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[math-fun] Yet more about A027623
by David Wilson 12 Jun '05

12 Jun '05

In MathWorld article "Ring", I find [There are] 52 rings of size p^3 for p = 2, and 53 rings of size p^3 for p > 2 (Ballieu 1947, Gilmer and Mott 1973; Dresden). which conflicts with A027623(27) = A037234(27) = 59. - David W. Wilson "Truth is just truth -- You can't have opinions about the truth." - Peter Schickele, from P.D.Q. Bach's oratorio "The Seasonings"

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[math-fun] A027623 is multiplicative?
by David Wilson 12 Jun '05

12 Jun '05

Ok, in http://www.algebra.uni-linz.ac.at/~noebsi/pub/smallrings.ps I find Theorem 1. [Sho38] Every finite ring R can be uniquely (up to isomorphism) decomposed into a direct sum of rings of prime power order. I assume this is analogous to the decomposition therem for finite abelian groups, and would imply A027623 is multiplicative. - David W. Wilson "Truth is just truth -- You can't have opinions about the truth." - Peter Schickele, from P.D.Q. Bach's oratorio "The Seasonings"

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[math-fun] A027623, is it multiplicative?
by David Wilson 12 Jun '05

12 Jun '05

A027623 gives the number of rings with n elements, and looks empirically to be multiplicative. Is there some reason to believe that if gcd(x,y) = 1, a(xy) = a(x)a(y)? Maybe a finite-ring analogue to Sylow's theorem for finite abelian groups? - David W. Wilson "Truth is just truth -- You can't have opinions about the truth." - Peter Schickele, from P.D.Q. Bach's oratorio "The Seasonings"

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[math-fun] Re: Cheesy note from Gosper
by James Propp 10 Jun '05

10 Jun '05

Bill Gosper wrote: > Mandelbrot calls this Cantor set > "Apollonian gasket", and wrote several years ago that its fractal > dimension remained unknown. People who like Apollonian gaskets should look at recent work of Jeff Lagarias and his collaborators (on the arXiv). The article that appeared in the Monthly ("Beyond the Descartes circle theorem" by Lagarias, Mallows, and Wilks; math.MG/0101066) would be a good place to start. A particularly nice new wrinkle of Lagarias' work is the notion of supergaskets: these don't exactly match the description > Then the distinct images of C under G cover C (except for a Cantor > set) with non-overlapping circles. (the complement isn't a Cantor set) but a supergasket does in a weaker sense "cover" the interior of the disk; more precisely, a supergasket extends a gasket to a dense family of circles, no two of which intersect in more than one point. See math.MG/0010302 . >Plotting the circles to high order [Thanks to the kind auspices of Stephen > M. Jones, the circular gasket is http://gosper.org/suds.gif, the strip is > http://gosper.org/stripcheese.gif What are those colored circles? (I see yellow, purple, green, blue, and red when I look at this picture closely.) Are they added just to give the cheese extra pungency? >Problem: >is there a regular expression (or even a subgroup) which generates each circle >only once? It's worth mentioning that the circles have unique representatives in a tree that generalizes the Stern-Brocot process. The construction for moving down the tree is a swap operation: given four mutually tangent circles, replace one of them (call it C) by the unique other circle C' that is also tangent to the other three. If you let the four initial "circles" be the x-axis, the line y = 1, and the circles of radius 1/2 centered at (0,1/2) and (1,1/2), then you get the gasket in the unit strip. If you refrain from swapping out the x-axis, then you just get the Ford circles, and the swap operation is essentially just the mediant operation (where the mediant of two reduced fractions a/b, c/d is (a+c)/(b+d)). Just as the Stern-Brocot process gives you a tree in which every rational number in [0,1] except for 0 and 1 occurs as a node in a binary tree, this "Descartes-Soddy swap process" gives you a tree in which every circle in the gasket in the region bounded by the four initial circles occurs once as a node. > maps the gasket onto the strip between the real axis and imag(z)=1, > with the circles now including the Ford circles: > > |z - n/d - i/d^2| = 1/d^2, for every reduced rational n/d. Looks to me like there's a factor of two missing here. (The big circles that fit between the real axis and imag(z)=1 have radius 1/2, not 1.) I've fallen foul of this particular factor of two in the past. I'm not sure whether it's best to use the strip of thickness 1 or the strip of thickness 2. In any case, it's noteworthy that for all such gaskets, if the radius of each of the four original circles is the reciprocal of an integer, and you do swap operations, every derived circle will also have its radius the reciprocal of an integer. (This follows from the Descartes relation on the curvatures of four mutually tangent circles, along with the fact that if a monic quadratic equation in one variable has integer coefficients, and it has one integer root, then the other root is also an integer.) Jim Propp

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[math-fun] Cheesy note from Gosper
by Richard Schroeppel 07 Jun '05

07 Jun '05

Date: Tue, 7 Jun 2005 10:01:49 -0700 (PDT) From: "R. William Gosper" <rwg(a)osots.com> cheesy old mail While looking for something else, Rich kindly exhumed one of my old blurtings: Received: from SWEATHOUSE.macsyma.com ([192.233.166.105]) by NEWTON.Macsyma.COM via INTERNET with SMTP id 267086; 28 Nov 1993 01:37:52-0500 Date: Sat, 27 Nov 1993 22:28-0800 From: Bill Gosper <rwg(a)NEWTON.macsyma.com> Subject: Zero calorie Swiss cheese To: math-fun(a)cs.arizona.edu Message-Id: <19931128062850.2.RWG(a)SWEATHOUSE.macsyma.com> Let C be the circle |z| = (1 + sqrt(3))/2 and let D be the disk |z| < (1 + sqrt(3))/2, and 2 i pi 1 ------ z - - 3 2 f1(z) := e z, f2(z) := (2 - sqrt(3)) ----- z + 1 be the generators of a group G of homographic transformations. Then the distinct images of C under G cover C (except for a Cantor set) with non-overlapping circles. No, the distinct images of D under G cover D (except for a Cantor set) with non-overlapping disks. Mandelbrot calls this Cantor set "Apollonian gasket", and wrote several years ago that its fractal dimension remained unknown. "2D suds" is more descriptive, although the bubbles are unrealistically round. Define three different gaskets: K := the open disk D minus the (proper) images of closure(D); O := closure(D) minus the (proper) images of D; and U := the union of all the circles. It might seem that K would be empty, and O = U, but both of these intuitions are wild underestimates. In fact, O = closure(U) = closure(K) = boundary(U) = boundary(K), yet U is of measure zero in both O and K. K + U = O, where "+" = disjoint union. Still, U, O, and K have the same Hausdorff dimension and graphical appearance. (The really correct graphical appearance would be completely invisible!) [Ford circle pix I found on the Web were too arty for clarity. The Book of Numbers covers them, but I can't find mine.] Plotting the circles to high order [Thanks to the kind auspices of Stephen M. Jones, the circular gasket is http://gosper.org/suds.gif, the strip is http://gosper.org/stripcheese.gif, and the Ford circle subset is http://gosper.org/ford10.gif .] entails several annoyances, e.g., a tendency toward "unfloatable" numbers: 1 --------------------------------------------- 72010600134783751 - 41575339372323900 sqrt(3) gives 1/.0, even with double precision. (Molly Brown's antithesis? Unthinkable.) Another problem is whittling the collisions down to "the distinct images of C". I.e., it appears that the number of distinct group elements divided by the number of distinct circles they generate goes to infinity with increasing word length. Crudely, the smaller the circle, the more words which generate it. E.g., abbreviate f1 as W, f2 as G, and f2^-1 as g. Then the only relation is W^3 = 1. But obviously WWC = WC = C. Less obviously, G^n WGC = WGC, G^n WWGC = WWGC, W^n GGC = GGC, GWggC = WGWWGGGC. Is there a name (e.g., "projected group") for this sort of structure? Problem: is there a regular expression (or even a subgroup) which generates each circle only once? (If distinct group elements do generate the same circle c, they do not do so pointwise, but rather are related by a continuous "permutation" of the points of c. Is this permutation of finite order?) Exercise: Exhibit an explicit point of K. (It's infinitely bigger than U, so how can you miss?-) JHC's method for the triangular gasket should work, if you can find his old mail. Another approach might start with finding that limit point on the Soddy spiral page of www.tweedledum.com/rwg/ . The rest of the message was i pi 5 i pi - ---- - ------ 6 6 (sqrt(3) + 1) e z + e f(z) := - ----------------------------------- (sqrt(3) + 1) z + 1 maps the gasket onto the strip between the real axis and imag(z)=1, with the circles now including the Ford circles: |z - n/d - i/d^2| = 1/d^2, for every reduced rational n/d. These are non-overlapping circles which cluster against the real axis from above. This strip gasket is simply generated by 1 f1(z) := z + 1, f2(z) := ----- z - i acting on |z - i/2| = 1/2. --rwg

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[math-fun] More Eric Numbers
by Eric Angelini 07 Jun '05

07 Jun '05

Here: http://angelink.be/?EricNumbers Best, E.

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[math-fun] A Mall for Mathemticians
by Richard Schroeppel 06 Jun '05

06 Jun '05

I got a note recently from Don Knuth. He had exhumed some old notes, on finite algebraic systems with a binary operator that satisfies the axiom B.AB = A. He called them Gropes. I briefly called them Babagroups, as a reminder of the axiom. The formal name is Semisymmetric Quasigroups. Don's note includes a short six page summary of Grope properties, including some constructions and the relationship with Steiner Triples. It's pretty dense. I haven't fully grokked it, but he doesn't seem to mention our group's discovery about the number of idempotents mod 3. The most interesting part was toward the end: He wrote a program to count the non-isomorphic systems, to contribute to Sloane's database. He found my numbers (thanks, Neil), hence the note. Meet me at 1,1,2,3,4,9,41,595,26620,3908953,1867918845. I'm the guy with purple hair. Rich

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RE: [math-fun] FW: Beautiful Zeta(3) formulas!
by Meeussen Wouter (bkarnd) 30 May '05

30 May '05

Harrumph indeed! Simply replacing n!^p /(n+w)!^p by 1/Product[n+k,{k,1,w}] makes it easy: Sum[n!^p/(n+w)!^p,{n,0,oo}] == HypergeometricPFQRegularized[{1,1 ..p+1 copies}, {1 + w, 1 + w,.. p copies}, 1] In[1]:= Table[Sum[1/(Product[n + k, {k, 1, w}])^p, {n, 0, \[Infinity]}] // FullSimplify, {w, 1, 7, 2}, {p, 3, 7, 2}] // InputForm // Timing Out[1]={2.781 Second, { {Zeta[3], Zeta[5], Zeta[7]} , {29/32 - (3*Zeta[3])/ 4, (-3*(-791 + 520*Zeta[3] + 160*Zeta[5]))/ 512, (-3*(-35731 + 21448*Zeta[3] + 8288*Zeta[5] + 1344*Zeta[7]))/ 4096} , {(5*(-2077 + 1728*Zeta[3]))/ 1327104, (-1186062547 + 781704000*Zeta[3] + 237634560*Zeta[5])/ 330225942528, (5*(-1404612185513 + 848813853888*Zeta[3] + 322757084160*Zeta[5] + 49203044352*Zeta[7]))/ 4565043429507072} , {(7*(389467 - 324000*Zeta[3]))/ 503884800000, (-7*(-784046040073 + 516756240000*Zeta[3] + 157075200000*Zeta[5]))/ 125382662553600000000, (156827579578950685957 - 94800261054346848000*Zeta[3] - 36021058152710400000*Zeta[5] - 5475321411840000000*Zeta[7])/584985350410076160000000000000} } } W. -----Original Message----- From: R. William Gosper [mailto:rwg@osots.com] Sent: maandag 30 mei 2005 9:10 To: math-fun(a)mailman.xmission.com Subject: Re: [math-fun] FW: Beautiful Zeta(3) formulas! rcs>Well, cute anyway. I assume these aren't news to some of our readers. I should hope not. >-----Original Message----- From: Number Theory List [mailto:NMBRTHRY@LISTSERV.NODAK.EDU] On Behalf Of Troy Kessler Sent: Friday, May 27, 2005 6:46 AM To: NMBRTHRY(a)LISTSERV.NODAK.EDU Subject: Beautiful Zeta(3) formulas! 29/32 -3*Zeta[3]/4 = sum n=0 to oo (n)!^3/(n+3)!^3 (mathematica can do) 5*(-2077+1728Zeta[3])/1327104 = sum n=0 to oo (n)!^3/(n+5)!^3 Mathematica gives an answer that is 5 lines long for this second sum. Harrumph. (c154) closedform(sum(n!^3/(n+5)!^3,n,1,inf)) Time= 1131 msec. 5 zeta(3) 1298221 (d154) --------- - --------- 768 165888000 (c156) closedform(sum(n!^3/(n+7)!^3,n,1,inf)) Time= 2163 msec. 935108917 7 zeta(3) (d156) --------------- - --------- 172832486400000 1555200 =============================== 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] FW: Beautiful Zeta(3) formulas!
by Schroeppel, Richard 30 May '05

30 May '05

Well, cute anyway. I assume these aren't news to some of our readers. Rich -----Original Message----- From: Number Theory List [mailto:NMBRTHRY@LISTSERV.NODAK.EDU] On Behalf Of Troy Kessler Sent: Friday, May 27, 2005 6:46 AM To: NMBRTHRY(a)LISTSERV.NODAK.EDU Subject: Beautiful Zeta(3) formulas! 29/32 -3*Zeta[3]/4 = sum n=0 to oo (n)!^3/(n+3)!^3 (mathematica can do) 5*(-2077+1728Zeta[3])/1327104 = sum n=0 to oo (n)!^3/(n+5)!^3 Mathematica gives an answer that is 5 lines long for this second sum. It can be shown that if 3 or 5 is replaced with a odd number the result is rational + rational *Zeta[3]. (Use Zeilberger in A=B)

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[math-fun] number theory question
by wouter meeussen 27 May '05

27 May '05

since 1/Product[(1 - 1/Prime[k]^w), {k, \[Infinity]}] equals Zeta[w], I wondered what this product becomes when specialising for either primes p with 4|(p+1) or primes q with 4|(q-1). Numerics at w=2 point to 1.690 and 1.335 aboutish. W.

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