January 2006 - mathfun@mailman.xmission.com

[math-fun] Re: mathematical cons
by James Propp 22 Jan '06

22 Jan '06

37 minutes into the movie, Addie (Tatum O'Neal's character) does the same con on a cotton-candy vendor, and this time it's clear what's going on (which is basically what I inferred).

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Re: [math-fun] Re: mathematical cons
by Daniel Asimov 22 Jan '06

22 Jan '06

Here's an ancient classic I suspect everyone has heard before. It isn't *exactly* a scam per se -- it's a story about a scam. Then again, the story itself is a scam. << Three men go to stay at a hotel, and the man at the desk charges them $30.00 for a room. They split the cost ten dollars each. Later the manager tells the desk man that he overcharged the men, that the actual cost should have been $25. The manager gives the bellboy $5 and tells him to give it to the men. The bellboy, however, decides to cheat the men and pockets $2, giving each of the men only one dollar. Now each man has paid $9 to stay in the room and 3 x $9 = $27. The bellboy has pocketed $2. $27 + $2.00 = $29 -- so where is the missing $1 ? >> (I remember how much this confused me when I first encountered it around age 7 -- foreshadowing the Fields medal that I was never to receive, many years later.) --Dan

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Re: [math-fun] Re: even Q(n)
by Christian G.Bower 21 Jan '06

21 Jan '06

So Dean Hickerson's post settles the issue for A000009(n) == 2 (mod 4) I submitted the c(n) sequence as A115323 with a generous supply of comments lifted from Dean's post. I wonder if a similar result can be derived for A001935(n) == 2 (mod 4) I created a function similar to Dean's c(n), I'll call it e(n): Let e(n) be the number of solutions of 8*n+1 = x^2 + 8*y^2, x>0. Like c(n), e(n) appears to have the property that e(n) == 2 (mod 4) iff A001935(n) == 2 (mod 4) If 8*n+1 is not a square or if sqrt(8*n+1) == 1 or 7 (mod 8), then A001935(n) == e(n) (mod 4). If sqrt(8*n+1) == 3 or 5 (mod 8) then A001935(n) == e(n) + 2 (mod 4). but I don't know enough about quadratic equations to prove this or to show that all n such that 8*n+1 is prime have a value of 2 or to show how this function can be calculated from the prime factorization. Christian

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RE: [math-fun] mathematical cons
by Andy Latto 20 Jan '06

20 Jan '06

There's the "I have a system to win at craps! I'll teach you the system, you play under my guidance, and you just have to pay 25% of your winnings! And if you lose, you pay me *nothing*!" I think that this one, along with a couple of related ones (possibly done with horse racing instead of craps) are in one of Feynman's books. More generally, you can think of casino gambling as such a scam. I think that most people who play blackjack in a casino cannot answer the following question: "So you get to split and to double down, and the dealer doesn't; you get paid 3-2 on blackjack, and the dealer doesn't. You can use whatever strategy you like, after seeing one of the dealer's cards, and the dealer has to use a simple fixed strategy. Why does the dealer have an advantage?" Where are you teaching this class? > -----Original Message----- > From: math-fun-bounces+andy.latto=pobox.com(a)mailman.xmission.com > [mailto:math-fun-bounces+andy.latto=pobox.com@mailman.xmission.com]On > Behalf Of James Propp > Sent: Thursday, January 19, 2006 11:34 PM > To: math-fun(a)mailman.xmission.com > Subject: [math-fun] mathematical cons > > > I'm teaching a course on quantitative reasoning for an audience > of nearly two hundred math-averse students, and one of the ways > I'm hoping to "sell the product" to them is to pitch the course > as a kind of self-defense art that helps you not get ripped off > by used car dealers, cell-phone companies, credit-card companies, > banks, etc. > > For instance, I'll talk about the scam wherein the scammer sends > free investment advice to 1024 people (half of whom get one piece > of advice and half of whom get a conflicting piece of advice), > then sends free advice to the 512 who got good advice on the > first round (again splitting his advice half-and-half), then > sends free advice to 256 people, and so on, and finally starts > to ask a small number of people to pay a lot of money, hoping > that they'll reason that someone who's been right so often must > be onto something. > > I'll also show the class the grifting scene in "Paper Moon". > > Can any of you think of other good mathematical cons, or good > resources for finding out about them? > > ("Ricky Jay's Big Book of Math Hustles" would be perfect if > such a book existed!) > > Jim > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

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Re: [math-fun] mathematical cons
by Daniel Asimov 20 Jan '06

20 Jan '06

<< There's the famous "Truth or Consequences" problem. You have three doors. . . . Minor historical correction: This problem came from the TV game show called "Let's Make a Deal" and is usually known by the name of the show's emcee, Monty Hall. --dan

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Re: [math-fun] mathematical cons
by Daniel Asimov 20 Jan '06

20 Jan '06

Jim writes: << I'm teaching a course on quantitative reasoning for an audience of nearly two hundred math-averse students, and one of the ways I'm hoping to "sell the product" to them is to pitch the course as a kind of self-defense art that helps you not get ripped off by used car dealers, cell-phone companies, credit-card companies, banks, etc. . . . . . . . . . >> Hmmm. There's a classic book by Darrell Huff called "How to Lie With Statistics" -- and not being misled by statistics is an important form of mathematical self-defense. Chain letters also come to mind, the kind that can expand exponentially and make the originators quite a lot of money while leaving most of the the nth-generation folks $25 poorer. (Aside to Thane: Does this kind of chain letter qualify as a form of a Ponzi scheme? Does Amway?) --Dan

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[math-fun] Re: even Q(n)
by Dean Hickerson 19 Jan '06

19 Jan '06

Christian G. Bower wrote: > The 1's had not been tallied yet, so I entered them as A114913. > Interesting thing is that the sequence is very similar to A111174 > > A114913 Values such that A114912(a(n))=1. Values such that > A000009(a(n))==2 (mod 4). That last line should be: A000009(n)==2 (mod 4). > 3, 4, 8, 10, 13, 14, 17, 18, 19, 24, 25, 28, 32, 39, 42, 43, 47, 48, 50, 52, > 54, 55, 62, 67, 69, 73, 74, 75, 76, 78, 83, 84, 87, 88, 89, 90, 95, 99, 101, > 103, 105, 108, 109, 112, 113, 118, 119, 123, 125, 127, 130, 132, 134, 138, > 140, 143, 144, 147, 149, 153, 154, 157 > > A111174 Numbers n such that 24*n + 1 is prime. > > 3, 4, 8, 10, 13, 14, 17, 18, 19, 24, 25, 28, 32, 39, 42, 43, 47, 48, 50, 52, > 54, 55, 62, 67, 69, 73, 74, 75, 78, 83, 84, 87, 88, 89, 90, 95, 99, 103, 105, > 108, 109, 112, 113, 118, 119, 123, 125, 127, 130, 132, 134, 138, 140, 143, > 144, 147, 153, 154, 157, 158, 162 (list) > > matching in the first 28 terms. A little more searching and it appears > that every number in A111174 is in A114913, but there are probably > infinitely many numbers in the latter not in the former. Yes, for example if 24*n+1 = 25p for some prime p, then n will be in A114913 but not in A111174; see below. > I looked a little at Dean Hickerson's 8/4/2005 posting about q(n) mod 64. > > I didn't fully understand it, but it does not appear to have an obvious > connection to prime number theory. Here's what I wrote about Q(n) mod 64: > I looked at Q(n) mod small powers of 2 a few years ago. I found that the > value of Q(n) mod 64 is determined by representations of 24n+1 by the binary > quadratic form x^2 + 24 y^2. Specifically: > > Let f(x,y) be determined by the values of x mod 192 and y mod 2: > > x mod 192 1 5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 > f(x,even) 1 51 55 29 5 15 27 25 41 11 31 53 45 39 3 49 > f(x,odd) 7 5 1 11 3 9 29 15 31 13 25 19 27 17 21 23 > > x mod 192 49 53 55 59 61 65 67 71 73 77 79 83 85 89 91 95 > f(x,even) 17 35 7 13 21 63 43 9 57 59 47 37 61 23 19 33 > f(x,odd) 23 21 17 27 19 25 13 31 15 29 9 3 11 1 5 7 > > x mod 192 97 101 103 107 109 113 115 119 121 125 127 131 133 137 139 143 > f(x,even) 33 19 23 61 37 47 59 57 9 43 63 21 13 7 35 17 > f(x,odd) 7 5 1 11 3 9 29 15 31 13 25 19 27 17 21 23 > > x mod 192 145 149 151 155 157 161 163 167 169 173 175 179 181 185 187 191 > f(x,even) 49 3 39 45 53 31 11 41 25 27 15 5 29 55 51 1 > f(x,odd) 23 21 17 27 19 25 13 31 15 29 9 3 11 1 5 7 > > Let b(n) be the sum of f(x,y) over all solutions of > > 24n+1 = x^2 + 24 y^2, x>0. (0) > > Then > > Q(n) == b(n) (mod 64). (1) If we're only interested in Q(n) mod 4, we can simplify this to: Let c(n) be the number of solutions of (0). If 24n+1 is not a square or if sqrt(24n+1) == 1 or 11 (mod 12), then Q(n) == c(n) (mod 4). If sqrt(24n+1) == 5 or 7 (mod 12) then Q(n) == c(n) + 2 (mod 4). For example, if n=51, then the solutions of (0) are (x,y) = (7,#7), (19,#6), (25,#5), (29,#4), (35,0) (where "#" means plus or minus), so c(51)=9. Since sqrt(24n+1) = 35 == 11 (mod 12), we should have Q(51) == c(51) == 1 (mod 4). And in fact Q(51) = 4097 == 1 (mod 4). What follows is implied by the arithmetic of Q(sqrt(-6)); I'll skip the details. If 24n+1 is prime, then there's exactly one solution of (0), so Q(n) == 2 (mod 4). More generally, we can determine c(n) from the prime factorization of 24n+1: Let 24n+1 = p_1^e_1 * ... * p_r^e_r * q_1^f_1 * ... * q_s^f_s, where the p_i's are distinct primes == 1, 5, 7, or 11 (mod 24) and the q_i's are distinct primes == 13, 17, 19, or 23 (mod 24). If some f_i is odd, then c(n) = 0. Otherwise, c(n) = (e_1 + 1) * ... * (e_r + 1). So c(n) == 2 (mod 4) iff all of the f_i's are even and all but one of the e_i's are even and the one e_i which is odd is == 1 (mod 4). Since Q(n) and c(n) are both odd if 24n+1 is a square, we can replace c by Q in this: Q(n) == 2 (mod 4) iff all of the f_i's are even and all but one of the e_i's are even and the one e_i which is odd is == 1 (mod 4). In particular, if 24n+1 = 25p for some prime p, then c(n) == 2 (mod 4), so there are infinitely many n in A114913 but not in A111174. Dean Hickerson dean(a)math.ucdavis.edu

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Re: [math-fun] even Q(n)
by Christian G.Bower 19 Jan '06

19 Jan '06

On 8/2/2005 Bill Gosper wrote to math-fun > George Andrews kindly informs me that any fixed power of 2 > divides almost all Q(n) (usually written q(n), to confuse > you with the nome). This got me interested in tallying which power of 2 divides q(n) (known in the EIS as A000009) so I sent in A114912: A114912 2^a(n) divides A000009(n) but 2^(a(n)+1) does not. 0, 0, 0, 1, 1, 0, 2, 0, 1, 3, 1, 2, 0, 1, 1, 0, 5, 1, 1, 1, 6, 2, 0, 3, 1, 1, 0, 6, 1, 8, 3, 2, 1, 6, 9, 0, 2, 3, 5, 1, 0, 2, 1, 1, 3, 11, 8, 1, 1, 6, 1, 0, 1, 10, 1, 1, 2, 0, 3, 6, 7, 2, 1, 9, 2, 3, 2, 1, 13, 1, 0, 5, 9, 1, 1, 1, 1, 0, 1, 3, 9, 2, 6, 1, 1, 6, 6, 1, 1, 1, 1, 11, 0, 5, 6, 1, 2, 8, 6, 1, 0, 1 The 0's of this sequences are the generalized pentagonal numbers A001318. The 1's had not been tallied yet, so I entered them as A114913. Interesting thing is that the sequence is very similar to A111174 A114913 Values such that A114912(a(n))=1. Values such that A000009(a(n))==2 (mod 4). 3, 4, 8, 10, 13, 14, 17, 18, 19, 24, 25, 28, 32, 39, 42, 43, 47, 48, 50, 52, 54, 55, 62, 67, 69, 73, 74, 75, 76, 78, 83, 84, 87, 88, 89, 90, 95, 99, 101, 103, 105, 108, 109, 112, 113, 118, 119, 123, 125, 127, 130, 132, 134, 138, 140, 143, 144, 147, 149, 153, 154, 157 A111174 Numbers n such that 24*n + 1 is prime. 3, 4, 8, 10, 13, 14, 17, 18, 19, 24, 25, 28, 32, 39, 42, 43, 47, 48, 50, 52, 54, 55, 62, 67, 69, 73, 74, 75, 78, 83, 84, 87, 88, 89, 90, 95, 99, 103, 105, 108, 109, 112, 113, 118, 119, 123, 125, 127, 130, 132, 134, 138, 140, 143, 144, 147, 153, 154, 157, 158, 162 (list) matching in the first 28 terms. A little more searching and it appears that every number in A111174 is in A114913, but there are probably infinitely many numbers in the latter not in the former. (I admit I didn't look that hard for an exception, only about 2000 or so.) I wonder if there might be an interesting theorem in number theory. I looked at the Gordon/Ono paper http://www.math.wisc.edu/~ono/reprints/018.pdf and the Alladi paper http://www.ams.org/tran/1997-349-12/ S0002-9947-97-01831-X/S0002-9947-97-01831-X.pdf that rwg mentioned. Alladi mentions that all the members of A114913 are of the form of a (generalized) pentagonal number + a square, but that's a much larger set, and is it even known that the 24n+1 primes satisfy that requirement. I looked a little at Dean Hickerson's 8/4/2005 posting about q(n) mod 64. I didn't fully understand it, but it does not appear to have an obvious connection to prime number theory. The Alladi paper also mentions A001935 (calling it g(n)). It speculates that g(n) has the same power of 2 property (while I believe Gordon/Ono proves it.) It also studies the 1 mod 2 and 2 mod 4 cases. So I submitted the power of 2 sequence as A115247. The 0's of that sequence (the odd values of A001935) are the the triangle numbers A000217. I submitted the 1's (the A001935 2 mod 4 values) as A115248 and found a similar result as the A114913 case: this time with the sequence A005123: A005123 (( primes = 1 mod 8 ) - 1)/8. 2, 5, 9, 11, 12, 14, 17, 24, 29, 30, 32, 35, 39, 42, 44, 50, 51, 54, 56, 57, 65, 71, 72, 74, 75, 77, 80, 84, 95, 96, 101, 107, 110, 116, 117, 119, 122, 126, 129, 131, 137, 141, 144, 149, 150, 152, 156, 161, 162, 165 (list) This time only the first 7 terms matched and then the next 8. More importantly, it appears that every term of A005123 is in A115248. Alladi mentions that each term of A115248 is a triangle number + a square. P.S. Anyone who studies sequences for a while realizes that partitions have been studied a lot. So it amazes me when I find a fairly basic partition sequence is not yet in the EIS. A001935 has 3 simple interpretations. Partitions into non multiples of 4. Partitions with no even part repeated. Partitions with no part repeated more than 3 times. Would you believe that none of the labeled versions of these are yet in the EIS. They will be soon though as I'm submitting them as A115275-A115277. Christian

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[math-fun] bernpoly root
by R. William Gosper 16 Jan '06

16 Jan '06

A calculation to first order gave 2 n (2 - 2) B 1 2 n x ~ - - ---------------------- 4 8 n (2 n - 1) E 2 n - 2 as a root of B_2n(x). This bottoms out at B_2n(x) ~ .8e-20 for n=25. But to first order, n 1 4 - 2 x ~ - - ---------- 4 2 n 2 %pi 4 or just 1 1 x ~ - - --------. 4 n 2 %pi 4 But these are vastly inferior, bottoming out at ~3e-10 for n=12. But they are almost exactly mutual negatives, so that their average, n 1 4 - 1 x ~ - - ---------- 4 2 n 2 %pi 4 manages to bottom out at ~ .5e-14 for n=19. So there was something fortuitously 2nd order about that initial calculation. In any case, we have the mild paradox that for large n, B_2n(1/4) -> oo, even as a root of B_2n(x) -> 1/4. --rwg PS, It seems to me that floating mantissas should display with the point to the right of the first digit if it's <5 (or maybe sqrt 10), and to the left if it's greater. This leaves the best exponent if the mantissa is discarded.

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[math-fun] David Blaine, 37
by Jud McCranie 16 Jan '06

16 Jan '06

A friend of mine was just watching David Blaine on TV. One of his tricks was to say "Pick a number between 1 and 50. Be sure it is composed of two digits; they have to be odd and not duplicates of each other." Well, that permits only 13, 15, 17, 19, 31, 35, 37, 39. The guy on TV and my friend chose 37, which is what Blaine guessed. Do people tend to choose 37 from this set?

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