Enigmatic patterns in Archimedean spirals The Next CEO of Stack OverflowDivide circle into 9 pieces of equal areaLength and breadth of a rectangle enclosed between two semi-circles of given radiiIs there a plane algebraic curve with just 3-fold rotational symmetry, but without reflection symmetry?While calculating the arctan (1.01236) do we have to do 2 steps or one step before Taylor series?“Secondary spirals” within Fermat spiralsWhy is this family of dynamical systems able to produce spirals and clusters of points?How to prove the shape of the “Flower of Venus”Visualizing rational numbers as multiplication graphsThe slope of $nx % m$The quadrature of the circle: comparing Archimedean and Ulam spirals

When airplanes disconnect from a tanker during air to air refueling, why do they bank so sharply to the right?

Whats the best way to handle refactoring a big file?

If the heap is initialized for security, then why is the stack uninitialized?

WOW air has ceased operation, can I get my tickets refunded?

Does the Brexit deal have to be agreed by both Houses?

Robert Sheckley short story about vacation spots being overwhelmed

How easy is it to start Magic from scratch?

How do I get the green key off the shelf in the Dobby level of Lego Harry Potter 2?

How can I get through very long and very dry, but also very useful technical documents when learning a new tool?

Anatomically Correct Mesopelagic Aves

Trouble understanding the speech of overseas colleagues

Example of a Mathematician/Physicist whose Other Publications during their PhD eclipsed their PhD Thesis

How to be diplomatic in refusing to write code that breaches the privacy of our users

Science fiction (dystopian) short story set after WWIII

How to write papers efficiently when English isn't my first language?

Horror movie/show or scene where a horse creature opens its mouth really wide and devours a man in a stables

Why do professional authors make "consistency" mistakes? And how to avoid them?

Go Pregnant or Go Home

Why is there a PLL in CPU?

How long to clear the 'suck zone' of a turbofan after start is initiated?

Customer Requests (Sometimes) Drive Me Bonkers!

% symbol leads to superlong (forever?) compilations

What is the difference between "behavior" and "behaviour"?

Why did we only see the N-1 starfighters in one film?



Enigmatic patterns in Archimedean spirals



The Next CEO of Stack OverflowDivide circle into 9 pieces of equal areaLength and breadth of a rectangle enclosed between two semi-circles of given radiiIs there a plane algebraic curve with just 3-fold rotational symmetry, but without reflection symmetry?While calculating the arctan (1.01236) do we have to do 2 steps or one step before Taylor series?“Secondary spirals” within Fermat spiralsWhy is this family of dynamical systems able to produce spirals and clusters of points?How to prove the shape of the “Flower of Venus”Visualizing rational numbers as multiplication graphsThe slope of $nx % m$The quadrature of the circle: comparing Archimedean and Ulam spirals










4












$begingroup$


Distributing the natural numbers as circles evenly along the Archimedean spiral yields surprising patterns when changing the radius of the circles: they cover more and more of the plane, finally covering it completely. But shortly before this happens, more or less obvious and intricate patterns shortly show up, and I'd like to understand them.



The numbers are arranged by these formulas with $hat k = fracsqrtk2$



$$x_alpha(k) = -hat kcos(alphacdot 2 picdot hat k)$$
$$y_alpha(k) = -hat ksin(alphacdot 2 picdot hat k)$$



– the distance between consecutive numbers along the spiral being controlled by the parameter $alpha$.



This is how the number spirals look like for $alpha = fracsqrt22, 1, sqrt2, 2,2 sqrt2,4, ldots = sqrt2^k$:



enter image description here
[Click image to enlarge.]



When the radius of the circles is enlarged until they almost cover the plane, one observes different spot patterns: a spiral, a cross with $8$-fold rotation symmetry, another cross with $4$-fold rotation symmetry and a series of horizontal stripes.



enter image description here
[Click image to enlarge.]



When we compare $alpha = 0.99, 1, 1.01$ we see that the straight cross for $alpha = 1$ is the limit of two bundles of eight spirals running in opposite directions.



enter image description here
[Click image to enlarge.]



Here is my question:




How can the cross with the 8-fold rotational symmetry (for $alpha=1$)
be explained? ("For $alpha=1$ there is a cross with a 8-fold
rotational symmetry, because ....")




Note that for $alpha=1$ the square numbers are aligned along the horizontal axis:



enter image description here




One final remark: The observed patterns come as a surprise (somehow) because "from the distance" the spiral looks almost like a set of concentric circles thus having seemingly total rotational symmetry. But in fact the spiral has no rotational symmetry at all (only for 360°). And the ever changing patterns are due to this fact: the concentric circles do show only patterns reflecting the density of numbers along them.



In this example the density is $10$ per $2pi$. Note how the triangle numbers (compared to the square numbers in the case of the spiral) are arranged along the horizontal line. Let $triangle(k) = frack(k+1)2$ take the values $0,1,3,6,10,15,dots$ We observe along the horizontal line $n(k) = 10triangle(k) + 1$ (to the right) and $m(k) = n(k) + 5k$ (to the left):



enter image description here



enter image description here




To see how the spiral patterns continue here for $alpha = 4sqrt2, 8, 8sqrt2$:



enter image description here



There's a little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 4.










share|cite|improve this question











$endgroup$
















    4












    $begingroup$


    Distributing the natural numbers as circles evenly along the Archimedean spiral yields surprising patterns when changing the radius of the circles: they cover more and more of the plane, finally covering it completely. But shortly before this happens, more or less obvious and intricate patterns shortly show up, and I'd like to understand them.



    The numbers are arranged by these formulas with $hat k = fracsqrtk2$



    $$x_alpha(k) = -hat kcos(alphacdot 2 picdot hat k)$$
    $$y_alpha(k) = -hat ksin(alphacdot 2 picdot hat k)$$



    – the distance between consecutive numbers along the spiral being controlled by the parameter $alpha$.



    This is how the number spirals look like for $alpha = fracsqrt22, 1, sqrt2, 2,2 sqrt2,4, ldots = sqrt2^k$:



    enter image description here
    [Click image to enlarge.]



    When the radius of the circles is enlarged until they almost cover the plane, one observes different spot patterns: a spiral, a cross with $8$-fold rotation symmetry, another cross with $4$-fold rotation symmetry and a series of horizontal stripes.



    enter image description here
    [Click image to enlarge.]



    When we compare $alpha = 0.99, 1, 1.01$ we see that the straight cross for $alpha = 1$ is the limit of two bundles of eight spirals running in opposite directions.



    enter image description here
    [Click image to enlarge.]



    Here is my question:




    How can the cross with the 8-fold rotational symmetry (for $alpha=1$)
    be explained? ("For $alpha=1$ there is a cross with a 8-fold
    rotational symmetry, because ....")




    Note that for $alpha=1$ the square numbers are aligned along the horizontal axis:



    enter image description here




    One final remark: The observed patterns come as a surprise (somehow) because "from the distance" the spiral looks almost like a set of concentric circles thus having seemingly total rotational symmetry. But in fact the spiral has no rotational symmetry at all (only for 360°). And the ever changing patterns are due to this fact: the concentric circles do show only patterns reflecting the density of numbers along them.



    In this example the density is $10$ per $2pi$. Note how the triangle numbers (compared to the square numbers in the case of the spiral) are arranged along the horizontal line. Let $triangle(k) = frack(k+1)2$ take the values $0,1,3,6,10,15,dots$ We observe along the horizontal line $n(k) = 10triangle(k) + 1$ (to the right) and $m(k) = n(k) + 5k$ (to the left):



    enter image description here



    enter image description here




    To see how the spiral patterns continue here for $alpha = 4sqrt2, 8, 8sqrt2$:



    enter image description here



    There's a little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 4.










    share|cite|improve this question











    $endgroup$














      4












      4








      4


      3



      $begingroup$


      Distributing the natural numbers as circles evenly along the Archimedean spiral yields surprising patterns when changing the radius of the circles: they cover more and more of the plane, finally covering it completely. But shortly before this happens, more or less obvious and intricate patterns shortly show up, and I'd like to understand them.



      The numbers are arranged by these formulas with $hat k = fracsqrtk2$



      $$x_alpha(k) = -hat kcos(alphacdot 2 picdot hat k)$$
      $$y_alpha(k) = -hat ksin(alphacdot 2 picdot hat k)$$



      – the distance between consecutive numbers along the spiral being controlled by the parameter $alpha$.



      This is how the number spirals look like for $alpha = fracsqrt22, 1, sqrt2, 2,2 sqrt2,4, ldots = sqrt2^k$:



      enter image description here
      [Click image to enlarge.]



      When the radius of the circles is enlarged until they almost cover the plane, one observes different spot patterns: a spiral, a cross with $8$-fold rotation symmetry, another cross with $4$-fold rotation symmetry and a series of horizontal stripes.



      enter image description here
      [Click image to enlarge.]



      When we compare $alpha = 0.99, 1, 1.01$ we see that the straight cross for $alpha = 1$ is the limit of two bundles of eight spirals running in opposite directions.



      enter image description here
      [Click image to enlarge.]



      Here is my question:




      How can the cross with the 8-fold rotational symmetry (for $alpha=1$)
      be explained? ("For $alpha=1$ there is a cross with a 8-fold
      rotational symmetry, because ....")




      Note that for $alpha=1$ the square numbers are aligned along the horizontal axis:



      enter image description here




      One final remark: The observed patterns come as a surprise (somehow) because "from the distance" the spiral looks almost like a set of concentric circles thus having seemingly total rotational symmetry. But in fact the spiral has no rotational symmetry at all (only for 360°). And the ever changing patterns are due to this fact: the concentric circles do show only patterns reflecting the density of numbers along them.



      In this example the density is $10$ per $2pi$. Note how the triangle numbers (compared to the square numbers in the case of the spiral) are arranged along the horizontal line. Let $triangle(k) = frack(k+1)2$ take the values $0,1,3,6,10,15,dots$ We observe along the horizontal line $n(k) = 10triangle(k) + 1$ (to the right) and $m(k) = n(k) + 5k$ (to the left):



      enter image description here



      enter image description here




      To see how the spiral patterns continue here for $alpha = 4sqrt2, 8, 8sqrt2$:



      enter image description here



      There's a little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 4.










      share|cite|improve this question











      $endgroup$




      Distributing the natural numbers as circles evenly along the Archimedean spiral yields surprising patterns when changing the radius of the circles: they cover more and more of the plane, finally covering it completely. But shortly before this happens, more or less obvious and intricate patterns shortly show up, and I'd like to understand them.



      The numbers are arranged by these formulas with $hat k = fracsqrtk2$



      $$x_alpha(k) = -hat kcos(alphacdot 2 picdot hat k)$$
      $$y_alpha(k) = -hat ksin(alphacdot 2 picdot hat k)$$



      – the distance between consecutive numbers along the spiral being controlled by the parameter $alpha$.



      This is how the number spirals look like for $alpha = fracsqrt22, 1, sqrt2, 2,2 sqrt2,4, ldots = sqrt2^k$:



      enter image description here
      [Click image to enlarge.]



      When the radius of the circles is enlarged until they almost cover the plane, one observes different spot patterns: a spiral, a cross with $8$-fold rotation symmetry, another cross with $4$-fold rotation symmetry and a series of horizontal stripes.



      enter image description here
      [Click image to enlarge.]



      When we compare $alpha = 0.99, 1, 1.01$ we see that the straight cross for $alpha = 1$ is the limit of two bundles of eight spirals running in opposite directions.



      enter image description here
      [Click image to enlarge.]



      Here is my question:




      How can the cross with the 8-fold rotational symmetry (for $alpha=1$)
      be explained? ("For $alpha=1$ there is a cross with a 8-fold
      rotational symmetry, because ....")




      Note that for $alpha=1$ the square numbers are aligned along the horizontal axis:



      enter image description here




      One final remark: The observed patterns come as a surprise (somehow) because "from the distance" the spiral looks almost like a set of concentric circles thus having seemingly total rotational symmetry. But in fact the spiral has no rotational symmetry at all (only for 360°). And the ever changing patterns are due to this fact: the concentric circles do show only patterns reflecting the density of numbers along them.



      In this example the density is $10$ per $2pi$. Note how the triangle numbers (compared to the square numbers in the case of the spiral) are arranged along the horizontal line. Let $triangle(k) = frack(k+1)2$ take the values $0,1,3,6,10,15,dots$ We observe along the horizontal line $n(k) = 10triangle(k) + 1$ (to the right) and $m(k) = n(k) + 5k$ (to the left):



      enter image description here



      enter image description here




      To see how the spiral patterns continue here for $alpha = 4sqrt2, 8, 8sqrt2$:



      enter image description here



      There's a little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 4.







      geometry arithmetic visualization






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 4 hours ago







      Hans-Peter Stricker

















      asked yesterday









      Hans-Peter StrickerHans-Peter Stricker

      6,74043995




      6,74043995




















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          Let me try. We see a pattern where there is particularly much space between the circles. This happens when the circles are lined up with each other, so four neighbouring circles make an approximate rectangle.



          In the $n$'th revolution starting at the positive $x$-axis, there are $(2n+1)^2-(2n-1)^2 = 8n$ circles. This means that we can divide the revolution into 8 parts with $n$ circles each, and the circles must line up with each other at each division (and nowhere else). This explains the pattern.



          Have you tried making a circular pattern with $8n$ numbers in each circle? I suspect the same pattern would emerge (if the distance between the concentric circles are chosen appropriately).



          This also explains the other symmetric patterns. We can work out that if $alpha = sqrt 2^c$ for $cle 3$, then the circles that lie just before the positive $x$-axis will be $k=2^-ccdot((2n+1)^2-1) = 2^2-ccdot ncdot(n+1)$, which implies that there will be $2^3-ccdot n$ circles in the $n$'th revolution. This means that the circles are lined up just below the positive $x$-axis, and we have $2^3-c$-fold rotational symmetry. This matches what we observe on your plots, which correspond to $-1 le c le 4$. ($c=4$ is not explained by my calculation). I believe that you can see the 16-fold symmetry in the first plot, if you squint a little.






          share|cite|improve this answer








          New contributor




          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$












          • $begingroup$
            Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago


















          1












          $begingroup$

          This answer complements the other one by Milten; it explains the linear patterns in the diagrams. First consider the case $alpha=2$, which corresponds to the function $z_k:=sqrtke^2pi isqrtk$ (it is different from OP's function by a half). Take different cases of $k$.



          (i) $k=n^2+m$. $$z_n=x_n+iy_n=sqrtn^2+m,e^2pi isqrtn^2+m=sqrtn^2+m,e^2pi i n(1+fracmn^2)^1/2approx ne^pi im/n$$ so $x_napprox n$, $y_napprox n(fracpi mn-fracm^3pi^36n^3)approxpi m-fracm^3pi^36n^2$, that is $$y_napprox pi m-fracAx_n^2to pi m$$ This explains the right-hand lines in OP's diagrams with $m=ldots,-1,0,1,2,ldots$.



          (ii) $k=n^2+n+m$. $$z_n=sqrtn^2+n+m,e^2pi isqrtn^2+n+mapprox (n+tfrac12)e^2pi ine^pi ie^pi i m/napprox-(n+tfrac12)e^pi im/n$$ This is similar to case (i) but reflected, so the lines appear on the left side.



          Now consider $alpha=2sqrt2$ which corresponds to $z_k:=sqrtke^2pi isqrt2k$. The same type of analysis gives $$z_napprox n,e^2pi im/n$$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago


















          1












          $begingroup$

          To give some visual sugar to user Milten's nice answer find here highlighted those numbers that make approximate rectangles. You'll easily see the number pattern and see the role that the number $8$ plays, giving typical triangle number like sequences, e.g. $a = 2,12,30,56,90,ldots$ with $a_k = 8triangle(k) + 2(k+1)$ or $b = 3,13,31,57,91,ldots$ with $b_k = a_k +1$, $k=0,1,2,ldots$.



          enter image description here



          For the sake of comparison the circular pattern with $8n$ numbers in each circle. The similar numbers make approximate rectangles:



          enter image description here




          What needs to be shown analytically is that for example for the numbers $a_k,b_k, a_k+1, b_k+1$ we have



          $$|p(a_k) - p(b_k)| approx |p(a_k+1) - p(b_k+1)|$$



          and



          $$|p(a_k) - p(a_k+1)| approx |p(b_k) - p(b_k+1)|$$



          with $p(k) = langle x(k), y(k)rangle$ and $x(k) = -hat kcos(2 picdot hat k)$,
          $y(k) = -hat ksin(2 picdot hat k)$ and $hat k = fracsqrtk2$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_k+1)$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_k+1)$), since the line between $p(a_k)$ and $p(b_k)$ is perpendicular to centre.
            $endgroup$
            – Milten
            17 hours ago











          • $begingroup$
            I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $fracpi8n$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though.
            $endgroup$
            – Milten
            17 hours ago










          • $begingroup$
            Can you give an analytical expression for this requirement, please? I would add it to my answer.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            Did it. But still there's the ugly gap between 23 and 9.
            $endgroup$
            – Hans-Peter Stricker
            16 hours ago










          • $begingroup$
            Hmm, we could use the dot product: $(p(a_k+1-p(a_k))cdot(p(a_k)-p(b_k)) approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_k+1) parallel p(a_k)$, which means they have the same angle. So we could write it as $2pi hat a_k+1 approx 2pi (hat a_k+1) iff sqrta_k+1 approx sqrta_k + 2$. (+1 since we go once around the spiral).
            $endgroup$
            – Milten
            16 hours ago












          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%2f3164391%2fenigmatic-patterns-in-archimedean-spirals%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Let me try. We see a pattern where there is particularly much space between the circles. This happens when the circles are lined up with each other, so four neighbouring circles make an approximate rectangle.



          In the $n$'th revolution starting at the positive $x$-axis, there are $(2n+1)^2-(2n-1)^2 = 8n$ circles. This means that we can divide the revolution into 8 parts with $n$ circles each, and the circles must line up with each other at each division (and nowhere else). This explains the pattern.



          Have you tried making a circular pattern with $8n$ numbers in each circle? I suspect the same pattern would emerge (if the distance between the concentric circles are chosen appropriately).



          This also explains the other symmetric patterns. We can work out that if $alpha = sqrt 2^c$ for $cle 3$, then the circles that lie just before the positive $x$-axis will be $k=2^-ccdot((2n+1)^2-1) = 2^2-ccdot ncdot(n+1)$, which implies that there will be $2^3-ccdot n$ circles in the $n$'th revolution. This means that the circles are lined up just below the positive $x$-axis, and we have $2^3-c$-fold rotational symmetry. This matches what we observe on your plots, which correspond to $-1 le c le 4$. ($c=4$ is not explained by my calculation). I believe that you can see the 16-fold symmetry in the first plot, if you squint a little.






          share|cite|improve this answer








          New contributor




          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$












          • $begingroup$
            Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago















          2












          $begingroup$

          Let me try. We see a pattern where there is particularly much space between the circles. This happens when the circles are lined up with each other, so four neighbouring circles make an approximate rectangle.



          In the $n$'th revolution starting at the positive $x$-axis, there are $(2n+1)^2-(2n-1)^2 = 8n$ circles. This means that we can divide the revolution into 8 parts with $n$ circles each, and the circles must line up with each other at each division (and nowhere else). This explains the pattern.



          Have you tried making a circular pattern with $8n$ numbers in each circle? I suspect the same pattern would emerge (if the distance between the concentric circles are chosen appropriately).



          This also explains the other symmetric patterns. We can work out that if $alpha = sqrt 2^c$ for $cle 3$, then the circles that lie just before the positive $x$-axis will be $k=2^-ccdot((2n+1)^2-1) = 2^2-ccdot ncdot(n+1)$, which implies that there will be $2^3-ccdot n$ circles in the $n$'th revolution. This means that the circles are lined up just below the positive $x$-axis, and we have $2^3-c$-fold rotational symmetry. This matches what we observe on your plots, which correspond to $-1 le c le 4$. ($c=4$ is not explained by my calculation). I believe that you can see the 16-fold symmetry in the first plot, if you squint a little.






          share|cite|improve this answer








          New contributor




          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$












          • $begingroup$
            Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago













          2












          2








          2





          $begingroup$

          Let me try. We see a pattern where there is particularly much space between the circles. This happens when the circles are lined up with each other, so four neighbouring circles make an approximate rectangle.



          In the $n$'th revolution starting at the positive $x$-axis, there are $(2n+1)^2-(2n-1)^2 = 8n$ circles. This means that we can divide the revolution into 8 parts with $n$ circles each, and the circles must line up with each other at each division (and nowhere else). This explains the pattern.



          Have you tried making a circular pattern with $8n$ numbers in each circle? I suspect the same pattern would emerge (if the distance between the concentric circles are chosen appropriately).



          This also explains the other symmetric patterns. We can work out that if $alpha = sqrt 2^c$ for $cle 3$, then the circles that lie just before the positive $x$-axis will be $k=2^-ccdot((2n+1)^2-1) = 2^2-ccdot ncdot(n+1)$, which implies that there will be $2^3-ccdot n$ circles in the $n$'th revolution. This means that the circles are lined up just below the positive $x$-axis, and we have $2^3-c$-fold rotational symmetry. This matches what we observe on your plots, which correspond to $-1 le c le 4$. ($c=4$ is not explained by my calculation). I believe that you can see the 16-fold symmetry in the first plot, if you squint a little.






          share|cite|improve this answer








          New contributor




          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$



          Let me try. We see a pattern where there is particularly much space between the circles. This happens when the circles are lined up with each other, so four neighbouring circles make an approximate rectangle.



          In the $n$'th revolution starting at the positive $x$-axis, there are $(2n+1)^2-(2n-1)^2 = 8n$ circles. This means that we can divide the revolution into 8 parts with $n$ circles each, and the circles must line up with each other at each division (and nowhere else). This explains the pattern.



          Have you tried making a circular pattern with $8n$ numbers in each circle? I suspect the same pattern would emerge (if the distance between the concentric circles are chosen appropriately).



          This also explains the other symmetric patterns. We can work out that if $alpha = sqrt 2^c$ for $cle 3$, then the circles that lie just before the positive $x$-axis will be $k=2^-ccdot((2n+1)^2-1) = 2^2-ccdot ncdot(n+1)$, which implies that there will be $2^3-ccdot n$ circles in the $n$'th revolution. This means that the circles are lined up just below the positive $x$-axis, and we have $2^3-c$-fold rotational symmetry. This matches what we observe on your plots, which correspond to $-1 le c le 4$. ($c=4$ is not explained by my calculation). I believe that you can see the 16-fold symmetry in the first plot, if you squint a little.







          share|cite|improve this answer








          New contributor




          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|cite|improve this answer



          share|cite|improve this answer






          New contributor




          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered yesterday









          MiltenMilten

          1335




          1335




          New contributor




          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          Milten is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.











          • $begingroup$
            Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago
















          • $begingroup$
            Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago















          $begingroup$
          Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument.
          $endgroup$
          – Hans-Peter Stricker
          17 hours ago




          $begingroup$
          Thanks for your nice answer. Please have a look at my own answer which just gives some visual sugar to your argument.
          $endgroup$
          – Hans-Peter Stricker
          17 hours ago












          $begingroup$
          You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
          $endgroup$
          – Hans-Peter Stricker
          12 hours ago




          $begingroup$
          You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
          $endgroup$
          – Hans-Peter Stricker
          12 hours ago











          1












          $begingroup$

          This answer complements the other one by Milten; it explains the linear patterns in the diagrams. First consider the case $alpha=2$, which corresponds to the function $z_k:=sqrtke^2pi isqrtk$ (it is different from OP's function by a half). Take different cases of $k$.



          (i) $k=n^2+m$. $$z_n=x_n+iy_n=sqrtn^2+m,e^2pi isqrtn^2+m=sqrtn^2+m,e^2pi i n(1+fracmn^2)^1/2approx ne^pi im/n$$ so $x_napprox n$, $y_napprox n(fracpi mn-fracm^3pi^36n^3)approxpi m-fracm^3pi^36n^2$, that is $$y_napprox pi m-fracAx_n^2to pi m$$ This explains the right-hand lines in OP's diagrams with $m=ldots,-1,0,1,2,ldots$.



          (ii) $k=n^2+n+m$. $$z_n=sqrtn^2+n+m,e^2pi isqrtn^2+n+mapprox (n+tfrac12)e^2pi ine^pi ie^pi i m/napprox-(n+tfrac12)e^pi im/n$$ This is similar to case (i) but reflected, so the lines appear on the left side.



          Now consider $alpha=2sqrt2$ which corresponds to $z_k:=sqrtke^2pi isqrt2k$. The same type of analysis gives $$z_napprox n,e^2pi im/n$$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago















          1












          $begingroup$

          This answer complements the other one by Milten; it explains the linear patterns in the diagrams. First consider the case $alpha=2$, which corresponds to the function $z_k:=sqrtke^2pi isqrtk$ (it is different from OP's function by a half). Take different cases of $k$.



          (i) $k=n^2+m$. $$z_n=x_n+iy_n=sqrtn^2+m,e^2pi isqrtn^2+m=sqrtn^2+m,e^2pi i n(1+fracmn^2)^1/2approx ne^pi im/n$$ so $x_napprox n$, $y_napprox n(fracpi mn-fracm^3pi^36n^3)approxpi m-fracm^3pi^36n^2$, that is $$y_napprox pi m-fracAx_n^2to pi m$$ This explains the right-hand lines in OP's diagrams with $m=ldots,-1,0,1,2,ldots$.



          (ii) $k=n^2+n+m$. $$z_n=sqrtn^2+n+m,e^2pi isqrtn^2+n+mapprox (n+tfrac12)e^2pi ine^pi ie^pi i m/napprox-(n+tfrac12)e^pi im/n$$ This is similar to case (i) but reflected, so the lines appear on the left side.



          Now consider $alpha=2sqrt2$ which corresponds to $z_k:=sqrtke^2pi isqrt2k$. The same type of analysis gives $$z_napprox n,e^2pi im/n$$






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago













          1












          1








          1





          $begingroup$

          This answer complements the other one by Milten; it explains the linear patterns in the diagrams. First consider the case $alpha=2$, which corresponds to the function $z_k:=sqrtke^2pi isqrtk$ (it is different from OP's function by a half). Take different cases of $k$.



          (i) $k=n^2+m$. $$z_n=x_n+iy_n=sqrtn^2+m,e^2pi isqrtn^2+m=sqrtn^2+m,e^2pi i n(1+fracmn^2)^1/2approx ne^pi im/n$$ so $x_napprox n$, $y_napprox n(fracpi mn-fracm^3pi^36n^3)approxpi m-fracm^3pi^36n^2$, that is $$y_napprox pi m-fracAx_n^2to pi m$$ This explains the right-hand lines in OP's diagrams with $m=ldots,-1,0,1,2,ldots$.



          (ii) $k=n^2+n+m$. $$z_n=sqrtn^2+n+m,e^2pi isqrtn^2+n+mapprox (n+tfrac12)e^2pi ine^pi ie^pi i m/napprox-(n+tfrac12)e^pi im/n$$ This is similar to case (i) but reflected, so the lines appear on the left side.



          Now consider $alpha=2sqrt2$ which corresponds to $z_k:=sqrtke^2pi isqrt2k$. The same type of analysis gives $$z_napprox n,e^2pi im/n$$






          share|cite|improve this answer









          $endgroup$



          This answer complements the other one by Milten; it explains the linear patterns in the diagrams. First consider the case $alpha=2$, which corresponds to the function $z_k:=sqrtke^2pi isqrtk$ (it is different from OP's function by a half). Take different cases of $k$.



          (i) $k=n^2+m$. $$z_n=x_n+iy_n=sqrtn^2+m,e^2pi isqrtn^2+m=sqrtn^2+m,e^2pi i n(1+fracmn^2)^1/2approx ne^pi im/n$$ so $x_napprox n$, $y_napprox n(fracpi mn-fracm^3pi^36n^3)approxpi m-fracm^3pi^36n^2$, that is $$y_napprox pi m-fracAx_n^2to pi m$$ This explains the right-hand lines in OP's diagrams with $m=ldots,-1,0,1,2,ldots$.



          (ii) $k=n^2+n+m$. $$z_n=sqrtn^2+n+m,e^2pi isqrtn^2+n+mapprox (n+tfrac12)e^2pi ine^pi ie^pi i m/napprox-(n+tfrac12)e^pi im/n$$ This is similar to case (i) but reflected, so the lines appear on the left side.



          Now consider $alpha=2sqrt2$ which corresponds to $z_k:=sqrtke^2pi isqrt2k$. The same type of analysis gives $$z_napprox n,e^2pi im/n$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 16 hours ago









          ChrystomathChrystomath

          1,858513




          1,858513











          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago
















          • $begingroup$
            You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
            $endgroup$
            – Hans-Peter Stricker
            12 hours ago















          $begingroup$
          You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
          $endgroup$
          – Hans-Peter Stricker
          12 hours ago




          $begingroup$
          You may want to have a look at this little Youtube video where you can see the number spiral continuously change for $alpha$ going from 1 to 8.
          $endgroup$
          – Hans-Peter Stricker
          12 hours ago











          1












          $begingroup$

          To give some visual sugar to user Milten's nice answer find here highlighted those numbers that make approximate rectangles. You'll easily see the number pattern and see the role that the number $8$ plays, giving typical triangle number like sequences, e.g. $a = 2,12,30,56,90,ldots$ with $a_k = 8triangle(k) + 2(k+1)$ or $b = 3,13,31,57,91,ldots$ with $b_k = a_k +1$, $k=0,1,2,ldots$.



          enter image description here



          For the sake of comparison the circular pattern with $8n$ numbers in each circle. The similar numbers make approximate rectangles:



          enter image description here




          What needs to be shown analytically is that for example for the numbers $a_k,b_k, a_k+1, b_k+1$ we have



          $$|p(a_k) - p(b_k)| approx |p(a_k+1) - p(b_k+1)|$$



          and



          $$|p(a_k) - p(a_k+1)| approx |p(b_k) - p(b_k+1)|$$



          with $p(k) = langle x(k), y(k)rangle$ and $x(k) = -hat kcos(2 picdot hat k)$,
          $y(k) = -hat ksin(2 picdot hat k)$ and $hat k = fracsqrtk2$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_k+1)$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_k+1)$), since the line between $p(a_k)$ and $p(b_k)$ is perpendicular to centre.
            $endgroup$
            – Milten
            17 hours ago











          • $begingroup$
            I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $fracpi8n$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though.
            $endgroup$
            – Milten
            17 hours ago










          • $begingroup$
            Can you give an analytical expression for this requirement, please? I would add it to my answer.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            Did it. But still there's the ugly gap between 23 and 9.
            $endgroup$
            – Hans-Peter Stricker
            16 hours ago










          • $begingroup$
            Hmm, we could use the dot product: $(p(a_k+1-p(a_k))cdot(p(a_k)-p(b_k)) approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_k+1) parallel p(a_k)$, which means they have the same angle. So we could write it as $2pi hat a_k+1 approx 2pi (hat a_k+1) iff sqrta_k+1 approx sqrta_k + 2$. (+1 since we go once around the spiral).
            $endgroup$
            – Milten
            16 hours ago
















          1












          $begingroup$

          To give some visual sugar to user Milten's nice answer find here highlighted those numbers that make approximate rectangles. You'll easily see the number pattern and see the role that the number $8$ plays, giving typical triangle number like sequences, e.g. $a = 2,12,30,56,90,ldots$ with $a_k = 8triangle(k) + 2(k+1)$ or $b = 3,13,31,57,91,ldots$ with $b_k = a_k +1$, $k=0,1,2,ldots$.



          enter image description here



          For the sake of comparison the circular pattern with $8n$ numbers in each circle. The similar numbers make approximate rectangles:



          enter image description here




          What needs to be shown analytically is that for example for the numbers $a_k,b_k, a_k+1, b_k+1$ we have



          $$|p(a_k) - p(b_k)| approx |p(a_k+1) - p(b_k+1)|$$



          and



          $$|p(a_k) - p(a_k+1)| approx |p(b_k) - p(b_k+1)|$$



          with $p(k) = langle x(k), y(k)rangle$ and $x(k) = -hat kcos(2 picdot hat k)$,
          $y(k) = -hat ksin(2 picdot hat k)$ and $hat k = fracsqrtk2$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_k+1)$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_k+1)$), since the line between $p(a_k)$ and $p(b_k)$ is perpendicular to centre.
            $endgroup$
            – Milten
            17 hours ago











          • $begingroup$
            I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $fracpi8n$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though.
            $endgroup$
            – Milten
            17 hours ago










          • $begingroup$
            Can you give an analytical expression for this requirement, please? I would add it to my answer.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            Did it. But still there's the ugly gap between 23 and 9.
            $endgroup$
            – Hans-Peter Stricker
            16 hours ago










          • $begingroup$
            Hmm, we could use the dot product: $(p(a_k+1-p(a_k))cdot(p(a_k)-p(b_k)) approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_k+1) parallel p(a_k)$, which means they have the same angle. So we could write it as $2pi hat a_k+1 approx 2pi (hat a_k+1) iff sqrta_k+1 approx sqrta_k + 2$. (+1 since we go once around the spiral).
            $endgroup$
            – Milten
            16 hours ago














          1












          1








          1





          $begingroup$

          To give some visual sugar to user Milten's nice answer find here highlighted those numbers that make approximate rectangles. You'll easily see the number pattern and see the role that the number $8$ plays, giving typical triangle number like sequences, e.g. $a = 2,12,30,56,90,ldots$ with $a_k = 8triangle(k) + 2(k+1)$ or $b = 3,13,31,57,91,ldots$ with $b_k = a_k +1$, $k=0,1,2,ldots$.



          enter image description here



          For the sake of comparison the circular pattern with $8n$ numbers in each circle. The similar numbers make approximate rectangles:



          enter image description here




          What needs to be shown analytically is that for example for the numbers $a_k,b_k, a_k+1, b_k+1$ we have



          $$|p(a_k) - p(b_k)| approx |p(a_k+1) - p(b_k+1)|$$



          and



          $$|p(a_k) - p(a_k+1)| approx |p(b_k) - p(b_k+1)|$$



          with $p(k) = langle x(k), y(k)rangle$ and $x(k) = -hat kcos(2 picdot hat k)$,
          $y(k) = -hat ksin(2 picdot hat k)$ and $hat k = fracsqrtk2$.






          share|cite|improve this answer











          $endgroup$



          To give some visual sugar to user Milten's nice answer find here highlighted those numbers that make approximate rectangles. You'll easily see the number pattern and see the role that the number $8$ plays, giving typical triangle number like sequences, e.g. $a = 2,12,30,56,90,ldots$ with $a_k = 8triangle(k) + 2(k+1)$ or $b = 3,13,31,57,91,ldots$ with $b_k = a_k +1$, $k=0,1,2,ldots$.



          enter image description here



          For the sake of comparison the circular pattern with $8n$ numbers in each circle. The similar numbers make approximate rectangles:



          enter image description here




          What needs to be shown analytically is that for example for the numbers $a_k,b_k, a_k+1, b_k+1$ we have



          $$|p(a_k) - p(b_k)| approx |p(a_k+1) - p(b_k+1)|$$



          and



          $$|p(a_k) - p(a_k+1)| approx |p(b_k) - p(b_k+1)|$$



          with $p(k) = langle x(k), y(k)rangle$ and $x(k) = -hat kcos(2 picdot hat k)$,
          $y(k) = -hat ksin(2 picdot hat k)$ and $hat k = fracsqrtk2$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 16 hours ago

























          answered 17 hours ago









          Hans-Peter StrickerHans-Peter Stricker

          6,74043995




          6,74043995











          • $begingroup$
            Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_k+1)$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_k+1)$), since the line between $p(a_k)$ and $p(b_k)$ is perpendicular to centre.
            $endgroup$
            – Milten
            17 hours ago











          • $begingroup$
            I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $fracpi8n$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though.
            $endgroup$
            – Milten
            17 hours ago










          • $begingroup$
            Can you give an analytical expression for this requirement, please? I would add it to my answer.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            Did it. But still there's the ugly gap between 23 and 9.
            $endgroup$
            – Hans-Peter Stricker
            16 hours ago










          • $begingroup$
            Hmm, we could use the dot product: $(p(a_k+1-p(a_k))cdot(p(a_k)-p(b_k)) approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_k+1) parallel p(a_k)$, which means they have the same angle. So we could write it as $2pi hat a_k+1 approx 2pi (hat a_k+1) iff sqrta_k+1 approx sqrta_k + 2$. (+1 since we go once around the spiral).
            $endgroup$
            – Milten
            16 hours ago

















          • $begingroup$
            Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_k+1)$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_k+1)$), since the line between $p(a_k)$ and $p(b_k)$ is perpendicular to centre.
            $endgroup$
            – Milten
            17 hours ago











          • $begingroup$
            I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $fracpi8n$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though.
            $endgroup$
            – Milten
            17 hours ago










          • $begingroup$
            Can you give an analytical expression for this requirement, please? I would add it to my answer.
            $endgroup$
            – Hans-Peter Stricker
            17 hours ago










          • $begingroup$
            Did it. But still there's the ugly gap between 23 and 9.
            $endgroup$
            – Hans-Peter Stricker
            16 hours ago










          • $begingroup$
            Hmm, we could use the dot product: $(p(a_k+1-p(a_k))cdot(p(a_k)-p(b_k)) approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_k+1) parallel p(a_k)$, which means they have the same angle. So we could write it as $2pi hat a_k+1 approx 2pi (hat a_k+1) iff sqrta_k+1 approx sqrta_k + 2$. (+1 since we go once around the spiral).
            $endgroup$
            – Milten
            16 hours ago
















          $begingroup$
          Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_k+1)$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_k+1)$), since the line between $p(a_k)$ and $p(b_k)$ is perpendicular to centre.
          $endgroup$
          – Milten
          17 hours ago





          $begingroup$
          Nice! In addition to your two approximate equations, we need that the (approximate) parallelogram formed by the four points have right angles. Otherwise we just have the normal situation. This is equivalent to $p(a_k)$ and $p(a_k+1)$ lying approximately on a radial line (or similarly for $p(b_k)$ and $p(b_k+1)$), since the line between $p(a_k)$ and $p(b_k)$ is perpendicular to centre.
          $endgroup$
          – Milten
          17 hours ago













          $begingroup$
          I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $fracpi8n$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though.
          $endgroup$
          – Milten
          17 hours ago




          $begingroup$
          I think we would get the cleanest pattern, if you make the circular pattern with $8n$ numbers per circle, but with the $n$'th circle rotated by $fracpi8n$, so that the spaces between the circles align with the axes. I still think the spiral will nicer to look at though.
          $endgroup$
          – Milten
          17 hours ago












          $begingroup$
          Can you give an analytical expression for this requirement, please? I would add it to my answer.
          $endgroup$
          – Hans-Peter Stricker
          17 hours ago




          $begingroup$
          Can you give an analytical expression for this requirement, please? I would add it to my answer.
          $endgroup$
          – Hans-Peter Stricker
          17 hours ago












          $begingroup$
          Did it. But still there's the ugly gap between 23 and 9.
          $endgroup$
          – Hans-Peter Stricker
          16 hours ago




          $begingroup$
          Did it. But still there's the ugly gap between 23 and 9.
          $endgroup$
          – Hans-Peter Stricker
          16 hours ago












          $begingroup$
          Hmm, we could use the dot product: $(p(a_k+1-p(a_k))cdot(p(a_k)-p(b_k)) approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_k+1) parallel p(a_k)$, which means they have the same angle. So we could write it as $2pi hat a_k+1 approx 2pi (hat a_k+1) iff sqrta_k+1 approx sqrta_k + 2$. (+1 since we go once around the spiral).
          $endgroup$
          – Milten
          16 hours ago





          $begingroup$
          Hmm, we could use the dot product: $(p(a_k+1-p(a_k))cdot(p(a_k)-p(b_k)) approx 0$. If we assume that $p(a_k)$ is perpendicular to $(p(a_k)-p(b_k))$ (since $a_k$ and $b_k$ are neighbours), this is equivalent to $p(a_k+1) parallel p(a_k)$, which means they have the same angle. So we could write it as $2pi hat a_k+1 approx 2pi (hat a_k+1) iff sqrta_k+1 approx sqrta_k + 2$. (+1 since we go once around the spiral).
          $endgroup$
          – Milten
          16 hours ago


















          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%2f3164391%2fenigmatic-patterns-in-archimedean-spirals%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

          Boston (Lincolnshire) Stedsbyld | Berne yn Boston | NavigaasjemenuBoston Borough CouncilBoston, Lincolnshire

          Ballerup Komuun Stääden an saarpen | Futnuuten | Luke uk diar | Nawigatsjuunwww.ballerup.dkwww.statistikbanken.dk: Tabelle BEF44 (Folketal pr. 1. januar fordelt på byer)Commonskategorii: Ballerup Komuun55° 44′ N, 12° 22′ O

          Serbia Índice Etimología Historia Geografía Entorno natural División administrativa Política Demografía Economía Cultura Deportes Véase también Notas Referencias Bibliografía Enlaces externos Menú de navegación44°49′00″N 20°28′00″E / 44.816666666667, 20.46666666666744°49′00″N 20°28′00″E / 44.816666666667, 20.466666666667U.S. Department of Commerce (2015)«Informe sobre Desarrollo Humano 2018»Kosovo-Metohija.Neutralna Srbija u NATO okruzenju.The SerbsTheories on the Origin of the Serbs.Serbia.Earls: Webster's Quotations, Facts and Phrases.Egeo y Balcanes.Kalemegdan.Southern Pannonia during the age of the Great Migrations.Culture in Serbia.History.The Serbian Origin of the Montenegrins.Nemanjics' period (1186-1353).Stefan Uros (1355-1371).Serbian medieval history.Habsburg–Ottoman Wars (1525–1718).The Ottoman Empire, 1700-1922.The First Serbian Uprising.Miloš, prince of Serbia.3. Bosnia-Hercegovina and the Congress of Berlin.The Balkan Wars and the Partition of Macedonia.The Falcon and the Eagle: Montenegro and Austria-Hungary, 1908-1914.Typhus fever on the eastern front in World War I.Anniversary of WWI battle marked in Serbia.La derrota austriaca en los Balcanes. Fin del Imperio Austro-Húngaro.Imperio austriaco y Reino de Hungría.Los tiempos modernos: del capitalismo a la globalización, siglos XVII al XXI.The period of Croatia within ex-Yugoslavia.Yugoslavia: Much in a Name.Las dictaduras europeas.Croacia: mito y realidad."Crods ask arms".Prólogo a la invasión.La campaña de los Balcanes.La resistencia en Yugoslavia.Jasenovac Research Institute.Día en memoria de las víctimas del genocidio en la Segunda Guerra Mundial.El infierno estuvo en Jasenovac.Croacia empieza a «desenterrar» a sus muertos de Jasenovac.World fascism: a historical encyclopedia, Volumen 1.Tito. Josip Broz.El nuevo orden y la resistencia.La conquista del poder.Algunos aspectos de la economía yugoslava a mediados de 1962.Albania-Kosovo crisis.De Kosovo a Kosova: una visión demográfica.La crisis de la economía yugoslava y la política de "estabilización".Milosevic: el poder de un absolutista."Serbia under Milošević: politics in the 1990s"Milosevic cavó en Kosovo la tumba de la antigua Yugoslavia.La ONU exculpa a Serbia de genocidio en la guerra de Bosnia.Slobodan Milosevic, el burócrata que supo usar el odio.Es la fuerza contra el sufrimiento de muchos inocentes.Matanza de civiles al bombardear la OTAN un puente mientras pasaba un tren.Las consecuencias negativas de los bombardeos de Yugoslavia se sentirán aún durante largo tiempo.Kostunica advierte que la misión de Europa en Kosovo es ilegal.Las 24 horas más largas en la vida de Slobodan Milosevic.Serbia declara la guerra a la mafia por matar a Djindjic.Tadic presentará "quizás en diciembre" la solicitud de entrada en la UE.Montenegro declara su independencia de Serbia.Serbia se declara estado soberano tras separación de Montenegro.«Accordance with International Law of the Unilateral Declaration of Independence by the Provisional Institutions of Self-Government of Kosovo (Request for Advisory Opinion)»Mladic pasa por el médico antes de la audiencia para extraditarloDatos de Serbia y Kosovo.The Carpathian Mountains.Position, Relief, Climate.Transport.Finding birds in Serbia.U Srbiji do 2010. godine 10% teritorije nacionalni parkovi.Geography.Serbia: Climate.Variability of Climate In Serbia In The Second Half of The 20thc Entury.BASIC CLIMATE CHARACTERISTICS FOR THE TERRITORY OF SERBIA.Fauna y flora: Serbia.Serbia and Montenegro.Información general sobre Serbia.Republic of Serbia Environmental Protection Agency (SEPA).Serbia recycling 15% of waste.Reform process of the Serbian energy sector.20-MW Wind Project Being Developed in Serbia.Las Naciones Unidas. Paz para Kosovo.Aniversario sin fiesta.Population by national or ethnic groups by Census 2002.Article 7. Coat of arms, flag and national anthem.Serbia, flag of.Historia.«Serbia and Montenegro in Pictures»Serbia.Serbia aprueba su nueva Constitución con un apoyo de más del 50%.Serbia. Population.«El nacionalista Nikolic gana las elecciones presidenciales en Serbia»El europeísta Borís Tadic gana la segunda vuelta de las presidenciales serbias.Aleksandar Vucic, de ultranacionalista serbio a fervoroso europeístaKostunica condena la declaración del "falso estado" de Kosovo.Comienza el debate sobre la independencia de Kosovo en el TIJ.La Corte Internacional de Justicia dice que Kosovo no violó el derecho internacional al declarar su independenciaKosovo: Enviado de la ONU advierte tensiones y fragilidad.«Bruselas recomienda negociar la adhesión de Serbia tras el acuerdo sobre Kosovo»Monografía de Serbia.Bez smanjivanja Vojske Srbije.Military statistics Serbia and Montenegro.Šutanovac: Vojni budžet za 2009. godinu 70 milijardi dinara.Serbia-Montenegro shortens obligatory military service to six months.No hay justicia para las víctimas de los bombardeos de la OTAN.Zapatero reitera la negativa de España a reconocer la independencia de Kosovo.Anniversary of the signing of the Stabilisation and Association Agreement.Detenido en Serbia Radovan Karadzic, el criminal de guerra más buscado de Europa."Serbia presentará su candidatura de acceso a la UE antes de fin de año".Serbia solicita la adhesión a la UE.Detenido el exgeneral serbobosnio Ratko Mladic, principal acusado del genocidio en los Balcanes«Lista de todos los Estados Miembros de las Naciones Unidas que son parte o signatarios en los diversos instrumentos de derechos humanos de las Naciones Unidas»versión pdfProtocolo Facultativo de la Convención sobre la Eliminación de todas las Formas de Discriminación contra la MujerConvención contra la tortura y otros tratos o penas crueles, inhumanos o degradantesversión pdfProtocolo Facultativo de la Convención sobre los Derechos de las Personas con DiscapacidadEl ACNUR recibe con beneplácito el envío de tropas de la OTAN a Kosovo y se prepara ante una posible llegada de refugiados a Serbia.Kosovo.- El jefe de la Minuk denuncia que los serbios boicotearon las legislativas por 'presiones'.Bosnia and Herzegovina. Population.Datos básicos de Montenegro, historia y evolución política.Serbia y Montenegro. Indicador: Tasa global de fecundidad (por 1000 habitantes).Serbia y Montenegro. Indicador: Tasa bruta de mortalidad (por 1000 habitantes).Population.Falleció el patriarca de la Iglesia Ortodoxa serbia.Atacan en Kosovo autobuses con peregrinos tras la investidura del patriarca serbio IrinejSerbian in Hungary.Tasas de cambio."Kosovo es de todos sus ciudadanos".Report for Serbia.Country groups by income.GROSS DOMESTIC PRODUCT (GDP) OF THE REPUBLIC OF SERBIA 1997–2007.Economic Trends in the Republic of Serbia 2006.National Accounts Statitics.Саопштења за јавност.GDP per inhabitant varied by one to six across the EU27 Member States.Un pacto de estabilidad para Serbia.Unemployment rate rises in Serbia.Serbia, Belarus agree free trade to woo investors.Serbia, Turkey call investors to Serbia.Success Stories.U.S. Private Investment in Serbia and Montenegro.Positive trend.Banks in Serbia.La Cámara de Comercio acompaña a empresas madrileñas a Serbia y Croacia.Serbia Industries.Energy and mining.Agriculture.Late crops, fruit and grapes output, 2008.Rebranding Serbia: A Hobby Shortly to Become a Full-Time Job.Final data on livestock statistics, 2008.Serbian cell-phone users.U Srbiji sve više računara.Телекомуникације.U Srbiji 27 odsto gradjana koristi Internet.Serbia and Montenegro.Тренд гледаности програма РТС-а у 2008. и 2009.години.Serbian railways.General Terms.El mercado del transporte aéreo en Serbia.Statistics.Vehículos de motor registrados.Planes ambiciosos para el transporte fluvial.Turismo.Turistički promet u Republici Srbiji u periodu januar-novembar 2007. godine.Your Guide to Culture.Novi Sad - city of culture.Nis - european crossroads.Serbia. Properties inscribed on the World Heritage List .Stari Ras and Sopoćani.Studenica Monastery.Medieval Monuments in Kosovo.Gamzigrad-Romuliana, Palace of Galerius.Skiing and snowboarding in Kopaonik.Tara.New7Wonders of Nature Finalists.Pilgrimage of Saint Sava.Exit Festival: Best european festival.Banje u Srbiji.«The Encyclopedia of world history»Culture.Centenario del arte serbio.«Djordje Andrejevic Kun: el único pintor de los brigadistas yugoslavos de la guerra civil española»About the museum.The collections.Miroslav Gospel – Manuscript from 1180.Historicity in the Serbo-Croatian Heroic Epic.Culture and Sport.Conversación con el rector del Seminario San Sava.'Reina Margot' funde drama, historia y gesto con música de Goran Bregovic.Serbia gana Eurovisión y España decepciona de nuevo con un vigésimo puesto.Home.Story.Emir Kusturica.Tercer oro para Paskaljevic.Nikola Tesla Year.Home.Tesla, un genio tomado por loco.Aniversario de la muerte de Nikola Tesla.El Museo Nikola Tesla en Belgrado.El inventor del mundo actual.República de Serbia.University of Belgrade official statistics.University of Novi Sad.University of Kragujevac.University of Nis.Comida. Cocina serbia.Cooking.Montenegro se convertirá en el miembro 204 del movimiento olímpico.España, campeona de Europa de baloncesto.El Partizan de Belgrado se corona campeón por octava vez consecutiva.Serbia se clasifica para el Mundial de 2010 de Sudáfrica.Serbia Name Squad For Northern Ireland And South Korea Tests.Fútbol.- El Partizán de Belgrado se proclama campeón de la Liga serbia.Clasificacion final Mundial de balonmano Croacia 2009.Serbia vence a España y se consagra campeón mundial de waterpolo.Novak Djokovic no convence pero gana en Australia.Gana Ana Ivanovic el Roland Garros.Serena Williams gana el US Open por tercera vez.Biography.Bradt Travel Guide SerbiaThe Encyclopedia of World War IGobierno de SerbiaPortal del Gobierno de SerbiaPresidencia de SerbiaAsamblea Nacional SerbiaMinisterio de Asuntos exteriores de SerbiaBanco Nacional de SerbiaAgencia Serbia para la Promoción de la Inversión y la ExportaciónOficina de Estadísticas de SerbiaCIA. Factbook 2008Organización nacional de turismo de SerbiaDiscover SerbiaConoce SerbiaNoticias de SerbiaSerbiaWorldCat1512028760000 0000 9526 67094054598-2n8519591900570825ge1309191004530741010url17413117006669D055771Serbia