January 2011 - mathfun@mailman.xmission.com

Re: [math-fun] FW: A conjectural series for 1/pi of a new type
by Bill Gosper 14 Jan '11

14 Jan '11

Simon> hello everybody, about that formula for 1/pi, it is nice , the coefficient a(k) is a binomial term The sequence a(k) is : 1, 1, 33, 97, 1729, 8001, 105441, 627873, 6989697, 48363649, 488206753, 3701949153, 35289342529, 283146701761, 2610495177057, 21695983405857, 196218339243777, 1667338615773441, 14917038493453089, 128562758660255073, 1143482133220664769, 9946278255903268929, 88205310329762729697, 771946983805271894433, 6837125121111415598721,... Which has this g.f. 1 -------------------- = 2 1/2 (1 - 2 x - 63 x ) Now, this has a closed binomial expression, of course anyhow, this series : can be programmed into : proc(k) local x; (x^2 + x + 16)^k; expand(%); coeff(%, x, k) end; and proc(k) (30*k + 7)*binomial(2*k, k)^2/(-256)^k;%*f(k); end; which gives a sequence : which converges to 24/Pi at a rate of approx. .25 digits per term. or in other words, 1000 terms = 250 digits of precision. best regards, Simon Plouffe ---------------- In matrixland, this identity is prod(matrix([-(k+1/2)^3/(8*(k+1)^3),-(k+1/2)^2/(16*(k+1)^2),30*(k+7/30)],[-63*k*(k+1/2)^2/(16*(k+1)^3),0,0],[0,0,1]),k,0,inf) = matrix([0,0,24/%pi],[0,0,0],[0,0,1]) [ 1 3 1 2 ] [ (k + -) (k + -) ] [ 2 2 7 ] [ - ---------- - ----------- 30 (k + --) ] inf [ 3 2 30 ] [ 24 ] /===\ [ 8 (k + 1) 16 (k + 1) ] [ 0 0 --- ] | | [ ] [ %pi ] | | [ 1 2 ] = [ ] | | [ 63 k (k + -) ] [ 0 0 0 ] k = 0 [ 2 ] [ ] [ - ------------- 0 0 ] [ 0 0 1 ] [ 3 ] [ 16 (k + 1) ] [ ] [ 0 0 1 ] which looks very much like a coordinate transformation of some very special case of the "3F2[1] Rosetta Stone". (http://gosper.org/stanfordn3.dvi) --rwg

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Re: [math-fun] geometric magic squares
by Lee Sallows 14 Jan '11

14 Jan '11

Hi Fred, Well no, not hard for me to construct. Because it was all done for me by an instrument-maker friend. And even he merely programmed the machine that milled the aluminium blocks. An interesting discovery afterwards was that fitting together the 3 pieces chosen from any row, column or diagonal was in several cases quite a challenge. So you get 8 such puzzles in one go. Here at home I have the puzzle built into a box like a cigar-box, with hinged lid. On opening the box you see a square board with 9 square holes cut in it, just like the 9 cells of a 3x3 magic square, each serving as a 'nest' for one of the nine pieces. But now here's the catch. The pieces are NOT arranged so as to form a magic square. As a piece of paper glued into the top of the box explains, the challenge is to rearrange the pieces within their nests so that they DO form a magic square. If you think about it, that is a frightfully difficult thing to do. Regards, Lee At 08:53 PM 1/14/2011 +0000, you wrote: >I liked your Introduction fig.15 --- was it hard to construct? WFL > >On 1/14/11, Lee Sallows <Lee.Sal(a)inter.nl.net> wrote: >> Hi Folks, >> >> Bored by magic squares? Me too. But a glance at >> http://www.GeomagicSquares.com/ will change all that for ever. Money back >> guarantee. >> >> Happy 2011 to all, >> >> Lee Sallows >> >> >> HURRY! HURRY! HURRY! SECONDHAND SATELLITES AT EXORBITANT PRICES! >> >> >> >> >> >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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[math-fun] geometric magic squares
by Lee Sallows 14 Jan '11

14 Jan '11

Hi Folks, Bored by magic squares? Me too. But a glance at http://www.GeomagicSquares.com/ will change all that for ever. Money back guarantee. Happy 2011 to all, Lee Sallows HURRY! HURRY! HURRY! SECONDHAND SATELLITES AT EXORBITANT PRICES!

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[math-fun] The mod 12 symptom?
by Dan Asimov 13 Jan '11

13 Jan '11

<< . . . including one occasion when he insisted that the number 6 was actually the number 18 . . . --Dan Those who sleep faster get more rest.

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[math-fun] FW: A conjectural series for 1/pi of a new type
by Schroeppel, Richard 13 Jan '11

13 Jan '11

I'll guess that some here would dispute that this series qualifies as "converges rapidly", but it seems worth passing along the challenge (to prove the conjecture.) Rich -----Original Message----- From: Number Theory List [mailto:NMBRTHRY@LISTSERV.NODAK.EDU] On Behalf Of Zhi-Wei SUN Sent: Thursday, January 13, 2011 8:35 AM To: NMBRTHRY(a)LISTSERV.NODAK.EDU Subject: A conjectural series for 1/pi of a new type Dear number theorists, On Jan. 2, 2011 I found a new type series for 1/pi. CONJECTURE (Zhi-Wei Sun). We have Sum_{k=0,1,2,...}(30k+7)binom(2k,k)^2*a_k/(-256)^k = 24/pi, where a_k=T_k(1,16) is the coefficient of x^k in (x^2+x+16)^k. I have included this conjecture in an article of mine (to be expanded later) available from http://arxiv.org/abs/1101.0600 Since a_k=T_k(1,16) ~ 0.75*9^k/sqrt(k*pi) as k tends to the infinity, we have binom(2k,k)^2*a_k/(-256)^k ~ 0.75(-9/16)^k/(k*pi)^{1.5} and hence the series in the conjecture converges rapidly. I have contacted several famous experts at pi-series or modular forms, they have never seen such a series for 1/pi before and none of them could prove the conjecture. It seems that all known methods used to prove Ramanujan-type series for 1/pi (including the current theory of modular functions and the WZ method) do not work for this curious series. Such a new series for 1/pi should be very rare! I consider the conjecture particularly difficult and very challenging. It might appeal for a powerful tool or a new theory. Any comments are welcome! Zhi-Wei Sun http://math.nju.edu.cn/~zwsun

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[math-fun] Jungle gyms and mathematics
by Dan Asimov 13 Jan '11

13 Jan '11

So it turns out that jungle gyms, patented in 1920 by an Illinois lawyer, were intended to help children understand Euclidean 3-space in the hope this would make it easier to eventually understand 4-space! Who knew? According to another site: << Math on the Playground ---------------------- [Sebastian] Hinton had in mind a game for children on the jungle gym. The jungle gym was a three-dimensional space. Hinton's game made physical the Cartesian coordinates that name points in space on three axes, x, y, and z. When [an ordered triple of numbers] was called out, the children would climb on the bars to find the named location. Seems like an odd game for a lawyer to invent, doesn't it? Not if you consider Hinton's ancestors, who included many mathematicians who contributed to higher-level algebra, including George Boole, remembered best for inventing Boolean algebra. >> --Dan Those who sleep faster get more rest.

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[math-fun] Google & Sudoku; challenge for sequence recognition
by Henry Baker 11 Jan '11

11 Jan '11

FYI -- Perhaps it's time for a sequence recognition app ? Any Google folks out there? Point your cellphone camera at a sequence (or number) & Google will tell you what the sequence or number is... http://www.telegraph.co.uk/technology/google/8251003/Google-solves-Sudoku.h… Google solves Sudoku A new version of Google's image recongition app, Goggles, recognises barcodes and adverts - and solves Sudoku Google's new version of its Goggles image recognition app can solve Sudoku puzzles instantly By Matt Warman, Consumer Technology Editor 6:30PM GMT 10 Jan 2011 The days of Sudoku could be numbered, thanks to Google. The puzzle that took the world by storm with its simple grid of digits and empty boxes can now be solved instantly by any mobile phone Â no matter what the level of difficulty. A newly updated mobile phone application, called Google Goggles, uses a phoneÂs camera to capture a picture of any Sudoku puzzle and then sends it back to Google. The companyÂs servers compute the answers and send an image of the completed puzzle back to the user. Depending on the speed of the phoneÂs internet connection, the results can be back in seconds. Phil McNeill, the Daily TelegraphÂs Puzzles Editor said the move was Âa sad dayÂ for puzzle fans. ÂWhen you finally solve a real mind-bender, it brings a rush of achievement. I hope Google's new tool won't diminish that feeling,Â he said. Google Goggles works by capturing images and sending them to the companyÂs central servers. They are then compared to the firmÂs vast web index, just as if they were a search term entered as text. The new version of Goggles means that, along with landmarks and other well known images, the software now also recognises adverts, barcodes and Sudoku. The applicationÂs main function, however, is for price comparison information and general web searching. Goggles is available on the iPhone and all Google Android phones.

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Re: [math-fun] Squares and factorials
by Phil Carmody 10 Jan '11

10 Jan '11

From: Dan Asimov > But I wanted to chime in that 24 is my choice for > most interesting integer, too. > > Partly because, among all the multiplicative groups of > rings > > G_n := (Z/nZ)*, > > 24 is the largest n for which k | n implies that > all elements of G_k are squares. In G_3, only 1 is a square, and 2 isn't. In fact, for any k | 24, G_k *only* has 1 as a square. Phil

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Re: [math-fun] Squares and factorials
by Dan Asimov 10 Jan '11

10 Jan '11

Adam Goucher wrote: << And [24 is] the largest n such that x² = 1 in Z_n, for all x coprime to n. (Followed by a lovely list of many more cheerful facts about the number 24.) Oy! Just got back.* Within 60 seconds of posting that erroneous nonsense about the elements of (Z/nZ)* being squares, I corrected it to the above, or more exactly to the below. But I was so rushed that I e-mailed it only to myself. Duh. --Dan _______________________________________________________________ * From studying Conway et al.'s book "The Symmetries of Things" with a friend. Just curious: Anyone else here have that book? << From: Dan Asimov <dasimov(a)earthlink.net> To: Dan Asimov <dasimov(a)earthlink.net> Subject: Re: [math-fun] Squares and factorials Date: Jan 9, 2011 11:03 AM << Partly because, among all the multiplicative groups of rings G_n := (Z/nZ)*, 24 is the largest n for which k | n implies that all elements of G_k are squares. >> I'm too rushed right now but I meant that all elements of G_k are square roots of 1. --Dan >> Those who sleep faster get more rest.

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Re: [math-fun] Exact rational 3D spacefill
by Bill Gosper 09 Jan '11

09 Jan '11

WFL>Please define your notation here: eg. does "8^6" mean edge 2^8 in 6-D space? Oops, sorry, strictly 3D. Each cube is essentially 8 of the next lower order. Several earlier such constructions of this nature had poor locality, compared to the classic Hilbert walk in 2-space. Have you looked at how well these curves perform in this respect? WFL In all the recursions I understand, consecutive eighths of the unit interval map to "consecutive" octants. Empirically, this seems to hold for Spirilbert as well. Re-ping me if I misunderstood your notion of locality. --rwg

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