October 2005 - mathfun@mailman.xmission.com

[math-fun] poly-anygons; count of chess positions
by Schroeppel, Richard 19 Oct '05

19 Oct '05

I decided to count polyanygons. These are kind of like polyominos, except you can use any regular polygons of side = 1, as long as there's no overlapping. A regular polygon of N sides has a score of N-2: triangles score 1, squares 2, etc. The score for a polyanygon is the total of its components. My counts (by hand): 1 2 3 9 20 75. These are with all symmetries (rotations, reflections) removed. (Confirmations wanted!) The 3 polyanygons of score 3 are the triamond (three regular triangles), a square stuck to a triangle (a house?), and the pentagon. A computer program to do this will be challenging. I think it will require working exactly with fairly high degree roots of 1. Like LCM(1...8) = 840. I made three decisions that affect the counting. (a) scoring: I might have used total area, but didn't. This would have split things into many more groups. (b) I might have used perimeter as the score, but didn't. My way of scoring is easier. This doesn't affect the counts much (so far). Note that higher scoring polyanygons can have partially offset edges, so the perimeter might not always be an integer. (c ) I regard the regular hexagon as different from the same shape made out of six triangles. It's also possible to make a dodecagon by adding a fringe or alternating triangles and squares to a regular hexagon. I think these are the only ambiguous cases. Again, the effect on the count is minor for small scores. I'm guessing that the limiting ratio Polyanygons(N+1)/ Polyanygons(N) is unbounded, but I have no good argument. ----- count of chess positions. The note from Guy Haworth points to Tim Krabbe's chess blog. http://www.xs4all.nl/~timkr/chess2/diary.htm Item 283 mentions "a serious approximation (see link <http://www.chessbox.de/Compu/schachzahl1b_e.html> ) of the number of possible chess positions is 2.28*10^46. " This is a little higher than our group's informal result from a few years ago; I think we got around 10^43. Unfortunately, the link seems broken. Rich

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[math-fun] Chess and Mathematics
by Guy Haworth 18 Oct '05

18 Oct '05

For those interested in the numeric aspects of chess ... Marc Bourzutschky and Yakov Konoval have broken Lewis Stiller's record for the 'deepest' known chess win, which was a KRNKNN position with depth = 243. This is a KRRNKRR win with depth = 290. KRRNKRR was recommended by Noam Elkies as worth investigating early. As Noam recommended several other endgames, it is possible that 290 and even 300 will be beaten in 7-man space, and we may not have long to wait. See http://www.xs4all.nl/~timkr/chess2/diary.htm, item 298 to play through the moves. guy

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[math-fun] Flags and antihistamines
by lkmitch＠att.net 18 Oct '05

18 Oct '05

The flag of the United States has 50 stars arranged in an alternating pattern: a row of six stars, then a row of five, etc. Alternatively, the star field can be viewed as a column of five stars, then a column of four, etc. This assemblage can also be seen as a four-by-five grid of stars interlaced within a five-by-six grid. See http://en.wikipedia.org/wiki/Image:Flag_of_the_United_States.svg for a good illustration. Since it is becoming allergy season for me, I happened to whip out my antihistamines and noticed that the 10 pills on a card are arranged in three rows: a row of three, a row of four, and a row of three. Or, in a one-by-four grid within a two-by-three grid. This got me wondering: what numbers can be represented as the total number of nodes (or stars or pills) where the nodes are laid out in two interlaced rectangular grids? For this problem, Im only considering the case where the numbers of rows for the two grids differ by exactly one, as do the numbers of columns (so a 1x4 in a 2x3 is ok, but not a 1x4 in a 3x6). This constraint makes it easy to alternate rows and columns to make adjacent nodes align diagonally instead of vertically. What I found more interesting than the numbers that can be represented is the set of numbers that cannot be represented: 1, 2, 3, 6, 9, 15, etc. (not in OEIS). 1 and 2 are trivial because there must be at least three rows. For every element after 2 (that Ive found) is a multiple of 3. The elements exhibit a decidedly non-regular pattern, and occur about half as frequently as primes. Can anyone point me to more information about this before I foolishly think that discovered something? Kerry Mitchell -- lkmitch(a)att.net www.fractalus.com/kerry

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[math-fun] Re: Triangulation
by Richard Guy 17 Oct '05

17 Oct '05

It depends on the arrangement. Suppose that the convex hull of the N nodes is a polygon with b boundary nodes, and hence b edges. Suppose that there are E edges and T triangles. Then 2E = 3T + b and Euler tells us that N + (T+1) = E + 2 Eliminate E to give T = 2N - b - 2 b is at least 3 (in which case do you count the boundary as a triangle of the triangulation?), so the max is 2N - 5. E.g., if N = 4, you either have b = 3 or b = 4, i.e., a triangle with a node and 3 triangles inside it, or a quadrangle chopped into 2 triangles. R. On Mon, 17 Oct 2005, Pfoertner, Hugo wrote: > SeqFans, > > my colleague Guido Dhondt just sent me the following question: > > -----Original Message----- > From: Dhondt, Guido , Dr. > Sent: Monday, October 17, 2005 11:18 > To: Pfoertner, Hugo > Subject: triangulation > >> What is the maximum number of triangles of a triangulation connecting N > nodes in a plane? >> >> (needed as upper memory bound for triangulation in CalculiX, > www.calculix.de) > > I tried to search OEIS for triangles & triangulation, triangles & maximal. > > http://www.research.att.com/projects/OEIS?Anum=A027610 > > seems to be related (Number of chordal planar triangulations; also number of > planar > triangulations with maximal number of triangles; ....) > > Any hints? > > Hugo Pfoertner > > =========================================================================== > REFERENCE of the administration: This E-Mail was delivered via InterNet. > Unencrypted InterNet-E-Mails could be read or falsified by unauthorized > persons. > --------------------------------------------------------------------------- > HINWEIS der Administration: Diese E-Mail wurde Ihnen ueber INTERNET > zugestellt. Unverschluesselte Internet-E-Mails koennten von Unbefugten > gelesen oder verfaelscht werden. > =========================================================================== >

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RE: [math-fun] Power Sum Conjecture
by dasimov＠earthlink.net 16 Oct '05

16 Oct '05

Rich writes: << I can't put my hand on the book, but I think Vaughn has shown that every sufficiently large number is of the form a^2+b^3+c^5. The key is that the sum 1/2 + 1/3 + 1/5 = 31/30 > 1, so the expected number of representations for a number N is, on average, K * N^1/30. >> and << This is false when the sum of 1/a + 1/b + 1/c < 1. The set of available sums < N is less than the product of the possible x,y,z values, N^(1/a) * N(1/b) * N^(1/c), so the average spacing is at least N ^ ( 1 - 1/a - 1/b - 1/c). The interesting cases are abc = 22* 233 234 235 236 244 333. >> A mere coincidence perhaps, but 1/a + 1/b + 1/c = R(a,b,c) > 1 <=> a sphere of Gaussian curvature K == 1 can be tiled by geodesic triangles whose angles are pi/a, pi/b, pi/c. (Just as R(a,b,c)=1 <=> the case of the plane with K==0, and R(a,b,c) < 1 <=> the case of the (hyperbolic) plane with K == -1). --Dan

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[math-fun] Sums of powers
by David Wilson 16 Oct '05

16 Oct '05

Under 10^8, it looks as if all n >= 0 not of the form a^2+b^3+c^4+d^5 for a,b,c,d >= 0 are 15 23 55 62 71 471 478 510 646 806 839 879 939 1023 1063 1287 2127 5135 6811 7499 9191 26471 which one would surmise is a finite list. all n <= 10^8 appear to be of the form a^2+b^3+c^4+d^5+e^6 -------------------------------- - David Wilson

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[math-fun] Power Sum Conjecture
by David Wilson 16 Oct '05

16 Oct '05

I have very little evidence to believe this, but i'll put it forward anyway For a,b,c >= 2, x,y,z >= 0, the gap between numbers of the form x^a + y^b + z^c is bounded. -------------------------------- - David Wilson

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[math-fun] Observation
by David Wilson 16 Oct '05

16 Oct '05

Is it true that every nonnegative n is of the form n = a^2 + b^3 + c^4 + d^5 + e^6 for nonnegative a,b,c,d,e? -------------------------------- - David Wilson

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Re: [math-fun] More Median Musings
by David Gale 16 Oct '05

16 Oct '05

Neil,Unfortunately I don't have access at the moment to the OEIS, but here is the whole story on the median sequence. I'll be interested to know whether and if so how it qualifies for inclusion. The integer m in N={1,2,..,n} is a MEDIAN of n if 1+..+m=m+..+n. I write the pair (n,m) . Example: (1,1) is a median pair >The integer m' in N={1,2,..,n} is a SUBMEDIAN of n if 1+..+m'=(m'+1)+..+n. I write the pair (n,m'). Example: (3,2) is a submedian pair. One wants to know which n have medians and submedians. Here are the first 7 pairs, alternately median and submedian. (1,1) med. (3,2) submed. (6,8) med. (20,14) submed. etc. (49,35) (119,84) (288,204) ...... Here is the fundemental recusion: If (n,m) is a median pair then(n+2m,m+n) is a submedian pair. If (n,m') is a submedian pair them (n+2m'+1,n+m'+1.) is a median pair. C'est tout. David >David, is this sequence: >For which n does the sequence (1,2, . ,n) have a perfect median? >, in the OEIS? If so that property should be noted there, >and if not the sequence should be added! >Similarly for your other sequence. Can you send me the first few terms? >You mentioned a Pellian equation. > >Thanks! > >Neil > > At 08:57 AM 10/6/2005, you wrote: >David, is this sequence: >For which n does the sequence (1,2, . ,n) have a perfect median? >, in the OEIS? If so that property should be noted there, >and if not the sequence should be added! >Similarly for your other sequence. Can you send me the first few terms? >You mentioned a Pellian equation. > >Thanks! > >Neil > > Neil J. A. Sloane > AT&T Shannon Labs, Room C233, > 180 Park Avenue, Florham Park, NJ 07932-0971 > Email: njas(a)research.att.com > Office: 973 360 8415; fax: 973 360 8178 > Home page: http://www.research.att.com/~njas/

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RE: [math-fun] 6x6 multiplicative magic squares
by Christian Boyer 15 Oct '05

15 Oct '05

Dear friends, It seems impossible to get better 6x6 multiplicative squares that my examples found two days ago. So: the smallest possible magic product for 6x6 multiplicative squares should be 25945920 = 2^6 x 3^4 x 5 x 7 x 11 x 13. Here is one of the numerous examples: 1 22 39 54 28 20 12 65 18 8 3 77 30 27 14 52 11 4 42 16 6 33 15 13 26 21 44 5 24 9 66 2 10 7 78 36 This example uses the smallest possible set of integers (1..78) for that magic product. But is interesting to note that it is possible to construct 6x6 multiplicative squares with a more compact set of integers (1..66), but with a... bigger magic product! Here is an example with magic product 39916800: 8 45 42 66 1 40 50 56 44 3 54 2 16 33 5 28 30 18 21 10 15 9 64 22 27 4 6 32 55 35 11 12 48 25 7 36 The 10 smallest magic products of 6x6 multiplication squares should be: 25945920 26611200 28828800 29937600 31449600 33264000 33929280 34594560 34927200 35380800 I have the 40 smallest 6x6 magic products that I will propose to Neil for his OEIS. Christian. -----Message d'origine----- De : Christian Boyer [mailto:cboyer@club-internet.fr] Envoyé : vendredi 14 octobre 2005 08:42 À : 'math-fun' Objet : RE: [math-fun] 6x6 multiplicative magic squares Latest (and good) news! I can construct a lot of 6x6 multiplicative squares. My current smallest 6x6 multiplicative magic square has a magic product more than 70 times smaller than the Borkovitz & Hwang square. The B&H square used 36 distinct integers, the biggest being 1,260. My current best 6x6 example uses also 36 integers (of course...), but the biggest integer is 78: an extremely compact set of integers! But I need more checking before to be sure that it will be impossible to get better 6x6 results, and will keep you informed. Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr(a)mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Christian Boyer Envoyé : lundi 10 octobre 2005 12:07 À : 'math-fun' Objet : [math-fun] 6x6 multiplicative magic squares After 3x3, 4x4, 5x5, what about 6x6? My algorithm needs to be better optimized to attack the question. I will work on it in the next days. The smallest published 6x6 seems to be an example constructed by Borkovitz & Hwang, with magic product = 2,000,376,000. My question: can somebody produce an example with a smaller product? Michael Kleber? Rich? Unfortunately the latin squares method, used in previous messages for 4x4 and 5x5, does not work for 6x6: the famous "36-officers problem" of Euler has no solution! Christian. _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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