February 2004 - mathfun@mailman.xmission.com

[math-fun] Five-field Kono...
by Guy Haworth 19 Feb '04

19 Feb '04

Marc Bourzutschky reports that: "Even counting stalemates as wins, Five Field Kono is rather drawish. More than 95% of the possible positions are draws, and the longest winning sequence is only 12 moves. The a-e symmetry can be applied to both the "even" and "odd" boards separately, reducing the number of combinations by a factor of four. I wrote a simple program with two bitvectors for win/loss/draw only, without taking into account the a-e symmetry. Even with more than 1 billion positions this only took a few minutes on a 2Ghz machine." Guy

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[math-fun] Conway's 5^3 blockpacking puzzle
by R. William Gosper 18 Feb '04

18 Feb '04

still works with the 2^3 replaced by a (fourteenth) 1x2x4. Which version is harder? --rwg

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[math-fun] another limit question
by Jon Perry 17 Feb '04

17 Feb '04

Everyone knows that the n-th triangular number squared is the sum of the first n cubes, but do you know that the n-th triangular number cubed divided by the sum of the first n fifth powers tends to 0.75? And also t(n)^4/sum of seventh powers tends to 0.5. And t(n)^5/sum of ninth powers tends to 5/16... t^6/sum_11 tends to 3/16, the next are 7/64 and 1/16, what is the general limit for t^k/sum_{2k-1}? Jon Perry perry(a)globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com

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[math-fun] P'r'aps not fun, but curious
by Richard Guy 15 Feb '04

15 Feb '04

In CMP 2003 #17 Dec19 2003 under 11D I find a paper by Liss, G'eza. First query: the accent is an apostrophe between the G and the e and surmounted by ' (acute accent). My Hungarian colleague isn't in this morning, so I can't get any further explanation. But the amazing thing is: when I search MathSciNet with Liss I get 3 hits: a William J. Liss, an Elizabeth D. Liss and the following, which agrees with the CMP entry in all but the K. MR1995985 Kiss, G. On the extremal Frobenius problem in a new aspect. Ann. Univ. Sci. Budapest. EÃ¶tvÃ¶s Sect. Math. 45 (2002), 139--142 (2003). 11D04 (11B13 11D85) References: 0 Reference Citations: 0 Review Citations: 0 {A review for this item is in process.} What's going on? Incidentally, the family name of my beautiful and talented neurosurgeon is Kiss. R.

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[math-fun] Re: Partial-Sum -> Some Primes
by Chris Landauer 14 Feb '04

14 Feb '04

hihi, all - i get different values from the original after a while (though my results agree with Fred Helenius' as far as he included them), but the original question appears to be intact (not just that the a[m] are all different, but whether they form a permutation of the positive integers) On Fri, 13 Feb 2004, Leroy Quet wrote: > Let a(1) = 1; > > Let a(m) be the lowest yet unpicked positive integer such that: > > sum{k=1 to m} k* a(k) > > is a prime. a[1] = 1, sum = 1 a[2] = 2, sum = 5 a[3] = 4, sum = 17 a[4] = 3, sum = 29 a[5] = 6, sum = 59 a[6] = 5, sum = 89 a[7] = 12, sum = 173 a[8] = 7, sum = 229 a[9] = 16, sum = 373 ****different from here on**** more soon, cal Chris Landauer Aerospace Integration Science Center The Aerospace Corporation cal(a)aero.org

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Re: [math-fun] Re: Partial-Sum -> Some Primes
by Fred W. Helenius 14 Feb '04

14 Feb '04

At 06:25 PM 2/13/04, Richard Guy wrote: >Also by hand, so please check > >1 2 4 3 6 5 12 7 20 9 8 21 24 10 16 25 18 ... > >the sequence of primes may be of interest > >(1) 5 17 29 59 89 173 229 409 499 587 839 >1151 1291 1531 1931 2237 ... R. That 20 should be a 16 (corresponding to the prime 373). Here are the first 100 terms of both sequences: 1, 2, 4, 3, 6, 5, 12, 7, 16, 9, 14, 13, 8, 11, 10, 17, 18, 15, 22, 24, 20, 23, 26, 25, 30, 28, 32, 33, 38, 29, 48, 27, 34, 44, 36, 19, 42, 47, 52, 35, 50, 21, 56, 53, 46, 39, 60, 45, 54, 31, 40, 37, 58, 49, 70, 55, 68, 61, 72, 43, 98, 41, 74, 57, 84, 59, 78, 64, 82, 63, 108, 51, 88, 76, 90, 75, 66, 79, 102, 69, 94, 67, 114, 73, 92, 80, 100, 62, 86, 87, 128, 77, 116, 65, 104, 71, 120, 97, 110, 81. 1, 5, 17, 29, 59, 89, 173, 229, 373, 463, 617, 773, 877, 1031, 1181, 1453, 1759, 2029, 2447, 2927, 3347, 3853, 4451, 5051, 5801, 6529, 7393, 8317, 9419, 10289, 11777, 12641, 13763, 15259, 16519, 17203, 18757, 20543, 22571, 23971, 26021, 26903, 29311, 31643, 33713, 35507, 38327, 40487, 43133, 44683, 46723, 48647, 51721, 54367, 58217, 61297, 65173, 68711, 72959, 75539, 81517, 84059, 88721, 92369, 97829, 101723, 106949, 111301, 116959, 121369, 129037, 132709, 139133, 144757, 151507, 157207, 162289, 168451, 176509, 182029, 189643, 195137, 204599, 210731, 218551, 225431, 234131, 239587, 247241, 255071, 266719, 273803, 284591, 290701, 300581, 307397, 319037, 328543, 339433, 347533. Every positive integer up to 82 appears in the first 100 terms; everything up to 922 appears in the first 1000. Note that, apart from a(1), odd-index terms must be even; even-index terms are occasionally even as well, so the odd values tend to arrive late. >On Fri, 13 Feb 2004, Leroy Quet wrote: > >> Let a(1) = 1; >> >> Let a(m) be the lowest yet unpicked positive integer such that: >> >> sum{k=1 to m} k* a(k) >> >> is a prime. -- Fred W. Helenius <fredh(a)ix.netcom.com>

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[math-fun] Re: Partial-Sum -> Some Primes
by Richard Guy 13 Feb '04

13 Feb '04

Also by hand, so please check 1 2 4 3 6 5 12 7 20 9 8 21 24 10 16 25 18 ... the sequence of primes may be of interest (1) 5 17 29 59 89 173 229 409 499 587 839 1151 1291 1531 1931 2237 ... R. On Fri, 13 Feb 2004, Leroy Quet wrote: > Let a(1) = 1; > > Let a(m) be the lowest yet unpicked positive integer such that: > > sum{k=1 to m} k* a(k) > > is a prime. > > Figured by hand, so very well could be wrong, I get: > > a(k) : 1, 2, 4, 3, 6, 5, 12, 7,... > > Question: > Is this a permutation of the positive integers? > > If so, its inverse-permutation is: > > 1, 2, 4, 3, 6, 5, 8,... > > (I get some EIS matches for the few terms I give for the inverse > permutation, but no matches for the primary sequence.) > > thanks, >

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[math-fun] RE: Five Field Kono
by Guy Haworth 13 Feb '04

13 Feb '04

.. and, trivially, there is a left-right symmetry to reduce the number of positions to 536,870,912 < 2^30. This is well within the range of existing endgame-table generators and doesn't even require 64-bit integers. I've forwarded the challenge for, at least, the brute force treatment. Guy

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Re: [math-fun] First player win?
by Richard Tucker 13 Feb '04

13 Feb '04

Richard Guy wrote: >Five Field Kono, {\it Symmetry plus}, {\bf22}(Autumn 2003) 16. Have I missed something, or can't either player force a draw just by leaving one of their pieces in their own starting area? I admit it wouldn't be very sporting style of play. ... or maybe to avoid this problem, moves _cannot_ be backwards? Which also raises the question of what happens if you have no move. Richard/

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[math-fun] police badge tiling problem
by Thane Plambeck 13 Feb '04

13 Feb '04

The cement squares in front of the San Francisco Sheriff's department have a motif on them that's meant to suggest a police badge, but it's less than satisfactory because the tiles produce eight rather than seven points where they meet up: http://plambeck.org/archives/2004_02.html#000065 So, what might be a better scheme for this? Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.plambeck.org

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