Further comments from Dean Hickerson (dean(AT)math.ucdavis.edu), Nov 01 2007 and Nov 10 2007 ---------------------------------------------------------------- I (njas) had written to Dean Hickerson saying: > Dean, In A014544, > there is a remark %C A014544 If m and n are in the sequence, so is m+n-1, since n-dissecting one cube in an m-dissection gives an (m+n-1)-dissection. 1, 8, 20, 38, 49, 51, 54 are in the sequence because of dissections corresponding to the equations 1^3=1^3, 2^3=8*1^3, 3^3=2^3+19*1^3, 4^3=3^3+37*1^3, 6^3=4*3^3+9*2^3+36*1^3, 6^3=5*3^3+5*2^3+41*1^3, and 8^3=6*4^3+2*3^3+4*2^3+42*1^3. Combining these facts gives the remaining terms shown, and all numbers > 47. It's been shown that no other numbers occur. > Do you know where it was so shown? Dean replied as follows, Nov 01 2007: And after rereading my old emails about this from 2002-2003, I'm not certain that it has been proved. According to Jud McCranie, the result is stated in the Gardner book (M. Gardner, Fractal Music, Hypercards and More: Mathematical Recreations from Scientific American Magazine. New York: W. H. Freeman, pp. 297-298, 1992). I had written: > I haven't been able to determine exactly which numbers should be in the > sequence. Unfortunately our library doesn't have Martin Gardner's book > "Fractal Music, Hypercards, and More: Mathematical Recreations from > Scientific American Magazine", which might answer the question. Do any > of you have a copy of that? And Jud replied: > Yes, I do. It gives the sequence as listed, and states that it was > established (one way or the other) for all values except n=54. Then it > gives 54, so 54 should be in the sequence, and all other terms are correct. Based on Eric Weisstein's web pages (see below), it appears that Problem E724 by H. Hadwiger in the Amer. Math. Monthly (vol. 53 (1946), 271) asked for the largest number not in the sequence and the partial solution by W. Scott (vol. 54 (1947), 41-42) showed that it was either 47 or 54. In 1977 D. Rychener and A. Zbinden independently found the 54-cube dissection. That seems to imply that Scott proved that 47 is not in the sequence, but he didn't need to settle the smaller cases to solve the original problem. I don't know for sure if those cases have been proved; it's possible that Gardner misstated what was known. (To show that the current version of the sequence is correct, we'd need proofs that there are no dissections into 31, 32, 42, 44, or 47 cubes; the other cases would follow from these.) So I'd suggest looking at the Gardner book and Scott's partial solution in the Monthly. (Unfortunately I don't have access to either of those.) Here's the relevant part of the Weissteiin link: > Cube Dissection > > A cube can be divided into n subcubes for only n=1, 8, 15, 20, 22, 27, 29, > 34, 36, 38, 39, 41, 43, 45, 46, and n>=48 (Sloane's A014544; Hadwiger 1946; > Scott 1947; Gardner 1992, p. 297). This sequence provides the solution to > the so-called Hadwiger problem, which asks for the largest number of > subcubes (not necessarily different) into which a cube cannot be divided > by plane cuts, and has the answer 47 (Gardner 1992, pp. 297-298). > > If m and n are in the sequence, so is m+n-1, since n-dissecting one cube > in an m-dissection gives an (m+n-1)-dissection. The numbers 1, 8, 20, 38, > 49, 51, 54 are in the sequence because of dissections corresponding to the > equations ... (see the OEIS entry for these) > Combining these facts gives the remaining terms in the sequence, and all > numbers >47, and it has been shown that no other numbers occur > (D. Hickerson). ... (irrelevant stuff) > REFERENCES: > Ball, W. W. R. and Coxeter, H. S. M. Mathematical Recreations and Essays, > 13th ed. New York: Dover, pp. 112-113, 1987. > > Cundy, H. and Rollett, A. Mathematical Models, 3rd ed. Stradbroke, > England: Tarquin Pub., pp. 203-205, 1989. > > Gardner, M. The Second Scientific American Book of Mathematical Puzzles & > Diversions: A New Selection. New York: Simon and Schuster, 1961. > > Gardner, M. "Block Packing." Ch. 18 in Time Travel and Other Mathematical > Bewilderments. New York: W. H. Freeman, pp. 227-239, 1988. > > Gardner, M. Fractal Music, Hypercards, and More: Mathematical Recreations > from Scientific American Magazine. New York: W. H. Freeman, pp. 297-298, 1992. > > Guy, R. K. "Research Problems." Amer. Math. Monthly 84, 810, 1977. > > Hadwiger, H. "Problem E724." Amer. Math. Monthly 53, 271, 1946. > > Honsberger, R. Mathematical Gems II. Washington, DC: Math. Assoc. Amer., > pp. 75-80, 1976. > > Hunter, J. A. H. and Madachy, J. S. Mathematical Diversions. New York: > Dover, pp. 69-70, 1975. > > Lonke, Y. "On Random Sections of the Cube." Discr. Comput. Geom. 23, > 157-169, 2000. > > Meier, C. "Decomposition of a Cube into Smaller Cubes." Amer. Math. > Monthly 81, 630-633, 1974. > > Scott, W. "Solution to Problem E724." Amer. Math. Monthly 54, 41-42, 1947. > > Sloane, N. J. A. Sequence A014544 in "The On-Line Encyclopedia of Integer > Sequences." > > Steinhaus, H. Mathematical Snapshots, 3rd ed. New York: Dover, pp. 168-169, > 1999. Eric also has a page, http://mathworld.wolfram.com/HadwigerProblem.html , which says: > Hadwiger Problem > > What is the largest number of subcubes (not necessarily different) into > which a cube cannot be divided by plane cuts? The answer is 47 (Gardner > 1992, pp. 297-298). The problem was originally posed by Hadwiger (1946), > and Scott (1947) showed that dissections were possible for more than 54 > subcubes. This left only 47 and 54 as possible candidates, and a dissection > into 54 pieces in 1977 independently by D. Rychener and A. Zbinden > resolved the solution as 47 (Guy 1977; Gardner 1992, p. 297). > > SEE ALSO: Cube Dissection, Cutting. [Pages Linking Here] > > REFERENCES: > Gardner, M. Fractal Music, Hypercards, and More: Mathematical Recreations > from Scientific American Magazine. New York: W. H. Freeman, 1992. > > Guy, R. K. "Research Problems." Amer. Math. Monthly 84, 810, 1977. > > Hadwiger, H. "Problem E724." Amer. Math. Monthly 53, 271, 1946. > > Scott, W. "Solution to Problem E724." Amer. Math. Monthly 54, 41-42, 1947. Dean Hickerson dean(AT)math.ucdavis.edu On Nov 10 2007, Dean wrote sent the following sequel to this message: I wrote: > Based on Eric Weisstein's web pages (see below), it appears that > Problem E724 by H. Hadwiger in the Amer. Math. Monthly (vol. 53 (1946), 271) > asked for the largest number not in the sequence and the partial solution > by W. Scott (vol. 54 (1947), 41-42) showed that it was either 47 or 54. > In 1977 D. Rychener and A. Zbinden independently found the 54-cube > dissection. It turns out that the Humboldt State University library has access to JSTOR, so I was able to read (and photograph with a digital camera) the E724 solution. My guess about what was asked and proved there is incorrect. First, the problem was by Fine and Niven, not Hadwiger. Second, the problem didn't specifically ask about 3-dimensional cubes, although the Editorial Note after the solution discusses that case. Here's what it says (if I haven't made any transcription errors): ---------- Start of quoted text ---------- Admissible Numbers E724 [1946,271]. Proposed by N. J. Fine, University of Pennsylvania, and Ivan Niven, Purdue University Define an n-admissible number k as one such that an n-dimensional cube may be subdivided into k cubes. Prove that for each n there exists an integer A_n such that all integers exceeding A_n are n-admissible. Solution by Fritz Herzog, Michigan State College. Let k be an n-admissible number, and let an n-dimensional cube be subdivided into k cubes. If one of these k cubes is further subdivided into a^n equal cubes, where a is any integer greater than unity, then we obtain a subdivision of the original cube into k + (a^n - 1) cubes. Thus, if k is n-admissible, so is k + (a^n - 1). Since unity is n-admissible, we conclude that all numbers of the form (1) 1 + p(a^n - 1) + q(b^n - 1), a>=2, b>=2, p>=0, q>=0, are n-admissible. Choose a and b so that (a^n - 1, b^n - 1) = 1, which is always possible, for instance by taking a=2, b=2^n-1. Then, by a well known fact in number theory (see, e.g., Polya-Szego, Aufgaben und Lehrsatze aus der Analysis, vol. 1, p. 4, Problem 26), every integer k>(a^n - 1)(b^n - 1) can be written in the form (1) in at least one way, and is, therefore, n-admissible. Also solved by Paul Bateman, J. B. Kelly, Leo Moser, William Scott, and the proposers. Editorial Note. The problem of finding the largest number not n-admissible appears not to be easy. For n=2 the answer is 5. Scott showed this by proving that the only numbers not 2-admissible are 2, 3, and 5. Following Herzog's suggestion of taking a=2, b=2^n-1, we find that all numbers greater than 2394 are 3-admissible. This was greatly improved upon by Scott, who showed that 1, 20, 38, 39, 49, 51, 61 are all 3-admissible. These numbers, along with those obtained by adding multiples of 7 to these, then constitute a set of 3-admissible numbers, and it is easily shown that every number greater than 54 is in this set. The question as to whether 54 is 3-admissible or not is still open. Suppose we define a number k to be strictly n-admissible if an n-dimensional cube may be subdivided into k different cubes. For some time the conjecture was that there are no numbers strictly 2-admissible. This conjecture was proved false in the paper The Dissection of Rectangles into Squares, by C. A. B. Smith, A. H. Stone, W. T. Tutte, Duke Mathematics Journal, Dec. 1940. This same paper proved that there are no numbers which are strictly 3-admissible. Bateman remarked that the proposed problem appeared in Ripley's "Believe It or Not" column. ---------- End of quoted text ---------- So it's not clear to me that Scott proved that 47, or any smaller number, is not 3-admissible. Perhaps Gardner's book says more about this; unfortunately the HSU library doesn't have it. Another place to look is Guy, R. K. "Research Problems." Amer. Math. Monthly 84, 810, 1977 which is mentioned on Eric Weisstein's website. Next time I'm at HSU, I'll take a look at it. Dean Hickerson dean@math.ucdavis.edu