June 2008 - mathfun@mailman.xmission.com

[math-fun] Smallest magic squares of powers
by Christian Boyer 01 Sep '08

01 Sep '08

Oct 7th, 2007: the smallest known magic square of 4th powers is now a 44x44 square constructed by Jaroslaw Wroblewski (Poland) and Hugo Pfoertner (Germany). Jaroslaw built the semi-magic square in June, then Hugo needed a very impressive computing job of 3 months, on 57 processors of a SGI computer http://www.sgi.com/products/remarketed/origin3000/, in order to find two magic diagonals in this square! A big progress, because the previous smallest known was a 243x243 square, coming from the 243x243 tetramagic square of Pan Fengchu (China). The best known results, and their first authors, are now: Magic square Smallest known ------------ --------------- Of squares 4x4 (Euler, 1770) Of cubes 9x9 (Boyer, 2006) Of 4th powers 44x44 (Wroblewski-Pfoertner, 2007) Of 5th powers 729x729 (from the pentamagic square of Li Wen, 2003) ON ANY POWER, WE DO NOT KNOW WHAT THE SMALLEST POSSIBLE MAGIC SQUARE IS! For example, as you know, we still do not know if a 3x3 magic square of squares is possible... 4x4, 5x5, 6x6, 7x7, and 8x8 magic squares of cubes are unknown... And the problem of the diagonals is extremely difficult, as proved by the impressive computing search of Hugo. It remains a big gap between the smallest known magic and semi-magic squares (=with or without magic diagonals) of 4th powers: 44x44 versus 4x4... Here is the 4x4 semi-magic square of 4th powers (Morgenstern-Boyer, 2006): 14101^4 938^4 931^4 37762^4 35866^4 20881^4 21038^4 13393^4 24806^4 30191^4 30418^4 9263^4 413^4 32026^4 31787^4 1106^4 Christian.

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[math-fun] Superfluous Multinomials
by wouter meeussen 04 Aug '08

04 Aug '08

dear funsters, when posting stuff like this, I presume it is generally known, or at least could be. Motivation: just fun and beauty. 1/. Intro consider the expansion of (a+b+c+d)^3 it is the sum over M(u,v,w,x) a^u b^v c^w d^x with positive or zero integers u+v+w+x=3 and M(u,v,w,x) is the multinomial u! v! w! x! /(u+v+w+x)! When we drop the multinomial coefficients, then we get the 'compositions' of 4 elements in groups of three. in Mma: Expand[(a+b+c+d)^3 /3! ]/. q_^n_ ->q^n n! gives a^3 + a^2*b + a*b^2 + b^3 + a^2*c + a*b*c + b^2*c + a*c^2 + b*c^2 + c^3 + a^2*d + a*b*d + b^2*d + a*c*d + b*c*d + c^2*d + a*d^2 + b*d^2 + c*d^2 + d^3 this is equivalent to Plus @@ Apply[Times,{a,b,c,d}^#& /@Compositions[3,4],{1}] 2/. Generalise replace the (a+b+c+d) above by Sum(j=1..n, t[ j ] ) , t stands for 'temporary', now replace t[j] with E^(2 Pi I j/n) and it's obvious that this sum equals zero. And that any power of zero is also zero. But now, dropping multinomials, we find: Table[ Plus@@ Apply[Times,(Exp[2 Pi I/n]^Range[n])^#& /@ Compositions[k,n],{1}],{n,5},{k,n}] and we find that it's also zero, except it's 1 when k equals n, that is, we take all n roots of (-1) and put it to the n-th (multinomial-less) power. 3/. an instructive typo : Exp[2 Pi I/k] instead of Exp[2 Pi I/n]: Table[ Plus@@ Apply[Times,(Exp[2 Pi I/k]^Range[n])^#& /@ Compositions[k,n],{1}],{n,5},{k,n}] equals Table[Ceiling[n/m],{n,9},{m,n}] alias A120885 That leads to this silly remark: Table[ Plus@@ (Exp[2 Pi I/k]^Range[n])^k ,{n,16},{k,n}] is integer, except for the following couples of {n,k} : {7, 5}, {8, 5}, {9, 7}, {10, 7}, {10, 8}, {11, 7}, {11, 8}, {11, 9}, {12, 5}, {12, 7}, {12, 8}, {12, 9}, {12, 10}, {13, 5}, {13, 8}, {13, 9}, {13, 10}, {13, 11}, {14, 8}, {14, 9}, {14, 10}, {14, 11}, {14, 12}, {15, 9}, {15, 10}, {15, 11}, {15, 12}, {15, 13}, {16, 7}, {16, 9}, {16, 10}, {16, 11}, {16, 12}, {16, 13}, {16, 14}, ... and, where's the system in *this* madness? Wouter.

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[math-fun] (no subject)
by rwg＠sdf.lonestar.org 08 Jul '08

08 Jul '08

If you split Ramanujan's 1/pi = 4F3[1/64] (search for Ramanujan in www.tweedledum.com/rwg/idents.htm) into two 3F2s, Mma 6.0 recognizes one of them, producing the identity expand(Hyper_F[P,Q]([1,1,3]/2,[1,1],1/64)=16/(21*%Pi) + (64*Elliptic_Kc(1/2*(1-3*Sqrt(7)/8))^2)/(21*%Pi^2)) 3 1 1 1 hyper_f ([-, -, -], [1, 1], --) = 3, 2 2 2 2 64 2 1 3 sqrt(7) 64 elliptic_kc (- - ---------) 2 16 16 ------------------------------ + ------ 2 21 %pi 21 %pi ~ 1.01336493939236. If we can find one more, we might get the whole contiguous family, since 3F2[z] obeys four term recurrences. Embarrassments: Mma 6.0 can't do FullSimplify[ FunctionExpand[ HypergeometricPFQ[{1, a, b}, {c, 3 - c + b + a},1]]] nor even the special case HypergeometricPFQ[{1,1,1},{5,5}/2,1] = 9 pi - 27, even though the general sum telescopes. My enhanced Macsyma also fails unless you first convert to Sigma notation (makesum). --rwg PS, overall, I'm favorably impressed with Mma 6.0, although I did lose a session when something in the help system cleared all my definitions.

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[math-fun] f(0,0)-f(0,1)-f(1,0)+f(1,1)
by James Propp 03 Jul '08

03 Jul '08

Is there an easy way to show that if f(x,y) is twice-differentiable on [0,1]x[0,1] with continuous second-order partial derivatives, then there must exist a point at which (d/dx)(d/dy) f(x,y) equals f(0,0)-f(0,1)-f(1,0)+f(1,1)? (I wonder if forums like math-fun, and sci.math in its heyday, are bad for a mathematician's brain. The above problem is probably a fairly standard exercise in advanced calculus, and in the old days, I would've had to wait until coffee hour to ask a colleague, by which time my own subconscious would have dredged up the needed facts and supplied the answer. Nowadays, I can just post the question and wait for someone in the 24-hour-a-day virtual coffee hour to post an answer. So my brain doesn't get as much exercise. Then again, my subconscious has more chew-toys to play with, thanks to *other* people's questions!) Jim Propp

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Re: [math-fun] Add to n its second smallest non-divider. Loop
by N. J. A. Sloane 30 Jun '08

30 Jun '08

There was a flurry of activity on this topic last week. I'm adding many of the resulting sequences to the OEIS - see A140481 onwards. Neil

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Re: [math-fun] no-4-in-plane problem
by N. J. A. Sloane 29 Jun '08

29 Jun '08

Ed, the other day you said: The no-3-in-line, in 2-space, is a well known problem (by Dudeney, 1917). http://www.research.att.com/~njas/sequences/A000769 I did find "No-three-in-line-in-3D" by<A0> Attila Por, David R. Wood http://citeseer.ist.psu.edu/por04nothreelined.html Me: I can't access the latter references (w/o spending money, that is). Do you know if the first few terms of that sequence have been computed? (Same question of course for your new version) Best regards Neil

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Re: [math-fun] Add to n its second smallest non-divider. Loop
by franktaw＠netscape.net 26 Jun '08

26 Jun '08

This got me thinking about, instead, simply adding tau(n) to n. It turns out that this sequence is in the OEIS: it's A064491. What I find interesting is the square members of this sequence. These are interesting because it is at these values that the sequence changes parity. (For anyone not familiar with this result: the divisors of any n fall naturally into pairs: d and n/d; only when n is a square there is one, the square root, that is unpaired -- or paired with itself -- in this way.) The squares in A064491, up to 100 million, are: 1,4,9,64,81,784,1521,29584,34225,132496,136161,4999696,5076009, 15492096,15547249 which are the squares of: 1,2,3,8,9,28,39,172,185,364,369,2236,2253,3936,3943 Neither of these is in the OEIS. Probably these sequences (the squares and square roots) are infinite, but I don't see any likelihood of proving it. For the sequence here -- n -> 2n + tau(n) -- it is likely that 1 is the only square in the sequence, which would make 1 and 3 the only odd numbers in the sequence. Again, I don't see any approach likely to actually prove this. One more note: there is a conjecture in A064491 that the sequence is asymptotic to c*n*log(n), with c=1.401.... (Conjecture by Benoit Cloitre, who is known for making conjectures on insufficient evidence.) I am almost certain that this conjecture is not correct. I would conjecture, on the contrary, that a(n) ~ n*log(n) (equivalently, there exists c_1 and c_2 s.t. c_1*n*log(n) < a(n) < c_2*n*log(n) for sufficiently large n), but that a(n) is not asymptotic to c*n*log(n) for any c. In particular, the slope of (a(n+1) - a(n)) / log(n) is different in areas where a(n) is even as compared with those where it is odd. Franklin T. Adams-Watters On Wed, Jun 25, 2008 at 17:57, <franktaw(a)netscape.net> wrote: This rule simply takes n to 2n + tau(n). On Wed, Jun 25, 2008 at 12:02, Hans Havermann <pxp(a)rogers.com> wrote: > < Add to n the n-th smallest number not dividing n > I think each term is twice the previous term plus the number of divisors of the previous term: {1, 3, 8, 20, 46, 96, 204, 420, 864, 1752, 3520, 7068, 14160, 28360, 56736, 113508, 227040, 454176, 908424, 1816944, 3633908, 7267828, 14535662,

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Re: [math-fun] Add to n its second smallest non-divider. Loop
by Eric Angelini 25 Jun '08

25 Jun '08

Could please someone compute a few terms more of: < Add to n the n-th smallest number not dividing n > (this is a kind of self-generalization of the rule herunder) I've found this first few terms: S = 1,3,8,20,46,96,... Best, É. (not in the OEIS though 1,3,8,20,46 gives 4 hits) -----Message d'origine----- De : Eric Angelini [mailto:Eric.Angelini@kntv.be] Envoyé : mercredi 25 juin 2008 13:07 À : math-fun; seqfan(a)ext.jussieu.fr Objet : Add to n its second smallest non-divider. Loop Hello MathFun & SeqFans, Rule: < Add to n its second smallest non-divider. Loop. > Let's start with n = 7, for instance Is 1 a divider of 7? yes 2 no 3 no --> then new n = 7+3 = 10 Is 1 a divider of 10? yes 2 yes 3 no 4 no --> then new n =10+4= 14 Is 1 a divider of 14? yes 2 yes 3 no 4 no --> then new n =14+4= 18 Is 1 a divider of 18? yes 2 yes 3 yes 4 no 5 no --> then new n =18+5= 23 ... etc. Sequence starting with 7 is: 7,10,14,18,23,... Sequence starting with 1 is: 1,4,9,13,16,21,... [not in the OEIS] Sequence starting with 2 is: 2,6,11,14,18,23,... [merges with "7-seq"] Sequence starting with 3 is: 3,7,10,14,... [merges with "7-seq"] Sequence starting with 5 is: 5,8,13,16,... [merges with "1-seq"] etc. We might map those sequences like this: 1--4--9---13--16--21--25--28--32--37--40 ... | | | 5--8---+ 29--+ | | 2--6--11--14--18--23--26--30------+ | | | 3--7--10--+ | 33--+ | 12--19--22--+ | | 15--+ | | 17--20--+ 24--31--34--38--42 ... | | 27--+ 35--+ What number starts the longest sequence containing 2008? Best, E. --- P.-S. This rule is not very productive: < Add to n its first smallest non-divider. Loop. > What about: < Add to n its third smallest non-divider. Loop. > Etc.

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[math-fun] Add to n its second smallest non-divider. Loop
by Eric Angelini 25 Jun '08

25 Jun '08

Hello MathFun & SeqFans, Rule: < Add to n its second smallest non-divider. Loop. > Let's start with n = 7, for instance Is 1 a divider of 7? yes 2 no 3 no --> then new n = 7+3 = 10 Is 1 a divider of 10? yes 2 yes 3 no 4 no --> then new n =10+4= 14 Is 1 a divider of 14? yes 2 yes 3 no 4 no --> then new n =14+4= 18 Is 1 a divider of 18? yes 2 yes 3 yes 4 no 5 no --> then new n =18+5= 23 ... etc. Sequence starting with 7 is: 7,10,14,18,23,... Sequence starting with 1 is: 1,4,9,13,16,21,... [not in the OEIS] Sequence starting with 2 is: 2,6,11,14,18,23,... [merges with "7-seq"] Sequence starting with 3 is: 3,7,10,14,... [merges with "7-seq"] Sequence starting with 5 is: 5,8,13,16,... [merges with "1-seq"] etc. We might map those sequences like this: 1--4--9---13--16--21--25--28--32--37--40 ... | | | 5--8---+ 29--+ | | 2--6--11--14--18--23--26--30------+ | | | 3--7--10--+ | 33--+ | 12--19--22--+ | | 15--+ | | 17--20--+ 24--31--34--38--42 ... | | 27--+ 35--+ What number starts the longest sequence containing 2008? Best, E. --- P.-S. This rule is not very productive: < Add to n its first smallest non-divider. Loop. > What about: < Add to n its third smallest non-divider. Loop. > Etc.

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Re: [math-fun] no-4-in-plane problem
by Dan Asimov 25 Jun '08

25 Jun '08

place -> planes (Duh.) --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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