June 2007 - mathfun@mailman.xmission.com

[math-fun] special functions
by Steven Finch 08 Jun '07

08 Jun '07

Hello! Let psi denote the digamma function and gamma=-psi(1). Everyone knows that sum_{n=1}^infnty 1/(n*(1+n*m)) = psi(1 + 1/m) + gamma and that psi(1 + x) + gamma = sum_{k=2}^infnty (-1)^k * zeta(k) * x^(k-1) for x close to zero. What can be said about sum_{n=1}^infnty log(n)/(n*(1+n*m)) and what is the corresponding Maclaurin series? Thank you, Steve Finch http://algo.inria.fr/bsolve/ > > > > > > > _________________________________________________________________ Like puzzles? Play free games & earn great prizes. Play Clink now. http://club.live.com/clink.aspx?icid=clink_hotmailtextlink2

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RE: [math-fun] Modular notation
by Dan Asimov 07 Jun '07

07 Jun '07

Fred wrote: << At 03:43 PM 6/7/2007, Schroeppel, Richard wrote: < I don't know the history: I don't think the extended notations are used in Hardy & Wright, except maybe for (mod 1) to mean fractional part, or for stuff with algebraic integers. > Actually Hardy & Wright do use this notation, and in exactly the same context as the Monthly article that Dan cited. Theorem 116 says "If p>3, and 1/i is the associate of i (mod p^2), then 1 + 1/2 + 1/3 + ... + 1/(p-1) == 0 (mod p^2)." (By "associate" they mean "multiplicative inverse".) >> Wait a sec -- the quote from Hardy & Wright looks as though it has no immediate connection with harmonic numbers H_n. Rather, it seems to be saying that if, in the ring Z/(p^2), we add the multiplicative inverses* of 1,2,...,p-1, then we get 0. (Otherwise, what does "i" mean here?) Maybe there is some clever way to connect this with harmonic numbers. But my first impression is that Hardy & Wright are not using the "mod" notation to refer to anything but integers (or integers mod N). --Dan __________________________________________________________ * Of course for p prime, not all members of Z/(p^2) have multiplicative inverses, but 1,2,...,p-1 do.

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RE: [math-fun] Modular notation
by Fred W. Helenius 07 Jun '07

07 Jun '07

At 03:43 PM 6/7/2007, Schroeppel, Richard wrote: >I don't know the history: I don't think the extended notations are used in Hardy & >Wright, except maybe for (mod 1) to mean fractional part, or for stuff with algebraic >integers. Actually Hardy & Wright do use this notation, and in exactly the same context as the Monthly article that Dan cited. Theorem 116 says "If p>3, and 1/i is the associate of i (mod p^2), then 1 + 1/2 + 1/3 + ... + 1/(p-1) == 0 (mod p^2)." (By "associate" they mean "multiplicative inverse".) -- Fred W. Helenius fredh(a)ix.netcom.com

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[math-fun] Northeastern University Researchers Solve Rubik's Cube in 26 Moves
by Henry Baker 07 Jun '07

07 Jun '07

FYI -- http://www.neu.edu/nupr/news/0407/rubik.html Northeastern University Researchers Solve Rubik's Cube in 26 Moves Northeastern University News (06/01/07) Northeastern University computer science professor Gene Cooperman and graduate student Dan Kunkle have proven that any configuration of a Rubik's cube can be solved in 26 moves. The research from Cooperman and Kunkle improves upon the previous record by one move. The use of 7 TB of distributed disk as an extension to RAM for holding large tables and the development of faster computer moves using mathematical group theory were keys to their research efforts. Once all of the configurations of a Rubik's cube were placed in a family of sets of configurations, the researchers focused on applying a single move to the family of cosets at once. They used the new mathematical group theory technique to simulate the result on a computer at a rate of 100,000,000 times per second. Computers at Teragrid and Northeastern were used. "Search and enumeration is a large research area encompassing many researchers working in different disciplines--from artificial intelligence to operations," says Coo perman. "The Rubik's cube allows researchers from different disciplines to compare their methods on a single, well-known problem."

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[math-fun] Modular notation
by Dan Asimov 07 Jun '07

07 Jun '07

In an article in the June-July 2007 Monthly, it is stated that for any prime p >= 5, the harmonic number H_(p-1) == 0 (mod p^2) (where == means a triple equals sign). Huh??? I thought? H_n can't be an integer! Must be a typo. Until I noticed more of the same kinds of apparent errors in the rest of the article. For the first case, H_4 = 1 + 1/2 + 1/3 + 1/4 = 25/12. So apparently x == j (mod N) is defined for rational numbers x, and means that the numerator of x (in lowest terms) == j (mod N) in the old-fashioned sense. Is it common notation to apply modulo to rationals? Of so, when did it begin to be used? --Dan

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[math-fun] HAKMEM addenda
by R. William Gosper 05 Jun '07

05 Jun '07

An PDP-1 hack probably deliberately omitted (e.g., from http://inwap.com/pdp10/hbaker/hakmem/hacks.html) but perhaps worth preserving: It was an 18 bit, 4K (!) machine with a 6 bit character code (FIO-DEC), and most of the printouts were three characters. E.g., in the assembler, usw = undefined symbol in word mce = macro capacity exceeded, and in Samson's music compiler, mts = measure too short agm = argument greedily masticates To type out the three characters in the AC, lio asa /load IO rcl 6s /rotate combined left 6 bits tyo asa, sas asa /skip if AC same as . jmp .-3 This was safe because the 6-bit order code for sas (not indirect) was an illegal FIO-DEC character (which actually jammed the Soroban typewriter?). Also, a most regrettable typo ruins a spectacular mate in 3: http://www.inwap.com/pdp10/hbaker/hakmem/games.html ITEM 70: A neat chess problem, swiped from "Chess for Fun and Chess for Blood", by Edward Lasker: white: pawns at QN3 and KN7, knight at QN4, bishop at KB7, king at QB2; (NO, the B should be KB8!) black: pawn at QN3, king at QR6. White mates in three moves.

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[math-fun] implausible-looking denestings
by R. William Gosper 05 Jun '07

05 Jun '07

One I may have sent earlier, but here rewritten entirely as nth roots: (c106)block([radexpand:false],sqrt(2160^(1/6)+4)= (-sqrt(10)+2*10^(1/6)+sqrt(6)+((5^(1/3)*(sqrt(5)+sqrt(3)))/(2^(1/6))))/3) 1/6 (d106) sqrt(2160 + 4) = 1/3 1/6 5 (sqrt(5) + sqrt(3)) - sqrt(10) + 2 10 + sqrt(6) + ------------------------ 1/6 2 --------------------------------------------------------- 3 (c107) dfloat(d106) (d107) 2.75596793350447d0 = 2.75596793350447d0 Still no sign of binomial^(1/n) = six or more real terms (n>0). My 30-yr old version of Robert Maas's implementation of Rich's "number recognizer" incompletely reduces large lattices in a manner suggestive of a short-by-1 coding bug. A multiday search with Mma's LatticeReduce has turned up numerous small counterexamples to my denester, at least some of which worked subsequent to its Nov04 latest edit, so I must have bent an autoload someplace. One such poser: 1/3 1/3 5/6 5/6 1/6 1/6 sqrt(39 2 - 34 3 ) = 2 3 - 3 2 sqrt(3) + sqrt(2) 3 whose sides seem to have disproportionate coefficients. --rwg MUSTARD SPINACH PHANTASMIC SURD

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[math-fun] Texas Hold 'em
by Thane Plambeck 04 Jun '07

04 Jun '07

Two cards are dealt to each of n>1 players in a game of Texas Hold 'em. For which values of n is it possible that all the two-card hands have an equal a priori probability of winning, assuming that everyone will stay in the hand until all the 5 common cards are dealt? -- Thane Plambeck tplambeck(a)gmail.com http://www.plambeck.org/ehome.htm

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[math-fun] Shortest sequence = add two terms, share no digit
by Eric Angelini 04 Jun '07

04 Jun '07

Hello math-fun, this sequence of integers (computed by Olivier Miakinen) has those properties: - starts with "1"; - increases monotonically; - sum of two consecutive terms doesn't share any digit with the two sumands; - stops after 29 terms. 1 2 9 11 22 93 111 222 933 1111 2222 9333 11111 22222 93333 111111 222222 933333 1111111 2222222 9333333 11111111 22222222 93333333 111111111 331393339 339112349 368595421 409182356 ** END ** Question: What would be the shortest such sequence? Best, É.

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Re: [math-fun] Symmetrical finite subsets of the plane
by Dan Asimov 03 Jun '07

03 Jun '07

Erich writes: << [I wrote:] > But what if, for each n -- say 1 <= n <= 100, > we ask specifically for the arrangement(s) of > n points in the plane having the largest-size > symmetry group? why isn't the answer n equally spaced points on a circle? >> This is clearly the right answer, and yet another confirmation that I tend to post too hastily. Sorry about that. Thus S_2(n) = 2n There are still many non-obvious questions in this area. What's the "best" arrangement(s) for the square torus T := R^2/Z^2 ? E.g., on T for n = 5 points, that tiling of T by 5 congruent squares gives an arrangement of 5 points -- the tiling's vertices -- whose full symmetry group is of order 20. Is there an arrangement of 5 points in T with a larger symmetry group? I suspect not. --Dan

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