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Fwd: [math-fun] A Mighty Number Falls: 2^1039 - 1
by Henry Baker 24 May '07

24 May '07

2^1039-1= 1159420574 0725730643698071 48876894640753899791 70201772498686835353882248385 9966756608 0006095408005179 47205399326123020487 44028604353028619141014409345 3512334712 7396798885022630 75752809379166028555 10550042581077117617761009413 7970787973 8061870084377771 86828680889844712822 00293520180607475545154137071 1023817 factors: 5585366661 9936291260 7492046583 1594496864 6527018488 6376480100 5234631985 3288374753 x 2075818194 6442382764 5704813703 5946951629 3970800739 5209881208 3870379272 9090324679 3823431438 8414483488 2534053344 7691122230 2815832769 6525376091 4101891052 4199389933 4109711624 3589620659 7216748116 1749004803 6597355734 0925320542 5523689 x 5080711 Checked by GNU Common Lisp >Date: Tue, 22 May 2007 23:03:04 -0700 >To: math-fun(a)mailman.xmission.com >From: Ray Tayek <rtayek(a)ca.rr.com> >Subject: [math-fun] A Mighty Number Falls: 2^1039 - 1 > >http://science.slashdot.org/science/07/05/22/1815208.shtml > >--- >vice-chair http://ocjug.org/

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[math-fun] A Mighty Number Falls: 2^1039 - 1
by Ray Tayek 23 May '07

23 May '07

http://science.slashdot.org/science/07/05/22/1815208.shtml --- vice-chair http://ocjug.org/

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[math-fun] Sharing the digits
by Eric Angelini 22 May '07

22 May '07

Hello Math-Fun, Add two consecutive terms of S: the result doesn't share any digit with the first nor the second term. Is this sequence finite? Best, É. S = 1,2,3,4,5,6,7,8,9,11,13,14,15,17,18,19,21,22,23,25... (to build the sequence, add the next integer which doesn't infringe the rule)

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[math-fun] Stupid digital question
by David Wilson 17 May '07

17 May '07

Let b >= 2, n >= 0. Let f(n,b) be the smallest nonnegative base-b numeral whose digital product is n. The value of f(b,0) and f(b,1) depend on whether we use the standard base-b representation of 0 ("0") or the purist representation ("", the null string). In the form case f(b,0) = "0" and f(b,1) = "1" while in the latter f(b,0) = "10" and f(b,1) = "" (whose digital product is technically 1). Either way, f(b,0) and f(b,1) are easy to compute and prove. The values of f(b,n) for n >= 2 are unaffected by this choice of representation. My stupid question is about computing f(b,n). In a recent program, I used the following algorithm: // Return the smallest base-b numeral whose digital product is n, or "-" if no such numeral. // Assumes purist base-b representation for simplicity function f(integer n, integer b) { // Handle zero if (n == 0) return "10"; // Start with null numeral numeral = ""; // For each base-b digit starting at the largest down to 2 for (integer digitValue = b-1; digitValue >= 2; digitValue--) { // While this digit divides n while (n%digitValue == 0) { // Divide digit out of n n /= digitValue; // Prepend base-b digit to numeral numeral = digit(digitValue, b) || numeral; } } // If n has factors that are not base-b digits, it is not a digital product if (n > 1) return "-"; // Return the numeral we built return numeral; } In effect, I went through the base-b digits in reverse order, each time dividing n by the digit as many times as possible, and took the dividing digits in reverse order as the ostensible smallest number with digital product n. Modulo probable errors in this hand-written pseudocode, is this algorithm correct? Does it actually return the smallest base-b numeral with digital product n? If so, how do we prove this? If not, does it work for some bases and not for others? Which bases does it work for? Does it work for base 10, as suggested by empirical testing I have done?

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Re: [math-fun] Fourier transforms (fitting exponentials)
by Chris Landauer 17 May '07

17 May '07

hihi, all - i remember a paper from one of the siam journals (probly numerical methods or something) from many years ago whose title was something like how not to fit an exponential the author(s?) describe i think 20 ways that are plausible, but do not work, and explain what the ill-conditioning means and even why (or at least that) it happens a lot i've run into this fitting some laser data to the energy transfer equations between different levels of different rare earths (not my problem, my favorite physicist friend's problem; she does the experiments and gets LOTSA data) more later, cal

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[math-fun] Wolfram 2,3 Turing Machine Research Prize
by Eric Angelini 16 May '07

16 May '07

Universal Turing Machine ? http://www.wolframscience.com/prizes/tm23/ É.

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Re: [math-fun] Torus tiling question
by James Propp 15 May '07

15 May '07

>Can someone prove the converse: > >"If T can be tiled by N congruent squares, then N is the sum >of 2 square numbers." . . . > >. . . or come up with a counterexample ??? > >--Dan This is a really nice question; has anyone made progress with it? For instance, can someone show that you can't tile a torus with 3 congruent squares? If nobody reports any progress with Dan's question, then I'll mention it on the domino forum (see jamespropp.org/ about-domino); I'm sure someone there will have some ideas. Jim

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[math-fun] Mathematical ringtones for your cellphone
by Henry Baker 14 May '07

14 May '07

Mathematical ringtones for your phone: http://www.cs4fn.org/rtones/

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[math-fun] Mayan Calendar and the Chinese Remainder Theorem
by wouter meeussen 13 May '07

13 May '07

playing with the Mayan calendar, cfr http://hermetic.nofadz.com/cal_stud/maya/chap1.htm I noticed that some web pages allow conversion of the Gregorian date to the Mayan Tzolkin-Haab date, but never give the inverse. Calculation of this inverse needs the Chinese Remainder Theorem: given integers u, v and w, find n so that mod(n,20)=u, mod(n,13)=v, mod(n,365)=w (with the added complication that 20 and 365 share a common factor of 5). Would the Mayan priests have had a trick to solve this (in Roman times)? for Mathematica fans: http://users.pandora.be/Wouter.Meeussen/mayan_calendar.nb enjoy, Wouter.

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Re: [math-fun] Torus tiling question
by Michael Reid 13 May '07

13 May '07

> T = R^2/Z^2 > "If T can be tiled by N congruent squares, then N is the sum > of 2 square numbers." . . . i think this is relatively easy to prove: lift such a tiling to a tiling of R^2 . it is easy to show that any tiling of R^2 by congruent squares is invariant under some translation, with length equal to the side length of the squares. let (x, y) be this translation. since the tiling is lifted from T , it is also invariant under (1, 0) and (0, 1) translations. the group of translational invariants must be a discrete subgroup of R^2 , so this means that (x, y) is commensurate with the lattice generated by (1, 0) and (0, 1) . in other words, some (non-zero) multiple of (x, y) is in the lattice ; say M(x, y) is that multiple. now we may write x = a/M , y = b/M for some integers a, b, M . the side length of the square is 1 / sqrt(N) , because N of them tile T . thus we have (a^2 + b^2) N = M^2 . from this equation, we see that any prime = 3 mod 4 must divide N with even multiplicity. therefore N is a sum of two squares. qed mike

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