August 2006 - mathfun@mailman.xmission.com

[math-fun] poly (regular) polyhedra
by Schroeppel, Richard 03 Aug '06

03 Aug '06

I've been working out counts of polypolyhedra by hand. (regular polyhedra, left-right mirror pairs count as only 1) volume 1 2 3 4 5 polytetrahedra 1 1 1 3 7 polycubes 1 1 2 7 ... polyoctahedra 1 1 3 11 polydodecahedra 1 1 3 16 polyicosahedra 1 1 4 33 I find the visualization is hard, and diagrams only help a little. Notes: 3 dodecahedra will fit around an edge, and 4 will fit around a vertex. 3 icosahedra will not fit around an edge, nor 4 around a vertex. 3 icosahedra will fit around a vertex. Rich

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Re: [math-fun] Apostol's Riemann song for your iPod
by Daniel Asimov 02 Aug '06

02 Aug '06

Fred writes: << Fifteen verses may found at http://www.ndsu.nodak.edu/instruct/isern/103/betsy.htm However, the tune is originally English and known as "Villikins / Villikens and his Dinah" >> A little more googling led me to learn that "Villikens and His Dinah" derives from an earlier tune, "Lord Ronald", lyrics said to be by Robert Burns (aka Child Ballad #12). The tune is suspected to have come from an even earlier italian tune. (Cf. white text box at http://www.worldburnsclub.com/expert/burns_songs_an_american_connection.htm .) But even the tune from Lord Ronald is quite different from the "Sweet Betsy from Pike" tune. In fact, "Billy Boy" is *another* descendant of "Lord Ronald". (Hey, it's the folk process.) --Dan

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Re: [math-fun] strange factoring in primes
by Daniel Asimov 02 Aug '06

02 Aug '06

Josh writes: << In addition to the two examples from previous messages (7 * 11 * 13 = 10^3 + 1 and 211 * 227 * 241 = 226^3 + 1), there are lots more products of three (not necessarily consecutive) primes that make one more than a cube. See http://www.research.att.com/~njas/sequences/A115403 . . . . . . . >> Wouldn't you just know it, that someone had to come along and think of almost my question, and ***exactly*** what Chris asked: << . . . [Dan] asked about consecutive primes p, q, r with p*q*r = n^3 + 1 for some integer n -- I thought i'd take a crack at it without requiring consecutive, . . . . . . >> Turns out the A115403 was contributed to the OEIS only last March 8, less than 5 months ago, (by one Zak Seidov). If we want to keep up, folks, we're gonna hafta start thinking just a little bit faster! --Dan

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[math-fun] sums of factors in GF(2)
by Marc LeBrun 02 Aug '06

02 Aug '06

Recently I've been trying to exhort the Sequence Phanatiques to compute analogs of the sum-of-prime-factors (with and without multiplicity) in other arithmetics, such as Gaussian integers, GF(2), etc. I was wondering, specifically about GF(2), summing (ie XORing) the prime factors of N with multiplicity: Noting that only the square-free part of N matters, since the square parts sum to 0... A. Aside from the perfect squares (eg 5) are there any other N that sum to 0? Can they be characterized? B. If some sum S occurs for any N then it occurs infinitely, for all the square multiples of N. Does every value of S occur?

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[math-fun] strange factoring in primes
by Chris Landauer 02 Aug '06

02 Aug '06

hihi, all - rich asked about consecutive primes p, q, r with p*q*r = n^3 + 1 for some integer n - i thought i'd take a crack at it without requiring consecutive, and found prety easily 211 * 227 * 241 = 226 ^ 3 + 1 = 11543177, which leadsd me to think that rich's expectation (that only 7*11*13 = 1001 works) is correct BTW, it is easy to show that the middle prime is n+1 more soon, cal

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Re: [math-fun] Re: sums of factors in GF(2)
by Daniel Asimov 02 Aug '06

02 Aug '06

Antti Karttunen wrote: << Marc LeBrun wrote: > Recently I've been trying to exhort the Sequence Phanatiques to > compute analogs of the sum-of-prime-factors (with and without > multiplicity) in other arithmetics, such as Gaussian integers, GF(2), > etc. > Don't worry, I will do the GF(2)[X] -thing, in my due time. . . . . . . >> Aha! I knew Marc couldn't be speaking of GF(2) per se, but (duh) I didn't know what he *was* referring to. Thanks, Antti. --Dan

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Re: [math-fun] strange factoring in primes
by Daniel Asimov 02 Aug '06

02 Aug '06

Chris, That's really cool. (But I just wanted to clarify that it was I who asked the question.) Regards, Dan ----------------------------------------------------------------------------- << hihi, all - rich asked about consecutive primes p, q, r with p*q*r = n^3 + 1 for some integer n - i thought i'd take a crack at it without requiring consecutive, and found prety easily 211 * 227 * 241 = 226 ^ 3 + 1 = 11543177, which leadsd me to think that rich's expectation (that only 7*11*13 = 1001 works) is correct BTW, it is easy to show that the middle prime is n+1 >>

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Re: [math-fun] Apostol's Riemann song for your iPod
by Daniel Asimov 02 Aug '06

02 Aug '06

Henry, Thanks for the link < http://olimu.com/Riemann/Song.htm > to John Derbyshire singing Tom Apostol's song, "Where are the Zeroes of Zeta(s) ?" Quite a nice little ditty. The tune was so familiar, but I couldn't place it at first. SO: Trivia question: What familiar old song uses the same tune? --Dan

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[math-fun] Kaplansky's Rat Theorem
by Bill Dubuque 02 Aug '06

02 Aug '06

Google Books reports Kaplansky has classified rodent tumors http://books.google.com/books?q=kaplansky%27s+theorem+rodent --Bill Dubuque

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[math-fun] a little more on strange factorizations
by Chris Landauer 02 Aug '06

02 Aug '06

hihi, all - first, apologies to dan, since i didn't notice that he'd asked the original question second, two more examples (with non-consecutive primes) - 421*443*463 = 442^2 + 1 = 86350889 7*13*19 = 12^3 + 1 = 1729 the first one is part of a family with (p+r)/2 = n, and the second is part of a family with p, q, r in arithmetic progression (i do not know if the families have any other members; i just divided the problem according to the value of (r + p - 2*n) / 2 which is a non-negative integer i have not explored the problem further than this (at least, not fruitfully 8-() more soon, cal

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