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[math-fun] Weak Skolem-Langford integers
by Eric Angelini 16 Aug '07

16 Aug '07

Hello Math-Fun & SeqFans, Could someone please compute for me all "weak Skolem-Langford" integers ? The sequence is finite (and ends, I guess, with 978416154798652002). http://www.research.att.com/~njas/sequences/A108116 Please find hereunder what I have found for integers of length 4, 6 and 8. Length 4: 2002 Length 6: 131003, 231213, 300131, 312132,420024 Length 8: 12132003, 14130043, 15120025, 23121300, 23421314, 25121005, 25320035, 30023121, 31213200, 31413004, 34003141, 40031413, 41312432, 45001415, 45121425, 45300435, 50012152, 51410054, 52002151, 52412154, 53002352, 53400354, 61310036, 62312136, 63001316, 63121326, 64200246. Lengths 10 to 18? Best, Éric A.

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Re: [math-fun] Tech question
by Dan Asimov 14 Aug '07

14 Aug '07

David writes: << Can someone give me the a few terms of the power series of the inverse of exp(x)-x at 0? Exact coefficients would be preferred if rational. >> The function f(x) := exp(x) - x has f'(0) = 0, so finv := f^(-1) does not possess a power series around 0 in the usual sense. There may be some ambiguity in the question. f(0) = 1, so is the desired series for the inverse function where its argument = 1 ??? Otoh, f(z) = 0 has a set of countably many solutions, symmetric about the x-axis, but with none of them *on* the axis. At any of these fixed points of exp(z), the derivative is nonzero and so the inverse function exists locally. Just as exp(z) has multiple branches of its inverse function, it's likely that exp(z) - z does the same. --Dan

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[math-fun] Tech question
by David Wilson 14 Aug '07

14 Aug '07

Can someone give me the a few terms of the power series of the inverse of exp(x)-x at 0? Exact coefficients would be preferred if rational.

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[math-fun] Stirling number p-adic valuation
by Fred lunnon 13 Aug '07

13 Aug '07

I mentioned earlier (see Letter from Grozny) that I'd been looking at \ord_p s(m, n), the maximum power of p dividing a Stirling number of the first kind; after much head-scratching, I've come up with the following rather pretty lower bound. Define a "natural ceiling" function by \nace(x) = \max(0, \ceil(x)), where \ceil(x) denotes the usual ceiling function. Theorem: for 1 <= m <= n, \ord_p s(m, n) >= \sum_{j > 0} \nace(m/p^j - n) . The proof is fairly straightforward, via induction on m; the summation terms are nonzero only for j <= [(\log(m) - \log(n))/\log(p)]. Though only bounding the order below, this expression has the advantage of monotonicity in both m and n; meaning it can be used for example to bound the remainder of the infinite p-adic sum for (thankyou, Don) "negaBell" numbers B(-n) = \sum_m |s(m, n)| . Furthermore, it often does give the exact exponent: for example when n = m (zero), or n = 1 (Legendre's (m-1)! result in a rather different form); indeed apparently whenever for some k, (n-1)p^k < m <= n p^k. Such a simple result must surely be known already: yet any references to s(m, n) divisibility I have so far encountered have been restrictive to the point of feebleness. Fred Lunnon

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[math-fun] Re: Death Valley marathon
by James Propp 11 Aug '07

11 Aug '07

Thanks to everyone who took the time to explain in more detail why the 7:1 split is right for both the Death Valley marathon version and the office building version. I wrote: >(I have a feeling that there's an >Arrow-like theorem lurking here that says that certain seemingly >compelling characteristics that fairness should have are actually >incompatible.) I totally retract that. There is in fact a complete first-order theory of this kind of fairness; it's called basic accounting, and the corresponding algorithm is called double- (or multiple-) entry bookkeeping. :-) Jim

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[math-fun] Death Valley marathon
by James Propp 10 Aug '07

10 Aug '07

>Either way, this problem is suitable for a kindergarten entrance exam. With arithmetic like 5 minus 8/3 = 7/3? You've got to be kidding! If instead of $8 to be divided there were $24 to be divided, so that we could avoid fractions, this problem might be suitable for discussion in grade school. But I think there are economic/ethical issues of what constitutes fairness, in addition to purely mathematical issues. Without a clearer discussion of this, I don't think the puzzle has a single "right" answer. (Though I suspect that the 7-to-1 split is preferable. Maybe consideration of related scenarios with more extreme numbers could make this clearer.) If I ran an elite grade school and wanted to use this problem or some variant as the basis for admission, I'd change the story so that the third party pays $24 instead of $8, so that the use of fractions could be avoided. I'd ask the students how they thought the money should be divided, and why. If their answer involved a $15 and $9 split, I'd explain the reasoning behind a $21 and $3 split; and if their answer involved a $21 and $3 split, I'd explain the reasoning behind a $15 and $9 split. The kids I'd most want to admit to the school would be those who could see that both answers make a kind of sense, and who find this cognitive dissonance both frustrating and enjoyable. ("Hey, that's weird; hey, this is fun!") Speaking of which: Anyone out there care to help me resolve my own cognitive dissonance? Because I'm still unsure which way of splitting the money should be called "fair"! (I have a feeling that there's an Arrow-like theorem lurking here that says that certain seemingly compelling characteristics that fairness should have are actually incompatible.) Jim

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[math-fun] Re: Death Valley marathon
by James Propp 10 Aug '07

10 Aug '07

Joshua Zucker writes: >I don't see the 15:9 split making any kind of sense, since the people >drank the original 8 units of water each too, so they GAVE 7:1 units >of water. What is the sense behind it? It seems to me to be the >solution you get if you just don't think through the problem >completely enough. Let's change the scenario: I invest $5,000,000 in a three-story office building and you invest $3,000,000. There are three tenants: Propp Enterprises, Zucker Enterprises, and Yoyodyne Incorporated, each of which occupies one floor. If the rent we charge Yoyodyne is $8,000 per month, how should you and I split it? Would you really settle for just $1,000 per month, even though you put up 3/8 of the money for the building? If this is different from the Death Valley marathon sccenario, what makes it different? (Maybe I'm being dense...) Jim

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[math-fun] negafibonacci numbers and the hyperbolic plane
by James Propp 10 Aug '07

10 Aug '07

Did anyone go to Knuth's talk at MathFest this past weekend? It sounded interesting, but I wasn't able to go and couldn't find a write-up on the web; I'm hoping one of you can summarize. Here's what's on the web: PI MU EPSILON J. SUTHERLAND FRAME LECTURE NEGAFIBONACCI NUMBERS AND THE HYPERBOLIC PLANE Donald E. Knuth, Stanford University, Professor Emeritus of the Art of Computer Programming Saturday, August 4, 8:00 pm - 8:50 pm All integers can be represented uniquely as a sum of zero or more "negative" Fibonacci numbers F-1 = 1, F-2 = -1, F-3 = 2, F-4 = -3, provided that no two consecutive elements of this infinite sequence are used. The NegaFibonacci representation leads to an interesting coordinate system for a classic infinite tiling of the hyperbolic plane by triangles, where each triangle has one 90 degree angle, one 45 degree angle, and one 36 degree angle. Jim Propp

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[math-fun] Invitation to connect on LinkedIn
by Marc LeBrun 09 Aug '07

09 Aug '07

Math, Cool! I'll see your Doostang and raise you a LinkedIn invite! Best regards, --Marc LeBrun Join Marc LeBrun's network: https://www.linkedin.com/e/isd/121323052/SaG5uDRP/ ------------------------------------------ "Increasingly, if you're not LinkedIn, you're left out." --Business 2.0 ------ (c) 2007, LinkedIn Corporation

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[math-fun] social networking invitations banned from Math-Fun
by Schroeppel, Richard 09 Aug '07

09 Aug '07

I'm assuming that Andy Latto and Marc LeBrun have been the victims of some social networking virus. I've marked their posts for moderation until this mischief passes. Rich

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