the sequence above. It forms a reasonable straight line with formula $log_{10}a(n)approx 0.9225+0.4806n$. I suspect with more terms in the sequence it would converge on $log_{10}a(n)=1+frac n2$ A plot is below
enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Are you saying the pattern is roughly $n^3* root 10$?
    $endgroup$
    – yolo
    Dec 16 '18 at 10:56












  • $begingroup$
    I plotted the number of decimals against the base $10$ log of the numbers in your series. The fit came out $0.9226+0.4806n$, so it goes up close to $sqrt {10}$ each time (the coefficient would be $0.5$). It is about $10^{1+frac n2}$
    $endgroup$
    – Ross Millikan
    Dec 16 '18 at 15:29










  • $begingroup$
    Can you get any closer?
    $endgroup$
    – yolo
    Dec 16 '18 at 16:05










  • $begingroup$
    I am sure with much more data the formula would converge on exactly what I gave. As I said, there is a fair amount of noise in the data because the remaining digits can vary so much.
    $endgroup$
    – Ross Millikan
    Dec 16 '18 at 16:13












  • $begingroup$
    probably shoulda mentioned this in the OP in the first place: but there seems to be some sort of pattern each time $n$ increases. Look at the pi values. They seem to have 5 extra 0s after diverging from pi. I'll post it.
    $endgroup$
    – yolo
    Dec 16 '18 at 16:17



















-1












$begingroup$

I did a quick check at the On-Line Encyclopedia of Integer Sequences® (OEIS®) at https://oeis.org/ and there was no know sequence in their database.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I don't think this is worth an answer. I did the same check myself and in finding nothing in the slightest I developed my opinion that the OP was unlikely to find a formula. I didn't post an answer.
    $endgroup$
    – Matt Samuel
    Dec 16 '18 at 11:56











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3041618%2fhelp-finding-the-formula-for-this-sequence-23-114-187-473-2792%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

As Jean-Claude Arbaut says, you are approximating $pi$ with $frac n2sin frac {2pi}n$. For $n$ large the argument of $sin$ is small and we can use the Taylor series. You are then computing $pi-frac n{12}left(frac {2pi}nright)^3+O(frac 1{n^4})$ The alternating series theorem says that the error of truncation is of the same sign as the first ignored term and smaller than it. You are looking for the times when the error gets smaller than the digits of $pi$ starting somewhere. For your first example, you need the error to be less than $0.04159265ldots$. The small gap between $114$ and $187$ comes because that digit of $pi$ is $1$. I am sure there is no easy way to compute the exact terms because they depend on the remainder after some number of digits, which can vary over a wide range. We can ask when $frac n{12}left(frac {2pi}nright)^3 lt 10^{-k}$, which comes when $ngt left(frac{12cdot 10^k}{8pi^3}right)^{1/3}$ and we would expect terms to be roughly multiplied by $sqrt {10} approx 3.162$



Added: I plotted $n$, the number of correct decimals, vs $log_{10}a(n)$ where $a(n)$ is the $n^{th}$ term from the sequence above. It forms a reasonable straight line with formula $log_{10}a(n)approx 0.9225+0.4806n$. I suspect with more terms in the sequence it would converge on $log_{10}a(n)=1+frac n2$ A plot is below
enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Are you saying the pattern is roughly $n^3* root 10$?
    $endgroup$
    – yolo
    Dec 16 '18 at 10:56












  • $begingroup$
    I plotted the number of decimals against the base $10$ log of the numbers in your series. The fit came out $0.9226+0.4806n$, so it goes up close to $sqrt {10}$ each time (the coefficient would be $0.5$). It is about $10^{1+frac n2}$
    $endgroup$
    – Ross Millikan
    Dec 16 '18 at 15:29










  • $begingroup$
    Can you get any closer?
    $endgroup$
    – yolo
    Dec 16 '18 at 16:05










  • $begingroup$
    I am sure with much more data the formula would converge on exactly what I gave. As I said, there is a fair amount of noise in the data because the remaining digits can vary so much.
    $endgroup$
    – Ross Millikan
    Dec 16 '18 at 16:13












  • $begingroup$
    probably shoulda mentioned this in the OP in the first place: but there seems to be some sort of pattern each time $n$ increases. Look at the pi values. They seem to have 5 extra 0s after diverging from pi. I'll post it.
    $endgroup$
    – yolo
    Dec 16 '18 at 16:17
















0












$begingroup$

As Jean-Claude Arbaut says, you are approximating $pi$ with $frac n2sin frac {2pi}n$. For $n$ large the argument of $sin$ is small and we can use the Taylor series. You are then computing $pi-frac n{12}left(frac {2pi}nright)^3+O(frac 1{n^4})$ The alternating series theorem says that the error of truncation is of the same sign as the first ignored term and smaller than it. You are looking for the times when the error gets smaller than the digits of $pi$ starting somewhere. For your first example, you need the error to be less than $0.04159265ldots$. The small gap between $114$ and $187$ comes because that digit of $pi$ is $1$. I am sure there is no easy way to compute the exact terms because they depend on the remainder after some number of digits, which can vary over a wide range. We can ask when $frac n{12}left(frac {2pi}nright)^3 lt 10^{-k}$, which comes when $ngt left(frac{12cdot 10^k}{8pi^3}right)^{1/3}$ and we would expect terms to be roughly multiplied by $sqrt {10} approx 3.162$



Added: I plotted $n$, the number of correct decimals, vs $log_{10}a(n)$ where $a(n)$ is the $n^{th}$ term from the sequence above. It forms a reasonable straight line with formula $log_{10}a(n)approx 0.9225+0.4806n$. I suspect with more terms in the sequence it would converge on $log_{10}a(n)=1+frac n2$ A plot is below
enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Are you saying the pattern is roughly $n^3* root 10$?
    $endgroup$
    – yolo
    Dec 16 '18 at 10:56












  • $begingroup$
    I plotted the number of decimals against the base $10$ log of the numbers in your series. The fit came out $0.9226+0.4806n$, so it goes up close to $sqrt {10}$ each time (the coefficient would be $0.5$). It is about $10^{1+frac n2}$
    $endgroup$
    – Ross Millikan
    Dec 16 '18 at 15:29










  • $begingroup$
    Can you get any closer?
    $endgroup$
    – yolo
    Dec 16 '18 at 16:05










  • $begingroup$
    I am sure with much more data the formula would converge on exactly what I gave. As I said, there is a fair amount of noise in the data because the remaining digits can vary so much.
    $endgroup$
    – Ross Millikan
    Dec 16 '18 at 16:13












  • $begingroup$
    probably shoulda mentioned this in the OP in the first place: but there seems to be some sort of pattern each time $n$ increases. Look at the pi values. They seem to have 5 extra 0s after diverging from pi. I'll post it.
    $endgroup$
    – yolo
    Dec 16 '18 at 16:17














0












0








0





$begingroup$

As Jean-Claude Arbaut says, you are approximating $pi$ with $frac n2sin frac {2pi}n$. For $n$ large the argument of $sin$ is small and we can use the Taylor series. You are then computing $pi-frac n{12}left(frac {2pi}nright)^3+O(frac 1{n^4})$ The alternating series theorem says that the error of truncation is of the same sign as the first ignored term and smaller than it. You are looking for the times when the error gets smaller than the digits of $pi$ starting somewhere. For your first example, you need the error to be less than $0.04159265ldots$. The small gap between $114$ and $187$ comes because that digit of $pi$ is $1$. I am sure there is no easy way to compute the exact terms because they depend on the remainder after some number of digits, which can vary over a wide range. We can ask when $frac n{12}left(frac {2pi}nright)^3 lt 10^{-k}$, which comes when $ngt left(frac{12cdot 10^k}{8pi^3}right)^{1/3}$ and we would expect terms to be roughly multiplied by $sqrt {10} approx 3.162$



Added: I plotted $n$, the number of correct decimals, vs $log_{10}a(n)$ where $a(n)$ is the $n^{th}$ term from the sequence above. It forms a reasonable straight line with formula $log_{10}a(n)approx 0.9225+0.4806n$. I suspect with more terms in the sequence it would converge on $log_{10}a(n)=1+frac n2$ A plot is below
enter image description here






share|cite|improve this answer











$endgroup$



As Jean-Claude Arbaut says, you are approximating $pi$ with $frac n2sin frac {2pi}n$. For $n$ large the argument of $sin$ is small and we can use the Taylor series. You are then computing $pi-frac n{12}left(frac {2pi}nright)^3+O(frac 1{n^4})$ The alternating series theorem says that the error of truncation is of the same sign as the first ignored term and smaller than it. You are looking for the times when the error gets smaller than the digits of $pi$ starting somewhere. For your first example, you need the error to be less than $0.04159265ldots$. The small gap between $114$ and $187$ comes because that digit of $pi$ is $1$. I am sure there is no easy way to compute the exact terms because they depend on the remainder after some number of digits, which can vary over a wide range. We can ask when $frac n{12}left(frac {2pi}nright)^3 lt 10^{-k}$, which comes when $ngt left(frac{12cdot 10^k}{8pi^3}right)^{1/3}$ and we would expect terms to be roughly multiplied by $sqrt {10} approx 3.162$



Added: I plotted $n$, the number of correct decimals, vs $log_{10}a(n)$ where $a(n)$ is the $n^{th}$ term from the sequence above. It forms a reasonable straight line with formula $log_{10}a(n)approx 0.9225+0.4806n$. I suspect with more terms in the sequence it would converge on $log_{10}a(n)=1+frac n2$ A plot is below
enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 21:16

























answered Dec 15 '18 at 23:49









Ross MillikanRoss Millikan

297k23198371




297k23198371












  • $begingroup$
    Are you saying the pattern is roughly $n^3* root 10$?
    $endgroup$
    – yolo
    Dec 16 '18 at 10:56












  • $begingroup$
    I plotted the number of decimals against the base $10$ log of the numbers in your series. The fit came out $0.9226+0.4806n$, so it goes up close to $sqrt {10}$ each time (the coefficient would be $0.5$). It is about $10^{1+frac n2}$
    $endgroup$
    – Ross Millikan
    Dec 16 '18 at 15:29










  • $begingroup$
    Can you get any closer?
    $endgroup$
    – yolo
    Dec 16 '18 at 16:05










  • $begingroup$
    I am sure with much more data the formula would converge on exactly what I gave. As I said, there is a fair amount of noise in the data because the remaining digits can vary so much.
    $endgroup$
    – Ross Millikan
    Dec 16 '18 at 16:13












  • $begingroup$
    probably shoulda mentioned this in the OP in the first place: but there seems to be some sort of pattern each time $n$ increases. Look at the pi values. They seem to have 5 extra 0s after diverging from pi. I'll post it.
    $endgroup$
    – yolo
    Dec 16 '18 at 16:17


















  • $begingroup$
    Are you saying the pattern is roughly $n^3* root 10$?
    $endgroup$
    – yolo
    Dec 16 '18 at 10:56












  • $begingroup$
    I plotted the number of decimals against the base $10$ log of the numbers in your series. The fit came out $0.9226+0.4806n$, so it goes up close to $sqrt {10}$ each time (the coefficient would be $0.5$). It is about $10^{1+frac n2}$
    $endgroup$
    – Ross Millikan
    Dec 16 '18 at 15:29










  • $begingroup$
    Can you get any closer?
    $endgroup$
    – yolo
    Dec 16 '18 at 16:05










  • $begingroup$
    I am sure with much more data the formula would converge on exactly what I gave. As I said, there is a fair amount of noise in the data because the remaining digits can vary so much.
    $endgroup$
    – Ross Millikan
    Dec 16 '18 at 16:13












  • $begingroup$
    probably shoulda mentioned this in the OP in the first place: but there seems to be some sort of pattern each time $n$ increases. Look at the pi values. They seem to have 5 extra 0s after diverging from pi. I'll post it.
    $endgroup$
    – yolo
    Dec 16 '18 at 16:17
















$begingroup$
Are you saying the pattern is roughly $n^3* root 10$?
$endgroup$
– yolo
Dec 16 '18 at 10:56






$begingroup$
Are you saying the pattern is roughly $n^3* root 10$?
$endgroup$
– yolo
Dec 16 '18 at 10:56














$begingroup$
I plotted the number of decimals against the base $10$ log of the numbers in your series. The fit came out $0.9226+0.4806n$, so it goes up close to $sqrt {10}$ each time (the coefficient would be $0.5$). It is about $10^{1+frac n2}$
$endgroup$
– Ross Millikan
Dec 16 '18 at 15:29




$begingroup$
I plotted the number of decimals against the base $10$ log of the numbers in your series. The fit came out $0.9226+0.4806n$, so it goes up close to $sqrt {10}$ each time (the coefficient would be $0.5$). It is about $10^{1+frac n2}$
$endgroup$
– Ross Millikan
Dec 16 '18 at 15:29












$begingroup$
Can you get any closer?
$endgroup$
– yolo
Dec 16 '18 at 16:05




$begingroup$
Can you get any closer?
$endgroup$
– yolo
Dec 16 '18 at 16:05












$begingroup$
I am sure with much more data the formula would converge on exactly what I gave. As I said, there is a fair amount of noise in the data because the remaining digits can vary so much.
$endgroup$
– Ross Millikan
Dec 16 '18 at 16:13






$begingroup$
I am sure with much more data the formula would converge on exactly what I gave. As I said, there is a fair amount of noise in the data because the remaining digits can vary so much.
$endgroup$
– Ross Millikan
Dec 16 '18 at 16:13














$begingroup$
probably shoulda mentioned this in the OP in the first place: but there seems to be some sort of pattern each time $n$ increases. Look at the pi values. They seem to have 5 extra 0s after diverging from pi. I'll post it.
$endgroup$
– yolo
Dec 16 '18 at 16:17




$begingroup$
probably shoulda mentioned this in the OP in the first place: but there seems to be some sort of pattern each time $n$ increases. Look at the pi values. They seem to have 5 extra 0s after diverging from pi. I'll post it.
$endgroup$
– yolo
Dec 16 '18 at 16:17











-1












$begingroup$

I did a quick check at the On-Line Encyclopedia of Integer Sequences® (OEIS®) at https://oeis.org/ and there was no know sequence in their database.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I don't think this is worth an answer. I did the same check myself and in finding nothing in the slightest I developed my opinion that the OP was unlikely to find a formula. I didn't post an answer.
    $endgroup$
    – Matt Samuel
    Dec 16 '18 at 11:56
















-1












$begingroup$

I did a quick check at the On-Line Encyclopedia of Integer Sequences® (OEIS®) at https://oeis.org/ and there was no know sequence in their database.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I don't think this is worth an answer. I did the same check myself and in finding nothing in the slightest I developed my opinion that the OP was unlikely to find a formula. I didn't post an answer.
    $endgroup$
    – Matt Samuel
    Dec 16 '18 at 11:56














-1












-1








-1





$begingroup$

I did a quick check at the On-Line Encyclopedia of Integer Sequences® (OEIS®) at https://oeis.org/ and there was no know sequence in their database.






share|cite|improve this answer









$endgroup$



I did a quick check at the On-Line Encyclopedia of Integer Sequences® (OEIS®) at https://oeis.org/ and there was no know sequence in their database.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 15 '18 at 22:38









Lorenz H MenkeLorenz H Menke

10411




10411












  • $begingroup$
    I don't think this is worth an answer. I did the same check myself and in finding nothing in the slightest I developed my opinion that the OP was unlikely to find a formula. I didn't post an answer.
    $endgroup$
    – Matt Samuel
    Dec 16 '18 at 11:56


















  • $begingroup$
    I don't think this is worth an answer. I did the same check myself and in finding nothing in the slightest I developed my opinion that the OP was unlikely to find a formula. I didn't post an answer.
    $endgroup$
    – Matt Samuel
    Dec 16 '18 at 11:56
















$begingroup$
I don't think this is worth an answer. I did the same check myself and in finding nothing in the slightest I developed my opinion that the OP was unlikely to find a formula. I didn't post an answer.
$endgroup$
– Matt Samuel
Dec 16 '18 at 11:56




$begingroup$
I don't think this is worth an answer. I did the same check myself and in finding nothing in the slightest I developed my opinion that the OP was unlikely to find a formula. I didn't post an answer.
$endgroup$
– Matt Samuel
Dec 16 '18 at 11:56


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3041618%2fhelp-finding-the-formula-for-this-sequence-23-114-187-473-2792%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Aardman Animations

Are they similar matrix