most recent 30 from http://mathoverflow.net 2013-05-22T02:49:45Z http://mathoverflow.net/feeds/question/114744 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114744/2d-visualization-of-sum-of-divisors-using-cantor-pairing joro 2012-11-28T08:52:42Z 2012-12-12T09:11:50Z

<p>Related to Gerhard's question about <a href="http://mathoverflow.net/questions/77794/ascii-prime-plots-and-prime-rich-quadratic-polynomials" rel="nofollow">ascii plots</a>. On the SeqFan mailing list <a href="http://list.seqfan.eu/pipermail/seqfan/2012-November/010494.html" rel="nofollow">was suggested</a> to plot an integer sequence this way:</p> <p>Let $F(x,y)= (x+y) (x+y+1)/2+y$ be the <a href="https://en.wikipedia.org/wiki/Cantor_pairing_function" rel="nofollow">Cantor pairing</a>. To plot an integer sequence $a(n)$, for a point $(x,y)$ compute $a(F(x,y))$ and assign color to the integer, e.g. in grayscale smaller is darker, for RGB/HSV there are other choices to map to color.</p> <p>When $a(n)=\sigma_0(n)$ where $\sigma_0(n)$ is the number of divisors of $n$, the 2D plot shows some structure (hopefully not caused by visual artifacts).</p> <blockquote> <p>Is there an explanation for the structure in the plot?</p> </blockquote> <p>Color plot of $\sigma_0(F(x,y))$, smaller is darker (grayscale is quite similar):</p> <p><img src="http://s16.postimage.org/9enmfsbyt/cantorpairing_sigma_0.png" alt="sigma_0 and cantor pairing"></p> <p>When examining the integer values there are some large diagonals indeed.</p>

http://mathoverflow.net/questions/114744/2d-visualization-of-sum-of-divisors-using-cantor-pairing/114856#114856 Aaron Meyerowitz 2012-11-29T08:00:59Z 2012-12-12T09:11:50Z

<p>Have you tried to find an explanation? </p> <p>The diagonals correspond to numbers $F(x,x+j).$ Every eighth one has $F(x,x+8k)=2x(x+1)+4(8k^2+4xk+3k).$ Since these are all multiples of $4$ that is already a boost. </p> <p>$F(x,x+1)=2(x+1)^2.$ This is the case $q=0$ of $F(x,x-(q^2-1))=2(x-\binom{q}{2}+1-q)(x-\binom{q}{2}+1).$ So these are all pretty composite and every $q$th member is a multiple of $2q^2.$ Probably you can prove that there are no other diagonals which factor algebraically. </p> <p>You will find the horizontal lines $F(\binom{j}{2}-1,y)$ worth examining.</p> <p>Your image also shows possible anti-diagonals $F(x,k-x)$ but I will leave that for someone else to examine (I did not immediately see anything).</p> <p><strong>A few later comments:</strong> Along any line (the ones easily seen are horizontal, vertical and slope $\pm 1$) the values are periodic $\mod p.$ Certain dark lines can be explained by verifying that no member can divide by a small prime. I seem to recall that the lines $F(x,x-(q^2-3))$ contain no multiples of $2,3,5$ and in some cases no multiples of any prime under $30$. Some of this shows in the graphic and some not as much. </p>