- mathfun@mailman.xmission.com - mailman.xmission.com

[math-fun] 999,999,999,999 places of pi on the wall, 999,999,999,999 places of pi;
by Richard Schroeppel 18 Jul '05

18 Jul '05

This looks cute. --Rich ----- Subject: [CCA] "Many Digits" Friendly Competition 2005 : Call for Participation From: Milad Niqui <milad(a)cs.ru.nl> Reply-To: manydigits(a)cs.ru.nl Date: Fri, 15 Jul 2005 20:57:26 +0200 Sender: owner-cca-list(a)fernuni-hagen.de [Apologies for multiple copies] THE "MANY DIGITS" FRIENDLY COMPETITION 2005 4 October 2005 Radboud University, Nijmegen, The Netherlands Call For Participation <http://www.cs.ru.nl/~milad/manydigits> In the Institute for Computing and Information Sciences at Radboud University Nijmegen we will organise a friendly competition/benchmark session for investigating the state of the art in the various implementations of exact real arithmetic. The event will be held on October 4th 2005, in conjunction with the Small TYPES workshop on Constructive analysis, types and exact real numbers <http://www.cs.ru.nl/fnds/typesreal/>. The event will be akin to those held in CCA00 <http://www.cs.swan.ac.uk/~csjens/pdf/resart.pdf> and CCA2002 <http://www.informatik.uni-trier.de/~mueller/cca2002/>. All the interactive software packages capable of handling arbitrary precision arithmetic on elementary functions are invited to participate. Moreover all libraries for handling arbitrary precision arithmetic in any programming language are welcome to participate provided that the participant writes a suitable interface/program. The participants can choose to attend the event at Nijmegen, or they can choose to participate remotely via Internet. Friendly Competition The friendly competition consists of giving a solution to a set of problems that are posed by the organisers shortly (few hours) before the event. The problems cover the minimum abilities of a system capable of handling the elementary real functions and are expected to ask for a large (>=10,000) number of digits. An example of a problem that might be in this set is to calculate the first 10,000 digits of sin(sin(1)). A set of `practice problems' can be found on the website of the competition <http://www.cs.ru.nl/~milad/manydigits/sample_questions.php>. The solutions consist of the input needed to generate the answers to the problems plus the answers themselves. The competition criteria will be the accuracy (correctness) of the solution and the time needed for producing the solution. Apart from this, we will also compare the other aspects of the solutions (such as memory usage and readability and length of the input code) to give a better comparison. On the day of the competition the solutions will be executed by participants on one specific machine at the event site, under either Debian/GNU Linux or Windows XP. A more detailed description of the setup and format of the competition can be found on the website of the competition <http://www.cs.ru.nl/~milad/manydigits/rules.php>. The outcome of this friendly competition is expected to give a comparison in performance of the systems that chose to participate for the selected problems. Remote Participation In case of remote participation, the participants will receive the set of problems by email. Then they will be given the opportunity to log into our local machine and perform their solution. Systems, Packages and Libraries Those who wish to participate in the Friendly Competition are expected to provide the software necessary for their solutions, so it can be run on the machine used for the competition. Schedule * August 26th deadline for registering for the competition * August 29th final list of registered participants on the web * September 23rd timings of the practice problems for the registered participants posted on the web * October 4th competition Registration If you are interested in participating in the event we will be happy to hear from you: 1.Team name, name and email of contact person(s) and the platform your system needs to run (GNU Linux or Windows XP). 2.Whether you plan to come to the meeting or whether you would like to participate remotely. We would prefer that participants (especially those whose system requires Windows) to send their intention to participate before August 26th, 2005. Please register by sending mail to <manydigits(a)cs.ru.nl>. Best Regards, Local Organisers: Milad Niqui and Freek Wiedijk

1 0

[math-fun] star polygon area
by R. William Gosper 15 Jul '05

15 Jul '05

(Very elementary stuff.) When n>2d, (d,n)=1, connecting every dth vertex of a regular n-gon makes an n-pointed star {n/d} consisting of d-2 concentric rings of kites surrounding a ring of isosceles triangles surrounding a regular n-gon. Working inward, successive rings of kites have winding numbers W=1,2,3,...; the triangles have W=d-1, and for the polygon, W=d. For unit edges, the area is %pi d cot(-----) n n n A(-) = ------------, d 4 where the component polygons are weighted by their winding numbers, i.e, W dx dy integrated over the whole plane. This correctly gives n n A(-----) = - A(-), n - d d i.e., mirror imaging negates the area by winding clockwise. The %pi d csc(-----) n circumradius = ----------, and the 2 %pi d cot(-----) n inradius = ----------. 2 For unit circumradius instead of unit sides, the area is %pi d sin(-----) n n n a(-) = ------------, d 2 which, for fixed n is clearly maximal for d=n/4. I.e., the star with greatest weighted area is the one with most nearly right-angled vertices, since the vertex angles are pi - 2 pi d/n. E.g., the star of Lakshmi (two squares) is maximal among the octagrams. Removing just the outermost edges that form the vertex angles leaves an {n/(d-1)} star polygon, if when (n,d-1)=g we interpret this as g {n/g,(d-1)/g} stars superposed. Interestingly, for such improper stars, continued edge deletion recovers the proper ones with lesser d. E.g., {9/4} -> {9/3} (star of Goliath = three triangles) -> {9/2}. This tells us all the interior angles. The circumradius formula for consecutive d provides the proportions of the inner stars, and hence the sizes of all the kites, etc. Note also that the area formulas work for improper stars: A((gn)/(gd)) = g A(n/d). --rwg I keep warm peanut butter in my pockets AND I VOTE

1 1

[math-fun] variants of the Kotani ant problem
by James Propp 15 Jul '05

15 Jul '05

The Kotani ant problem asks us to consider distances between points on the surface of a cube where the distance between two points is the length of the shortest path between them that stays on the surface. (See Dick Hess' article "Kotani's Ant Problem" in the book "Puzzler's Tribute: A Feast for the Mind".) Has anyone considered analogues of the Kotani ant problem for polyhedra other than cubes and rectilinear blocks, e.g., the other Platonic solids? (I'm looking for a research project for some students, and this one came to mind as something that doesn't require any post-high-school background.) Jim Propp

1 0

[math-fun] Applied graph theory
by Michael Kleber 14 Jul '05

14 Jul '05

This is more frustrating than it rationally ought to be: http://home.cwru.edu/~jnt5/Planarity/ --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.

2 1

[math-fun] FW: divisibility conjectures
by Schroeppel, Richard 09 Jul '05

09 Jul '05

This seems like it might appeal to some folks here. Rich rcs(a)cs.arizona.edu -----Original Message----- From: Number Theory List on behalf of Don Davis Sent: Fri 7/8/2005 8:38 PM To: NMBRTHRY(a)LISTSERV.NODAK.EDU Subject: divisibility conjectures Some conjectures about divisibility by 2 have arisen in my work in algebraic topology. The first one involves Stirling numbers of the second kind. It would give a nice form for homotopy groups of some important topological spaces. We write nu(-) for the exponent of 2 in an integer. Conj 1: If j \ge n and L is sufficiently large (e.g. L>3n/2), then nu(j! S(2^L + n -1, j)) \ge n-1 + nu([n/2]!). It is not difficult to remove the 2^L and the n-1 on the RHS, obtaining the following conjecture, which implies Conj 1. Conj 2: If j>m, then nu(sum ((j choose 2k) k^m)) \ge nu([(m+1)/2]!). Instead of dealing with expressions k^m, one might prefer to work entirely with binomial coefficients. We have the following conjecture, which implies Conj 2. Our sums are taken over all sensible values of k. Conj 3: nu( sum ((j choose (4k+2)) (k choose i))) \ge nu([j/2]!) - i - nu(i!). This can be interpreted as a coefficient in a power series, of which it appears that not only the coefficient in question is highly divisible, but also all subsequent coefficients. We obtain the following conjecture, which implies Conj 3. Conj 4: nu( coeff( (1+x)^j / (1-x^4)^(i+1), x^k) ) \ge nu([j/2]!) - i - nu(i!) for k > j-4i-4. It is quite dramatic that the coefficient of x^(j-4i-4) will be odd, but then apparently all the rest are highly divisible. It is enough to prove this when j = 4a, and probably when i+1 is a 2-power. Experimenting with related power series, by removing the 4's, we are led to the following conjecture, which is closely related to Conj 4. My feeling is that this last conjecture is one which others might have seen, and that a proof of it would probably lead to a proof of Conj 4. Conj 5: If 2^t > 2d, then nu( coeff( ((1+x)/(1-x))^(2^t) (1+x)^(2d), x^k) ) \ge t + d + nu(d!) + 1 if k > 2d. Just to clarify...the power series in Conj 5 equals (1+x)^(2^t + 2d) / (1-x)^(2^t). If anyone on this listserve can point to any literature related to these conjectures, or offer suggestions toward a proof, I would be most appreciative. If you can prove one, we can discuss an appropriate way to publish it. Don Davis dmd1(a)lehigh.edu http://www.lehigh.edu/~dmd1

1 0

Re: [math-fun] Lattice / finite group question
by dasimov＠earthlink.net 08 Jul '05

08 Jul '05

Gene writes: << No. Inversion through the origin is always a lattice isometry, so the isometry group must have even order. >> Aha -- good point. Are there further restrictions? --Dan

3 2

[math-fun] Self-erasure surviving numbers
by Eric Angelini 08 Jul '05

08 Jul '05

[Eric A.] much too affirmative, wrote a few days ago: > No SESN > 10 begins with "1" -- see why? > No SESN > 299 begins with "2", etc. > > The sequence is finite, thus. > > Last term? ... [Michael A. Cleverly] objects that 114 *is* a SESN: 11(4)1(1)4(1)14(1)1(4)114(1) ^^ ^ ^ ^^ !!! So I might be wrong about the finitude of the sequence, sorry. And I'll have to add a few more terms to the seq. in the OEIS... Best, É.

1 0

[math-fun] Lattice / finite group question
by dasimov＠earthlink.net 07 Jul '05

07 Jul '05

Let G be any finite group. Questions: 1) Is there always some n such that there is some n-dimensional lattice L whose isometry group (keeping the origin fixed) is isomorphic to G ? Call such an isometry group I_0(L). 2) If so, is there always some n and some lattice L that is a (not nec. proper) subgroup of the integer lattice Z^n, such that I_0(L) is isomorphic to G ? ------------------------------------------ Note: It's easy to see that if n = #(G), then the symmetric group S_n is a *subgroup* of I_0(Z^n), and hence, so is G. Close, but not quite what we are looking for. ------------------------------------------ (We define a n-dimensional lattice to be any discrete subgroup -- isomorphic to the group Z^n -- of the abelian group and Euclidean space R^m, necessarily for some m >= n.) --Dan

2 1

[math-fun] "Self-erasure surviving numbers"
by Eric Angelini 07 Jul '05

07 Jul '05

Hello Seqfan and math-fun, Take an integer like 36, for example. Concatenate an infinite amount of copies. You get: 363636363636363636363636363636... - read the leftmost digit -"3"-, - jump *over* 3 digits (to the right), land on (3) and erase it: 3636(3)6363636363636363636363636... ^ - read the leftmost unread digit, jump only visible digits, erase: 3636(3)6363(6)36363636363636363636... ^^ - repeat until you see a substring like this [...(a)36(b)...] [(a) and (b) are erased digits - "36" is the integer you are, testing]: bingo, you have found a "Self-erasure survi- ving number" (SESN): "36" is such a number: 3636(3)63(6)3(6)36(3)(6)3(6)36363636363636... ^^^^ ^^ .. <-- hit This erasing technique gives sometimes quite complicated pat- terns. "16", for instance, is not a SESN -- but it takes a while to see: 16(1)616(1)61(6)1(6)(1)6(1)(6)1(6)1(6)1(6)1(6)(1)6(1)616(1)61(6)1(6) ^^ ^^^ ^^ ^ ^ ^ ^ ^ ^ ^ ^^^ ^^ ^ |_______________________________________________| recurrent pattern The first SESN I have found by hand are: 0 1 2 3 4 5 6 7 8 9 10 20 23 24 25 26 27 28 29 30 32 36 37 38 39 40 42 ... [BTW, reading "0" means erasing the closest visible digit immediately to the right] No SESN > 10 begins with "1" -- see why? No SESN > 29 begins with "2", etc. The sequence is finite, thus. Last term? And what about recurrent patterns: do all integers behave like that? Could some strings be definitely "chaotic"? I guess no... Best, É.

5 4

[math-fun] "Self-erasure surviving numbers" (2)
by Eric Angelini 07 Jul '05

07 Jul '05

> No SESN > 29 begins with "2", etc. ... Oooops, mean: No SESN > 299 begins with "2", etc. Best, É.

1 0