January 2014 - mathfun@mailman.xmission.com

[math-fun] Largest n-fold tri-triangular numbers
by Allan Wechsler 15 Jan '14

15 Jan '14

For today's foray into Waring theory ... Consider, please, OEIS sequence A002636, the number of ways of expressing n as the sum of 3 triangular numbers. (For the present purposes, 0 counts as a triangular number.) By inspecting the list, I conjecture that 53 is the largest number to have only one tri-triangular representation: 53 = 28 + 15 + 10. Is this very very hard to prove? The largest number I could find with only two tri-triangular representations is 194. The entries (53,194) are enough to see easily that this sequence -- if it is well-defined -- is not in OEIS. I conjecture that for every n in N1, there is a largest k such that A002636(k) = n. How hard is *this* to prove? It looks like when n = 3, k = 470; when n = 4, k = 788; and when n = 5, k = 1730.

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[math-fun] Bit Reversal Puzzle
by Tom Rokicki 15 Jan '14

15 Jan '14

For an unsigned 64-bit integer, define br(x) to be the "bit reversal" of x: bit positions 0 and 63 are swapped, 1 and 62 are swapped, and so on. What's the fastest way in generic C code to compute br(a) < br(b) [Don't spoil this if you've seen it already, please.] -tom -- -- http://cube20.org/ -- http://golly.sf.net/ --

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[math-fun] High Moon (sic)
by Bill Gosper 15 Jan '14

15 Jan '14

Here (SF peninsula) and now, the (nearly full) Moon is much nearer zenith than the Sun ever gets at this latitude. What is the highest latitude at which the Moon can reach zenith? Same answer for full Moon? --rwg

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Re: [math-fun] We can't have mental gymnastics without a floor exercise.
by Bill Gosper 14 Jan '14

14 Jan '14

Gareth> On 14/01/2014 14:02, Bill Gosper wrote: > Not bad. But these four terms simplify to Julian's (very > unhomely) two term solution privately reported at 23:24 PST. > > I'm a little surprised FullSimplify fails this. It's obvious how to lose one term (move the constant 1 inside one of the floors); how do you get it down to 2? -- g Sum[Floor[y + k/n], {k, 0, n - 1}] == Floor[n*y] (Mma (FunctionExpand) developers: N.B.) --rwg

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Re: [math-fun] We can't have mental gymnastics without a floor exercise.
by Bill Gosper 14 Jan '14

14 Jan '14

Gareth> Half-integers are rounded to the nearest even integer. Exercise: Write Round using only Floor, +, -, *, /, ... SPOILER SPACE ... [...] ... SPOILER SPACE ...[ 0. I assume we are allowed constants too. 1. So we can (very familiarly) almost do it with floor(x+1/2). This rounds up all half-integers, whereas we need to round half of them down instead. 2. So it's wrong by the sum of an infinite stream of delta functions. 3. We can get an infinite stream of deltas, but not quite the right ones, by considering floor(x)+floor(-x) which is -1 except at integers where it's 0. Call this G(x). 4. So -1-G(x) is 0 except at integers where it's -1. 5. We need something that's mostly 0 except at 1/2, 5/2, ... so the correction we need is -1-G(x/2-1/4). Putting it together, we get floor(x+1/2) - floor(x/2-1/4) - floor(1/4-x/2) - 1. Two minutes of Python confirms this (well, actually two minutes of Python originally showed that I'd made a sign error, but I fixed it up). -- g Not bad. But these four terms simplify to Julian's (very unhomely) two term solution privately reported at 23:24 PST. I'm a little surprised FullSimplify fails this. --rwg

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Re: [math-fun] We can't have mental gymnastics without a floor exercise.
by Bill Gosper 14 Jan '14

14 Jan '14

WFL> On 1/14/14, Bill Gosper <billgosper(a)gmail.com> wrote: On 2013-12-06, at 9:36 AM, Adam P. Goucher wrote: Half-integers are rounded to the nearest even integer. Exercise: Write Round using only Floor, +, -, *, /, and of course, \[NotHumpDownHump]. --rwg (And please stay off the nonEuclidean parallel bars.) And what the San-Fairy-Ann does a \[NotHumpDownHump] look like --- assuming it can be persuaded to remain stationary for a period extended sufficiently to permit visual identification? <WFL Two squat Ωs humping belly-to-belly, under a slash to be more salacious. I can't find a UTF8(!), but unslashed is ≎. It's code 53 in http://www.dessci.com/en/support/mathtype/tech/encodings/mathmat3.htm "not geometrically equivalent". --rwg WFL> SPOILER . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x -> 2[(2 x + 1)/4] WFL

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Re: [math-fun] We can't have mental gymnastics without a floor exercise.
by Bill Gosper 14 Jan '14

14 Jan '14

WFL> On 1/14/14, Bill Gosper <billgosper(a)gmail.com> wrote: On 2013-12-06, at 9:36 AM, Adam P. Goucher wrote: Half-integers are rounded to the nearest even integer. Exercise: Write Round using only Floor, +, -, *, /, and of course, \[NotHumpDownHump]. --rwg (And please stay off the nonEuclidean parallel bars.) And what the San-Fairy-Ann does a \[NotHumpDownHump] look like --- assuming it can be persuaded to remain stationary for a period extended sufficiently to permit visual identification? <WFL Two squat Ωs humping belly-to-belly, under a slash to be more salacious. I can't find a UTF8(!), but unslashed is ≎. It's code 53 in http://www.dessci.com/en/support/mathtype/tech/encodings/mathmat3.htm "not geometrically equivalent". Non WFL>SPOILER . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . x -> 2[(2 x + 1)/4] < WFL No, Round[1] ≠ 0. So far, Julian has sent the only correct solution, albeit a bit homely. --rwg

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[math-fun] Dual palindromes
by Keith F. Lynch 13 Jan '14

13 Jan '14

My program for finding dual binary-ternary palindromes rolled over to 105 bits yesterday as expected, then immediately jumped to 67 trits. I terminated it. So the current ceiling is 3^66, which is about 2^104.6, 10^31.5, or e^72.5. It's about 800 million times higher than Giovanni Resta's ceiling of 2^75, and 4 times higher than the dual palindrome I discovered last week. In other words, a 10th dual binary-ternary palindrome, if it exists -- and I'm sure it does -- would be larger than 3^66. A correction to my previous message: Where I said "about 1.8 (3^10/2^9)" the 9 should be 15. In the same message I said that I suspected that the statistics on dual binary *balanced* ternary palindromes would be the same as on dual binary ternary palindromes. I gain a factor of two in one place (the middle digit can be plus or minus 1), but lose it in another (the leading digit must be +1; for standard ternary it can be 1 or 2). The actual palindromes would be different (though I had noticed that 1 and 6643 are palindromes in balanced ternary as well as in binary and standard ternary), but I figured that the long range probabilities ought to be exactly the same. But I've taken my first look at binary balanced ternary palindromes, and it looks like they're much more common. Does anyone have any idea how this can be? What am I overlooking? The first few are 1, 7, 73, 511, 5461, 6211, 6643, 306601, 432331, 471655, 6815731, 7036267, 21583141. (Not in OEIS.) I was feeling lazy, so to "calculate" these I wrote a one-line "program": cut -f 2 -d\ b006995.txt b134027.txt | sort -n | uniq -c | grep \ 2\ | cut -c 6-99 | sort -n (This made use of two files downloaded from OEIS, of course.)

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[math-fun] We can't have mental gymnastics without a floor exercise.
by Bill Gosper 13 Jan '14

13 Jan '14

On 2013-12-06, at 9:36 AM, Adam P. Goucher wrote: Half-integers are rounded to the nearest even integer. Exercise: Write Round using only Floor, +, -, *, /, and of course, \[NotHumpDownHump]. --rwg (And please stay off the nonEuclidean parallel bars.)

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[math-fun] Navier Stokes Clay $1M prize problem solved?
by Warren D Smith 12 Jan '14

12 Jan '14

http://bnews.kz/en/news/post/180213/ this if true is a major breakthrough. I have often thought this problem might be within reach. It says: Full Article of Muhtarbay Otelbaev was published in "Mathematical Journal" 13,4 (2013) 50 http://www.math.kz/index.php/ru/513 leading to http://www.math.kz/images/journal/2013-4/Otelbaev_N-S_21_12_2013.pdf pages 5-104. Only the final page is in English, rest Russian.

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