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[math-fun] Variation on Dan's puzzle
by Eugene Salamin 24 Apr '06

24 Apr '06

Let f(n) be the number of prime factors of n, and let g(n) be the number of distinct prime factors of n, e.g. f(12)=f(2.2.3)=3, g(12)=2. Let F(n) = (-1)^f(n), G(n) = 2^g(n). Then: sum(F(d) G(n/d), d divides n) = 1. Or, the other way around, F(n) = 2^f(n), G(n) = (-1)^g(n), and the result is also true. Now let F(n) = (1/2)^f(n), G(n) = (1/2)^g(n), and again the same conclusion is true. In fact, let F(n) = a^f(n), G(n)=b^g(n), a+b=1, and the conclusion continues to hold. Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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[math-fun] 6x6 pandiag. multiplic. squares ARE NOT impossible!
by Christian Boyer 24 Apr '06

24 Apr '06

A better 6x6 pandiagonal multiplicative square. With always the 2x2 and 3x3 sub-squares feature. 14 2100 63 350 84 1575 4410 15 980 90 735 20 50 588 225 98 300 441 126 525 28 3150 21 700 490 60 2205 10 2940 45 450 147 100 882 75 196 Using 2^ * 3^ * 5^ * 7^ integers, versus 2^ * 3^ in the previous example. Max nb = 4,410, vs 46,656. P = 8.58 E+13, vs 1.02 E+14. Christian.

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Re: [math-fun] 6x6 pandiag. multiplic. squares ARE NOT impossible!
by dasimov＠earthlink.net 24 Apr '06

24 Apr '06

Christian writes: << Dan, you are right, there is a great relationship between additive and multiplicative squares (and cubes), but not always a direct relationship. >> I may have been less than clear. Considering only conditions on rows & columns, all I was saying is that, suppose p_k is the largest prime divisor of any entry in a MuMaSq: Then if each entry p_1^e_1 * . . . * p_k^e_k (all e_j >= 0) is replaced by its tuple of exponents (e_1,...,e_k), the result is an AdMaSq -- not of mere integers, but of k-tuples of integers. Or if you prefer, k ordinary magic squares (possibly with repeated entries), one for each j in {1,...,k}, each created by replacing each entry in the MuMaSq by its jth exponent, e_j. And conversely, though extra care must be taken to assure that all final entries be distinct. --Dan

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[math-fun] 6x6 pandiag. multiplic. squares ARE NOT impossible!
by Christian Boyer 23 Apr '06

23 Apr '06

In his http://www.maa.org/editorial/mathgames/mathgames_11_07_05.html column, our friend Ed Pegg Jr. wrote: "The latin squares method, used for 4×4 and 5×5 squares, does not work for the 6×6 square due to the famous "36-officers problem" of Euler. Because of this, a 6×6 pandiagonal multiplicative magic square may be impossible. Can anyone prove that, or find an example?" Here is an example! 7776 3 3888 96 243 48 2 46656 4 1458 64 2916 2592 9 1296 288 81 144 486 192 972 6 15552 12 32 729 16 23328 1 11664 162 576 324 18 5184 36 All the rows, columns, diagonals AND broken diagonals give the same product: 101,559,956,668,416 And best, it is a "most-perfect" pandiagonal magic square, à la Kathleen Ollerenshaw. All 2x2 sub-squares have always the same product: 2,176,782,336 My feeling was the same than Ed, I thought that such a square was impossible... I wrote the same "Impossible?" than Ed in my recapitulative table at www.multimagie.com/English/Multiplicative.htm A new update will be necessary! And I have also first examples of pandiagonal multiplicative squares of orders 8, 9, 12, 15, 16. Best regards. Christian.

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Re: [math-fun] 6x6 pandiag. multiplic. squares ARE NOT impossible!
by dasimov＠earthlink.net 23 Apr '06

23 Apr '06

Michael Kleber's latest brilliant analysis of Christian's new multiplicative magic square -- in terms of bit planes -- is quite reminiscent of (what I think it) Conway & Coxeter's analysis of additive magic squares as it appears in (some of) the most recent edition(s) of W.W. Rouse Ball's "Mathematical Recreations and Essays" (updated by Coxeter). Let N := the set of nonnegative integers. Then {additive magic squares of elements of N^k for some k} is in 1-1 correspondence with {multiplicative magic squares} -- just letting the components of v in N^k correspond to prime exponents. I was wondering if there could be a multiplicative magic square that was somehow not a consequence of additive ones . . . but now I see that's impossible. --Dan

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Re: [math-fun] Square thirds
by Schroeppel, Richard 22 Apr '06

22 Apr '06

Jon Perry asks: >> Which shapes do have 2 or more representations by dissections into k congruent pieces? This looks hard to me. Even establishing uniqueness or non-existence of such dissections is hard, and very much a special-case argument for each K and for each polygon. --Rich

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[math-fun] dissections of regular polygons into congruent pieces
by Michael Reid 21 Apr '06

21 Apr '06

i have seen the the term "geometric dissection" used to mean dissection into closed pieces, each of which is the closure of its interior and such that the only overlap between is along the boundary, and the term "set-theoretic dissection" used to mean a partition into disjoint sets (as in banach-tarsky). besides considering dissections os a square into congruent pieces, we can consider dissections of a regular k-gon into congruent pieces. the case k = 4 is obviously special! for which N is it possible to (geometrically) dissect a regular hexagon into N congruent pieces? there are some "obvious" values of N that are possible, and i have found dissections for a lot of values of N that are not "obvious". the same for a regular pentagon: for which N is it possible to (geometrically) dissect a regular pentagon into N congruent pieces? as before, there are some "obvious" cases, although fewer. i have found some dissections for non-"obvious" values, although these are also scarce. i haven't yet looked at impossibility proofs; these are likely to be prohibitively difficult. mike

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Re: [math-fun] Puzzle
by dasimov＠earthlink.net 21 Apr '06

21 Apr '06

Franklin corrects the worst possible typo in my puzzle: What should be proved is indeed that Zeta_q(s) = 1/Zeta_sqf(s) (Thanks, Franklin!) Corrected puzzle follows: -------------------------------------------------------------------------------- For n in Z+, let q(n) := the total number of prime factors n has (distinct or not) = sum of the exponents in n's prime factorization. Let Zeta_q(s) := Sum{n=1..oo} (-1)^q(n)/(n^s) for Re(s) > 1. Let Zeta_sqf(s) := Sum{n=1..oo, n squarefree} 1/(n^s) for Re(s) > 1. ------------------------------------------------------------------------------------ Puzzle: Prove: Zeta_q(s) = 1/Zeta_sqf(s) (for Re(s) > 1). (Don't worry about convergence.) (Note: A := B just signifies that A will stand for B.) --Dan

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[math-fun] Puzzle
by dasimov＠earthlink.net 21 Apr '06

21 Apr '06

For n in Z+, let q(n) := the total number of prime factors n has (distinct or not) = sum of the exponents in n's prime factorization. Let Zeta_q(s) := Sum{n=1..oo} (-1)^q(n)/(n^s) for Re(s) > 1. Let Zeta_sqf(s) := Sum{n=1..oo, n squarefree} 1/(n^s) for Re(s) > 1. Puzzle: Prove: Zeta_q(s) = Zeta_sqf(s) (for Re(s) > 1). (Don't worry about convergence.) (Note: A := B just signifies that A will stand for B.) --Dan

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Re: [math-fun] What use are primes?
by Phil Carmody 21 Apr '06

21 Apr '06

From: Henry Baker <hbaker1(a)pipeline.com> > > One of my intuitions re numbers is that analysts worry about > the high-order bits of numbers, while number theorists worry > about the low-order bits. For example, some quick compositeness > tests look only at the low-order digits of numbers. Name two such tests please? "Is the number even" comes to mind (assuming that when you say 'digits' after you've previously said 'bits' that you did in fact mean 'bits' again), but I'm not really getting a second test. Would you care to demonstrate using the numbers 0xFFFCA5000000000000000000000000000000000000000000000000000000000000000001 0xFFFC8B000000000000000000000000000000000000000000000000000000000000000001 ? A "but I did actually mean digits" wiggle should be avoided. > The usual > GCD algorithm worries about reducing the length of the numbers, > but the so-called "binary algorithm" (see Knuth if you need to > know more about this) worries about reducing the number of > non-zero bits in the numbers. No it doesn't. Unless you'd like to explain to me how gcd(0b10010001,0b100000101) = 0b11101 involves reducing the number of non-zero bits in the numbers. It simply removes an extremal set bit, not caring what happens to any other bits. In particular, the lowest set bit. The traditional method ineffectually (depending on quite which version you use) attempts to remove a top set bit. (The whole binary algorithm can be summarised as "get rids of 0s, then get rid of the lowest 1, lather, rinse, repeat".) Phil () ASCII ribbon campaign () Hopeless ribbon campaign /\ against HTML mail /\ against gratuitous bloodshed [stolen with permission from Daniel B. Cristofani] __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com

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