March 2005 - mathfun@mailman.xmission.com

[math-fun] (Fwd) [NOBNET 4625] Mathematical Games
by Bernie Cosell 09 May '05

09 May '05

I know that in addition to your likely being interested in this, I know that several of you have contributed to Mathematical Games over the years and so might want to get in touch with Ed... /Bernie\ ------- Forwarded message follows ------- Date: Mon, 8 Nov 2004 06:30:37 -0800 (PST) From: ed pegg <ed(a)mathpuzzle.com> Subject: [NOBNET 4625] Mathematical Games To: nobnet(a)iijnet.or.jp Martin Gardner just recently celebrated his 90th birthday. Also, maa.org is *finally* releasing the entire collection of Mathematical Games on CD rom next month. There were frightening amounts of copyright issues to work out, but they have been resolved. Catalog Code: TDG/FW04 ISBN 0-88385-545-3 List Price: $54.95 This is all 15 Mathematical Games books on a CD-Rom. I'd like to do a "behind the scenes" look at Mathematical Games for an article. Martin did lots, but he avidly used a lot of help from hundreds of people. I'd like to write an article about the background help. If you ever assisted in one of Martin's columns, I'd like to hear from you -- a note about what you did. I'll use these notes in my article. Ed Pegg Jr http://www.maa.org columnist http://www.mathpuzzle.com ------- End of forwarded message ------- -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

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Re: [math-fun] better posed problem? (small surprise)
by Phil Carmody 07 May '05

07 May '05

From: "Meeussen Wouter (bkarnd)" <wouter.meeussen(a)vandemoortele.com> > > it is a small surprise that, for w = Exp[ 2 Pi I /30 ] alias 1^(1/30) , > w+w^7+w^8+w^18+w^19+w^25 = 0 > no cancellation of symmetric pairs, triples or quintuples in sight, > but w+w^7+w^19+w^25 = -w^13 (* four out of five *) > and w^8+w^18 = -w^28 = +w^13 (* two out of three *) > > could be called "cancellation by missing parts" ;-)) As (a) {0,6,12,18,24} cancels, (b) {0,10,20} cancels, and (c) {0,15} cancels. {5,15,25} cancels, so {5,6,12,18,24,25} cancels. The largest gap (mod 30) between terms is that between 25 and 5, so {0,1,7,13,19,20} is the symmetry-less set with minimal largest member that cancels. You've exploited the gap of 6 between 18 and 24 instead. 2*p*q is easily so minimised - any takers for zeta_210? Phil When inserting a CD, hold down shift to stop the AutoRun feature In the Device Manager, disable the SbcpHid device. http://www.cs.princeton.edu/~jhalderm/cd3/ __________________________________ Do you Yahoo!? Yahoo! Small Business - Try our new resources site! http://smallbusiness.yahoo.com/resources/

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[math-fun] derangements, three, shorter
by Meeussen Wouter (bkarnd) 30 Mar '05

30 Mar '05

conjecture (up to w=11) mod( !(3^w k + 1) , 3^(w+2) ) == 0 (w>1 ; all k>1) W. --- gibberish beyond this line --- =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.

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[math-fun] Re: Sums of Primes and Squares (fwd)
by Richard Guy 30 Mar '05

30 Mar '05

Here's an answer I've just made. Can any munster or sequester comment on the finiteness or otherwise of A064233 ? Best to all, R. ---------- Forwarded message ---------- Date: Wed, 30 Mar 2005 13:07:06 -0700 (MST) From: Richard Guy <rkg(a)cpsc.ucalgary.ca> To: Althoefer <320045884400-0001(a)t-online.de> Cc: Richard Guy <rkg(a)cpsc.ucalgary.ca> Subject: Re: Sums of Primes and Squares I'm not sure if I can be of much help. A19 of UPINT doesn't quite cover it. But let me tackle (ii):- If m^2 = n^2 + p, then necessarily n = m-1 and p = 2m-1 so all numbers [(p+1)/2]^2 are expressible as [(p-1)/2]^2 + p. On the other hand, if m is not of the form (p+1)/2 then m^2 is not equal to r^2 + p for any r, because p = m^2 - r^2 = (m-r)(m+r) requires m-r = 1 and then p = m+r = 2m-1 which we've assumed to be false. So the squares of 5, 8, 11, 13, 14, 17, 18, 20, 23, 25, 26, 28, 29, ... are not the sum of a square and a prime. I.e., almost all squares aren't. As for (i), sequence A064233 doesn't give any references, but I'd guess that the sequence is as likely to be infinite as finite -- good question which may find its way into future editions of UPINT. The answer for (iii) is presumably the same as that for (i). Let me know if you learn more. R. ID Number: A064233 URL: http://www.research.att.com/projects/OEIS?Anum=A064233 Sequence: 1,2,5,10,13,25,31,34,37,58,61,64,85,91,121,127,130,169,196, 214,226,289,324,370,379,400,439,526,529,571,625,676,706,730, 771,784,829,841,991,1024,1089,1225,1255,1351,1414,1444,1521, 1549,1681,1849,1906,1936,2116 Name: Numbers which are not the sum of a prime number and a nonzero square. Example: 5 = 1+4 or 2+3; a prime and a square do not appear together in either sum Math'ca: Complement[ Table[ n, {n, 1, 10000} ], Union[ Flatten[ Table[ Prime[ i ] + j^2, {i, 1, 1230}, {j, 1, 100} ] ] ] ] See also: Adjacent sequences: A064230 A064231 A064232 this_sequence A064234 A064235 A064236 Sequence in context: A003654 A047617 A018571 this_sequence A051952 A103188 A064392 Keywords: easy,nonn,nice Offset: 0 Author(s): Axel Harvey (axe(AT)cam.org) Sep 22 2001 Extension: More terms from Vladeta Jovovic (vladeta(AT)Eunet.yu), Robert G. Wilson v (rgwv(AT)rgwv.com) and Felice Russo (felice.russo(AT)katamail.com) Sep 23 2001 On Wed, 30 Mar 2005, Althoefer wrote: > Dear Prof. Guy, > > can you please tell me, what is known about the > following questions: > > (i) Are there infinitely many natural numbers which > are NOT the sum of a prime > and a square? (For instance, 130 is such a number; > 129=113+16 is not.) > > (ii) Are there infinitely many squares of natural > numbers which are NOT the sum > of a prime and a square? (For instance, 625 is such a > number.) > > (iii) Are there infinitely many natural numbers, which > are both not squares > numbers and not the sum of a prime and a square? > > My conjecture is that all three answers should be NO. > > Thanks in advance for your reply. > Sincerely yours, Ingo Althofer.

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[math-fun] Re: Galois-actions and star-polytopes
by James Propp 30 Mar '05

30 Mar '05

>> Is there anything analogous for any higher-dimensional star-polytopes that >> would enable one to view them as Galois conjugates of convex polytopes? > >Yes. For instance, in 3 dimensions, the dodecahedron is conjugate to >the great stellated dodecahedron; the icosahedron is conjugate to the >great icosahedron; and the small stellated dodecahedron is conjugate >to the great dodecahedron. (You can get pictures of all of these at >http://www.mathconsult.ch/showroom/unipoly/ ) > >There's also a version in 4 dimensions; I don't know about higher >dimensions. Does it make sense to ask similar questions about regular tilings of the hyperbolic plane (and higher-dimensional hyperbolic space)? This will only make sense if the vertices of such a tiling are algebraic. And that depends on what model of hyperbolic space you use. In 2 dimensions, you can use the complex structure on the inside of the unit disk, but for higher-dimensions it's probably most natural to use the parametrization of H^n by lines through R^{n+1} that intersect a hyperboloid. So then we just have to clarify what we mean by a Galois action on RP^n. There's probably a slick way to say things that I'm not thinking of right now, but if you stick to the part of RP^n where all coordinates are non-zero, then it's clear what to do (if not clear how best to describe it): e.g., in RP^2, the image of (x:y:z) should be (x':y':z') where the (well-defined) real numbers x'/y', y'/z', and z'/x' are the Galois-images of x/y, y/z, and z/x. Jim Propp

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[math-fun] binary words and compositions
by Emeric Deutsch 30 Mar '05

30 Mar '05

I am interested in various bijections between binary words of length n and compositions of n+1. Many thanks. Emeric

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[math-fun] CORRECTION Re: Are these infinite groups simple?.
by Daniel Asimov 29 Mar '05

29 Mar '05

A friend pointed out to me << In place of "the automorphisms that fix all but a finite number of elements" you want "the automorphisms that fix a subspace of finite codimension." >> The corrected questions, with slightly better notation: ------------------------------------------------------------------------------------------------ By analogy with the finite groups PSL(n,p) (n >=2), which are simple except for PSL(2,2) = S_3 and PSL(2,3) = A_4, (and, PSL(n,p) is uniquely determined by (n,p) except for the isomorphic pair PSL(3,2) = PSL(2,7)): For any prime p, let F(p) be the finite field of size p. Let F(p)^oo be the vector space of countably infinite dimension over F(p). Let G(oo,p) be the group of automorphisms Aut(F(p)^oo). This has a normal subgroup: those automorphisms that fix a subspace of finite codimension. Call this subgroup GL_f(oo,p). The quotient group Q(p) = GL(oo,p)/GL_f(oo,p) still has a normal subgroup (of size p-1), namely the nonzero scalars Sc(oo,p) = {kI in Q(p) : k in (F(p)*}, where I is the identity on Q(p). So, let PQ(p) denote Q(p)/Sc(oo,p). --------------------------------------------------------------------------------------------------- Q1. Is there any chance that PQ(p) is simple for some primes p, or maybe all of them? Q2. Is there some reason to believe that PQ(p) is isomorphic to PQ(p') always implies p=p' ? --Dan

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[math-fun] Are these countably infinite groups simple?
by Daniel Asimov 29 Mar '05

29 Mar '05

Given the field F_p of prime order p, we can form the group PSL(n,p) = SL(n,p)/{+-I}, where SL(n,p) is the group of vector space automorphisms of (F_p)^n having determinant = 1. Ignoring n=1, since PSL(1,p) is the trivial group: It's known that PSL(n,p) is simple for all n >= 2 and p prime, except for the two cases (n,p)=(2,2) and (n,p)=(2,3). Also, PSL(n,p) is isomorphic to PSL(n',p') implies (n,p)=(n',p'), except for one case: PSL(2,3) is isomorphic to PSL(7,2). ------------------------------------------------------------------------------------------ QUESTIONS: What about n = oo ? Can we find an analogous simple group? I don't know how to define the determinant of an arbitrary automorphism of the vector space over F_p of countable dimension. Let's call this (F_p)^oo. Its group of automorphisms GL(oo,p) is well-defined. This has a normal subgroup: those automorphisms that fix all but a finite number of elements. Let's call this GL_f(oo,p). The quotient group Q_p = GL(oo,p)/GL_f(oo,p) still has a normal subgroup, namely the nonero scalars Sc(oo,p) = {[kI] in Q_p : k in (F_p)*}, so let PQ(p) = Q_p/Sc(oo,p) Q1. Is there any chance that PQ(p) is simple for some primes p, or maybe all of them? Q2. Is there some reason to believe that PQ(p) is isomorphic to PQ(p') always implies p=p' ? --Dan

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[math-fun] dendroteleology
by R. William Gosper 28 Mar '05

28 Mar '05

Nobody got > PS, "square" tree rings: Replace one side of a square with a semicircle to make a figure with two corners and three straight sides. Near me is a large, sawn-off tree stump whose outer rings are routinely round, but about halfway in gradually deform to the aforementioned shape, with sides quite straight and corners quite sharp. Explain. Near the ground the tree originally had a nearly horizontal branch the same thickness as the trunk. When both were about 6 in. thick, the branch was cut off nearly vertically, fairly near the trunk, which continued thicken to over 2 ft, swallowing the stub of the branch. The horizontal section happens to pass right through the traces of the stub. --rwg paternalistic antiparticles

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[math-fun] derangements divisible by 3
by wouter meeussen 28 Mar '05

28 Mar '05

given the many alternatives of writing (!n), cfr. OEIS A000166, and considering there is a nice way to find the powers of p in n! by writing n in base p (if I remember correctly), it is funny to observe the powers of 3 in !n. I get, using a[ n ] for (!n) and p as representing 3 (to avoid immediate evaluation in Mathematica): {Mod[a[1 + k*p], k*p]==0, Mod[a[1 + k*p^2], k*p^3]==0, Mod[a[1 + k*p^2], k*p^4]==0, Mod[a[1 - p^2 + k*p^3], -p^5 + k*p^6]==0, Mod[a[1 - p^2 - p^3 + k*p^4], -p^6 - p^7 + k*p^8]==0, Mod[a[1 - p^2 - p^3 + k*p^5], -p^7 - p^8 + k*p^10]==0, Mod[a[1 - p^2 - p^3 + k*p^6], -p^8 - p^9 + k*p^12]==0, ......... } and I wonder where these lead to. W.

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