November 2002 - mathfun@mailman.xmission.com

[math-fun] Re: Books on their way
by Richard Guy 22 Nov '02

22 Nov '02

David Wilson wanted you all to get this message. Best to everyone, including David. R. On Fri, 22 Nov 2002, David Wilson wrote: > There were several people on math-fun/seqfan who expressed interest in my > books. Please relay to them that I have sold the books to you, that I did > so > because of the good price, the convenience of selling the whole lot at once, > and the knowledge that the valuable books would enter an archive. > > - David Wilson

1 0

Re: [math-fun] Multiply-gendered humanitarians
by Dan Hoey 22 Nov '02

22 Nov '02

About Bingo, I wrote: > << > There are at most five equivalence classes of lines: center horizontal(1), > off-center horizontal(4), center vertical(1), off-center vertical(4), and > diagonal(2). Is the distribution for diagonals the same as for the center > horizontal? > >> And Dan A. responded: > ... If the only question is Find the probability P(X;n) that line X > is filled after n draws, then there are only 2 relevant categories > for X: those that need 5 matching draws, and those that need only 4. I was a little fuzzy on the question last night. Try this: I want to find P_1(X;n), the probability that line X is filled on draw n _and_ that no line was filled after n-1 draws. The other question I'm asking is P_k(X;n), the probablity that line X is filled on one of k boards on draw n, and that no line was filled on any of the boards before that. Now what are the equivalence classes for X? Dan Hoey Hoey(a)AIC.NRL.Navy.Mil

1 0

Re: [math-fun] 316912650057057350374175801344000001
by Warut Roonguthai 22 Nov '02

22 Nov '02

Robert Harley found this and several other prime factors of Googolplex+1; see http://pauillac.inria.fr/~harley/bits.html

1 0

Re: [math-fun] Multiply-gendered humanitarians
by asimovd＠aol.com 22 Nov '02

22 Nov '02

Dan H. writes: << ... We assume that each [Bingo] board is equally likely, though that can't be true--with over 552 septillion possible boards, only a small fraction ... have ever been printed! ... >> Certainly only a small fraction have ever been printed, but that doesn't exclude the possiblity that they are random. Chances are they aren't, but it wouldn't be that hard to hook up a randomizer to the printer so that each board is the result of the appropriate randomization process. (Hmmm, you may have suggested a likely way to win at Bingo in the long run...) << There are at most five equivalence classes of lines: center horizontal(1), off-center horizontal(4), center vertical(1), off-center vertical(4), and diagonal(2). Is the distribution for diagonals the same as for the center horizontal? >> If the only question is Find the probability P(X;n) that line X is filled after n draws, then there are only 2 relevant categories for X: those that need 5 matching draws, and those that need only 4. The number of possible sequences of n draws is D_n = 75! / (75-n)!, and given X, the number of winning sequences among these for one of the 8 lines X needing 5 matches (for n >= 5) is W_5 = 5! * ( n! / (5! (n-5)!) ) = n! / (n-5)!, so the winning probability is the quotient W_5 / D_n = P_5 = n! (75-n)! / (75! (n-5)!) = n(n-1)(n-2)(n-3)(n-4) / (75*74*...*(75-n+1)), and likewise for one of the 4 lines X needing only 4 matches, W_4 / D_n = P_4 = n(n-1)(n-2)(n-3) / 75*74*...*(75-n+1). --Dan A.

2 1

Re: [math-fun] Sums of four squares
by asimovd＠aol.com 21 Nov '02

21 Nov '02

Hi, Gene. Thanks for the answer. (I hope to see the proof sometime. I wonder if there's some use for quaternions here...) << The number, r4(n), of integer solutions (w,x,y,z) of w^2 + x^2 + y^2 + z^2 = n, for positive integer n, is 8 times the sum of the divisors of n which are not divisible by 4. Expressed another way, r4(n)/8 is multiplicative, and r4(p^n)/8 = 1 + p + p^2 + ... + p^n for p an odd prime, r4(2^n)/8 = 1 + 2 for n>0. >> This formula, which I'm seeing for the first time, is amazingly interesting! --Dan

4 3

[math-fun] Multiply-gendered humanitarians
by Dan Hoey 21 Nov '02

21 Nov '02

Thanks to Ed Pegg for reminding me of the somewhat more difficult questions about BINGO. The game of BINGO comes in various flavors. In the one I know, each player is supplied with a 5x5 board. The first column is filled with five different numbers drawn at random from {1,2,...,15}, the second column from {16,...,30}, and so forth, except that the center square is marked out. A random number generator supplies numbers from 1 to 75, and when a number appears on a board, the player marks out that square. The first player to get five marks in a row vertically, horizontally, or diagonally wins. We assume that each board is equally likely, though that can't be true--with over 552 septillion possible boards, only a small fraction of them have ever been printed! Also I don't know whether the bingo-printing industry prints new boards all the time. It's possible they have a relatively small set of boards they print over and over. Do any funsters know? A second question: After N numbers have been drawn, what's the probability you have five in a row? I don't have an easy answer, but somehow I suspect that in the intersection of math funsters and generating funcsters there should be one. And a question that Ed passed on from Paul Stephens: Suppose we have a side bet on which of the 12 lines will be filled by on the winning board (and in a small number of cases, whether the winner has multiple lines). What are the odds on this bet? How does it vary as a function of the number of players? There are at most five equivalence classes of lines: center horizontal(1), off-center horizontal(4), center vertical(1), off-center vertical(4), and diagonal(2). Is the distribution for diagonals the same as for the center horizontal? I'm almost sure of that, but somehow not satisfied. Is the board symmetry group of order 32, or larger? Does this interact in any interesting way with S_75? Dan Hoey(a)AIC.NRL.Navy.Mil

1 0

[math-fun] Eight squares
by Eugene Salamin 21 Nov '02

21 Nov '02

The number of representations of n as a sum of 8 squares also is expressed in terms of elementary number theoretic functions. n d 3 r8(n) = 16 (-1) sum (-1) d . d|n __________________________________________________ Do you Yahoo!? Yahoo! Mail Plus Powerful. Affordable. Sign up now. http://mailplus.yahoo.com

1 0

[math-fun] bounced submissions for Math-Fun
by Richard Schroeppel 21 Nov '02

21 Nov '02

Several list members have seen their Math-Fun postings held for moderator approval. I've approved all of these, so they are eventually posted, but typically with a half-day delay. The rule is that list members get to post without approvals or delays. Unfortunately, the predicate for "is a list member" seems dumb: it looks like the "From" address on the message must exactly match the subscription address. So if you are subscribed as joe(a)fermat.math.univ.edu, but you post your message from joe(a)gauss.math.univ.edu, the message is treated as a non-member posting and dumped in a box for me to approve. Each time this happens, I add the new address to a white-list of "extra posting addresses that are ok". Eventually we'll have most of the alternate addresses, and this should become only a minor nuisance. If you want to change your subscription address, the only way I know of is to unsubscribe and resubscribe. You might want to resubscribe first, since I also have to approve new members, and then desubscribe later. If the web interface is too unfriendly, drop me a note and I'll handle it. On the bright side, I've discarded an average of .5 - 1 spam/day that was headed for the mailing list. These are forwarded from the old math-fun address; I've left that open while folks adjust to the new address. Rich rcs(a)cs.arizona.edu

1 0

Re: [math-fun] Sums of three squares
by Henry Baker 21 Nov '02

21 Nov '02

>From "Representations of Integers as Sums of Squares" by Emil Grosswald, Springer, 1985: n is representable with 3 squares iff n is not of the form 4^a (8k+7). R3(n) is # of _primitive_ solutions; r3(n) is # of all solutions. r3(n) = sum(d^2|n, R3(n/d^2)) R3(n) was determined by Gauss, although not in the following form. Let h = number of classes of primitive binary quadratic forms, corresponding to the discriminant D=-n if n=3 (mod 8), D=-4n if n=1,2,5,6 (mod 8), and let d_1=1/2, d_3=1/3, d_n=1 otherwise. Then R3(n) = 12 h d_n, if n=1,2,5,6 (mod 8), 24 h d_n, if n=3 (mod 8). Also, if n is squarefree and (r/n) is the Jacobi symbol, R3(n) = 24 sum(r=1,[n/4],(r/n)) if n=1 (mod 4), 8 sum(r=1,[n/2],(r/n)) if n=3 (mod 8). Don't ask me if I can follow the proof -- I can barely understand the statement. I assume that there are no simpler formulations, else Grosswald would have shown them. ----- At 12:21 PM 11/20/02 -0800, Eugene Salamin wrote: >There is a nice way to calculate the number r2(n), >n>0, of integer solutions (x,y) of > > x^2 + y^2 = n. > >Factor n as > > n = p1^a1 p2^a2 ... q1^b1 q2^b2 ... 2^c, > >where the p's are primes == 1 mod 4, and the q's are >primes == 3 mod 4. Then > > r2(n) = if any b is odd, then 0 > else, 4 (1 + a1) (1 + a2) ... . > >Is there a similar formula for the number r3(n) of >integer solutions (x,y,z) of > > x^2 + y^2 + z^2 = n ?

1 0

[math-fun] Sums of four squares
by Eugene Salamin 21 Nov '02

21 Nov '02

The number, r4(n), of integer solutions (w,x,y,z) of w^2 + x^2 + y^2 + z^2 = n, for positive integer n, is 8 times the sum of the divisors of n which are not divisible by 4. Expressed another way, r4(n)/8 is multiplicative, and r4(p^n)/8 = 1 + p + p^2 + ... + p^n for p an odd prime, r4(2^n)/8 = 1 + 2 for n>0. __________________________________________________ Do you Yahoo!? Yahoo! Mail Plus Powerful. Affordable. Sign up now. http://mailplus.yahoo.com

1 0