Thank you, Bob. Me, my, son, and my wife (who has a sister farming in
New Zealand) all think that Peter Loly is doing wonderful work.
Just a few hours ago, I submitted two sequences citing him, one of them
being as below.<br>
<br>
<pre><font size="4">%I A126276<br>%S A126276 18, 504, 5200, 31500, 136710, 471968, 1378944<br>%N A126276 Moment of inertia of all magic cubes of order n.<br>%C A126276 Loly and Rogers show that the moment of inertia of a magic
<br>cube is n^3(n^3 + 1)(n^2 - 1)/12. In effect, they demonstrate that <br>magic cubes have the same inertial form as a spherical top.<br>Loly investigated the "physical" properties of magic squares--treating <br>
the numbers of each such square as physical quantities. If the integers <br>are consecutive numbers from 1 to n^2, the square is said to be of nth <br>order. The magic sum itself is given by n(n^2 + 1)/2. Suppose you <br>
interpret the numbers as masses. You can then determine a magic square's <br>moment of inertia about a given axis of rotation. For any specific case, <br>you obtain the moment of inertia, In, of a magic square of order n about
<br>an axis at right angles to its center by summing mr^2 for each cell, <br>where m is the number centered in a cell and r is the distance of the <br>center of that cell from the center of the square measured in units of the
<br>nearest neighbor distance.<br>You find that the moment of inertia, I_z, about the square's center (an <br>axis at right angles to the square) is twice the moment of inertia <br>about an axis of rotation along the center row or column.
<br>Such an analysis can be extended to magic cubes. A magic cube consists <br>of n^3 numbers, arranged so that each row, column, and main diagonal <br>give the same sum. In the case, the magic constant is n(n^3 + 1)/2.<br>
%H A126276 Ivars Peterson, <a <br>href="<a href="http://www.sciencenews.org/articles/20060701/mathtrek.asp" target="_blank">http://www.sciencenews.org/articles/20060701/mathtrek.asp</a>">Magic <br>
Square Physics</a>. Science News online, Week of July 1, 2006; Vol. <br>170, No. 1<br>%H A126276 Peter Loly, <a <br>href="<a href="http://home.cc.umanitoba.ca/%7Eloly/MathGaz.pdf" target="_blank">
http://home.cc.umanitoba.ca/~loly/MathGaz.pdf</a>">The invariance of the <br>moment of inertia of magic squares</a>, Mathematical Gazette <br>88(March 2004):151-153<br>%H A126276 Adam Rogers and Peter Loly, <a
<br>href="<a href="http://www.cupj.ca/0302_rotational.pdf" target="_blank">http://www.cupj.ca/0302_rotational.pdf</a>">Rotational sorcery: The <br>inertial properties of magic squares and cubes</a>. Canadian
<br>Undergraduate Physics Journal 3(No. 2):25, 2005.<br>%F A126276 a(n) = (n^3)*(n^3 + 1)*(n^2 - 1)/12.<br>%Y A126276 Cf. A126276 Moment of inertia of all magic cubes of order n<br>%O A126276 2,1<br>%K A126276 ,easy,nonn,
<br>%A A126276 Jonathan Vos Post (<a href="http://us.f551.mail.yahoo.com/ym/Compose?To=jvospost2@yahoo.com&YY=11658&y5beta=yes&y5beta=yes&order=down&sort=date&pos=1&view=a&head=b">jvospost2@yahoo.com
</a>), Dec 23 2006<br>RH <br>RA <a href="http://192.20.225.32">192.20.225.32</a><br>RU <tt><br>RI </tt></font></pre>
<br><br><div><span class="gmail_quote">On 12/24/06, <b class="gmail_sendername">Bob Barbour</b> <<a href="mailto:bbarbour@unitec.ac.nz">bbarbour@unitec.ac.nz</a>> wrote:</span><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
Hi Jonathan and seqfans,<br><br>Seasons Greeting from New Zealand.<br>Peter Loly and his team have done substantial work<br>enumerating magic squares.<br><br>loly at cc dot umanitoba dot ca<br><br>should get to him.<br><br>
Regards,<br>Bob<br><br>>>> "Jonathan Post" <<a href="mailto:jvospost3@gmail.com">jvospost3@gmail.com</a>> 12/24/06 9:46 PM >>><br>Dear Frank,<br><br>It does look as if I imposed an extra constraint on constructing
<br>semiprime<br>magic squares. Interesting that it did not overconstrain and prevent<br>solution.<br><br>So now we have five open problems:<br><br>(1) How many different semiprime nxn magic squares are there with<br>maximum
<br>element < k;<br><br>(2) What is the sequence of row sums (magic numbers) of my semiprime<br>magic<br>squares whose elements are in arithmetic sequence;<br><br>(3) Tony Noe's question: what is the smallest nxn semiprime magic square
<br>where all elements are relatively prime;<br><br>(4) My modification of the previous: what is the smallest nxn semiprime<br>magic square where all elements in the same row, column, or diagonal are<br>relatively prime;<br>
<br>(5) Dr. Geoffrey Landis' question: what is the smallest semiprime magic<br>square whose magic number is semiprime?<br><br>Making a simple problem hard seems to be one of my specialties;)<br><br>Happy holidays,<br>
<br>Jonathan Vos Post<br><br><br><br>On 12/23/06, <a href="mailto:franktaw@netscape.net">franktaw@netscape.net</a> <<a href="mailto:franktaw@netscape.net">franktaw@netscape.net</a>> wrote:<br>><br>> The smallest prime magic square is:
<br>><br>> 17 89 71<br>> 113 59 5<br>> 47 29 101<br>><br>> We can just double that, to get:<br>><br>> 34 178 142<br>> 226 118 10<br>> 94 58 202<br>><br>> which is quite a bit smaller than your example.
<br>><br>> Even if you want to exclude squares where the entries have a common<br>> divisor, your example looks suspiciously large to me. And looking at<br>> A096003, it looks like you are assuming that your magic consists of
<br>> numbers in arithmetic progression. That is not a valid assumption.<br>><br>> Franklin T. Adams-Watters<br>><br>><br>> -----Original Message-----<br>> From: <a href="mailto:jvospost3@gmail.com">
jvospost3@gmail.com</a><br>><br>> Using A096003 and A097824, here is the smallest all-semiprime magic<br>> square, which I just discovered today:<br>><br>> ============<br>> 1139 635 995<br>> 779 923 1067
<br>> 851 1211 707<br>> ============<br>><br>> In numerical order, these entries are:<br>> 635 = 5*127<br>> 707 = 7*101<br>> 779 = 19*41<br>> 851 = 23*37<br>> 923 = 13*71<br>> 995 = 5*199<br>
> 1067 = 11*97<br>> 1139 = 17*67<br>> 1211 = 7*173<br>><br>> As I just noted in a submission to Prime Curios: 2769 is the sum of<br>> any row, column, or diagonal. Curiously, 2769 = 3 * 13 * 71, all three
<br>> of which are primes when reversed (but that is too "base" for us<br>here).<br>><br>> Using A096003(16) = 28213, and A097824(16) = 354, one likewise has<br>the<br>> smallest semiprime magic square of order 4.
<br>><br>> Using A096003(25) = 2012771, and A097824(25) = 9600, one likewise<br>has<br>> the smallest semiprime magic square of order 5.<br>><br>> I'm not sure if it's better to submit the finite full sequence 1139,
<br>> 635, 995, 779, 923, 1067,<br>> 851, 1211, 707 for the order 3, and similarly for the order 4 and<br>> order 5; or give the sequence of row sums as a function of order n; or<br>> what.<br>><br>> I don't know the order 6 or above.
<br>><br>> I also know the smallest order 3 of the 3-almost prime magic<br>squares,<br>> and so forth, but these become "less" for most seqfans, I suspect.<br>><br>> I admit to not yet having written down the smallest semiprime magic
<br>> cbe, but the same method can be adapted...<br>><br>> This is, new, is it not?<br>><br>> Happy Hanukkah, Merry Christmas, etcetera,<br>><br>> -- Jonathan Vos Post<br>><br>><br>><br>><br>
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<br></blockquote></div><br>