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[math-fun] RE: Help extending equation -- sequence related to pentagonal numbers
by Richard Guy 28 Dec '05

28 Dec '05

In this connexion, it may be that Conway's relation twixt the triangular and pentagonal numbers is not as well known as it deserves to be. Divide the triangular numbers by 3 (when you can exactly): 0 1 3 6 10 15 21 28 36 45 55 66 78 91 105 120 136 153 ... 0 - 1 2 - 5 7 - 12 15 - 22 26 - 35 40 - 51 ... - - + + - - + + - - + ... and you get the pentagonal numbers in pairs, one of positive rank and the other negative. Then append signs according as the pair have the same (+) or opposite (-) parity. Then Euler's pentagonal number theorem is easy to remember: p(n-0)-p(n-1)-p(n-2)+p(n-5)+p(n-7)-p(n-12)-p(n-15)++-- =0^n where p(n) is the partition function, the left side terminates before the argument becomes negative, and 0^n = 1 if n = 0 and = 0 if n > 0. E.g., p(0)=1, p(7)=p(7-1)+p(7-2)-p(7-5)-p(7-7)+0^7=11+7-2-1+0=15. This, of course, is how Major Percy MacMahon did his calculations up to p(200), against which Hardy & Ramanujan were able to check their famous formula (too large to go in this margin, but see the BoN again). There are pictures for the relations (see The BoN) a a a a a a a a a a a a a a a b a a b a a b a a a b b c c c b c c b b a a b b a a a b b '''' c c c b b c c c b b c c c b b b c c c c b b c c c c b b b c c c c c b b b where the pentagonal numbers are well known (?) to be made of three triangles, one of which differs in rank by one from the other two, more or less according as the pentagon is of pos or neg rank: 0+0+1=1 1+1+0=2 1+1+3=5 3+3+1=7 3+3+6=12 6+6+3=15 6+6+10=22 ... + - + - + - + x y y y z x x x x y y z x x y y z z x x x y z y x x z x x y z z x x x y z z z x x y y x z x x z z y y x z z x x x z z z x x x x z z z z [deform these carefully into regular pentagons!] Best to all for 2006, R. On Tue, 27 Dec 2005, Andrew Plewe wrote: > Point taken concerning the algebra -- I'll work it out. I'd like to come up > with a generalized equation similar to that for A005449, or (n(3n+1))/2, > which will give all values for the desired sequence with f(x), x = 0 as the > "starting" point. Incidentally, I find these sequences interesting because > I ran into them while doing the following simple transformation (add the > first two terms and redistribute over the remaining terms, then sum the > result): > > 1 + 1 + 2+3 = 3+4 = 7 > 1 + 2 + 3+4+5 = 4+5+6 = 15 > 1 + 3 + 4+5+6+7 = 5+6+7+8 = 26 > > which can be described by 7 + 8(x) + (3 * tri(x-1), starting with x = 0. > > And: > > 2 + 1 + 2+3+4 = 3+4+5 = 12 > 2 + 2 + 3+4+5+6 = 4+5+6+7 = 22 > 2 + 3 + 4+5+6+7+8 = 5+6+7+8+9 = 35 > > which can be described by 12 + 10(x) + (3 * tri(x-1)), starting with x = 0. > > Anyway, thanks for you help! > > -Andrew Plewe- > > -----Original Message----- > From: Marc LeBrun [mailto:mlb@fxpt.com] > Sent: Thursday, December 22, 2005 7:03 PM > To: seqfan > Subject: Re: Help extending equation -- sequence related to pentagonal > numbers > > I'm not 100% sure what you are asking, but maybe this will help: > > You have a bunch of statements of the form > > "Equation to produce sequence from Kth term: u + v(x) + (3 * tri(x-1))" > > So let's restate this using a general parameterized formula we'll define: > f[u,v](x) = u + v x + 3 tri(x-1). > > (You usually make this statement for K=4--except the first time, when you > use K=3 for some reason--but no matter...) > > You give a few "starting values" and then you're basically saying that > everything after that is f(1), f(2), f(3) and so on. > > But, at least for all those I checked, your formula applied to x<1, that is > f(0), f(-1), f(-2) and so on, seems to give the "starting values" as well. > > So I'll leap to the conclusion that what you want is the formula, a(N) > defined so that it that "starts" at the first term (ie N=1), rather than f > which "starts" at the Kth term. > > This is easy: because a(K)=f(1), a(K+1)=f(2), and so on, it's simply an > offset, so a(N)=f(N-K+1). > > So now all we have to do is substitute x=N-K-1 to get the formula for a(N): > > a(N) = f(N-K+1) = u + v (N-K-1) + 3 tri(N-K-2) > > I'll leave simplifying the algebra to you. > > By the way, I remark that these a(N) are just quadratics in N, and so their > properties are well understood (eg 3-term recurrence, rational 2nd-degree > generating functions, etc), so you probably should only trouble to submit > them if they have some other especially interesting property (such as being > pentagonal or whatever). > > ======== > > At 05:14 PM 12/22/2005, Andrew Plewe wrote: >> My new sequence follows from extending a group of partial equations for >> other sequences related to pentagonal/triangle numbers in the OEIS. To >> demonstrate the pattern, I've listed the sequences in the OEIS with >> their corresponding partial generating equations. I'd like to find a.) >> an equation that fully describes my new sequence, and b.) a general >> method for deriving equations for other similar sequences. "tri", here, >> represents the function for generating triangle numbers ((n * (n + 1))/2: >> >> A005449 = 0, 2, 7, 15, 26, 40, etc. Equation to produce sequence from >> third term: 7 + 8(x) + (3 * tri(x-1)) >> >> A000326 = 0, 1, 5, 12, 22, 35, etc. Equation to produce sequence from >> fourth term: 12 + 10(x) + (3 * tri(x-1)) >> >> A045943 = 0, 3, 9, 18, 30, 45, etc. Equation to produce sequence from >> fourth term: 18 + 12(x) + (3 * tri(x-1)) >> >> A095794 = 1, 6, 14, 25, 39, 56, etc. Equation to produce sequence from >> fourth term: 25 + 14(x) + (3 * tri(x-1)) >> >> And my new sequence: >> >> A000001 = 3, 10, 20, 33, 49, 68, etc. Equation to produce sequence >> from fourth term: 33 + 16(x) + (3 * tri(x-1)) >> >> >> As you can see there is a pattern to these generating equations. >> Looking at the equations already in the database for the sequences >> listed above, I was unable to determine a general method for generating >> equations which would fully describe sequences derived in this manner. >> I'm not sure if it's o.k. to submit equations that partially describe a >> sequence to the OEIS, so I'd like to find one which fully describes my >> sequence before submitting it. Any help is appreciated. Thanks! > > > > >

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[math-fun] Happy New Eban!
by Marc LeBrun 26 Dec '05

26 Dec '05

Here's a traditional holiday sampling of factoids about 2006 gleaned from the On-Line Encyclopedia of Integer Sequences: Among other things, 2006 is the... A022664 coefficient of x^14 in Product (1-n*x^n)^4. A030015 9th element in the binomial transform of 1,2,3,5,7,11,13,17,19,23,.... A035939 number of partitions of 38 into parts not of the form 7k, 7k+3 or 7k-3. (Also number of partitions with at most 2 parts of size 1 and differences between parts at distance 2 are greater than 1.) A066967 total of the odd parts in all partitions of 16. A078392 sum of the GCDs of each partition of 25. A080149 number for which both 2006^2+1 and 2006^2+3 are prime. (Hardy and Littlewood conjectured that the set of such twin primes is infinite.) A085945 number of subsets of {1,2,3,4,5,6,7,8,9,10,11} with relatively prime elements. A102628 number of partitions of 27 whose number of parts is relatively prime to 27. A006933 the 23rd "eban" number (names in English containing the letter "e" are banned). (The next "eban year" won't be until 2030.)

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: [math-fun] Squares on a Cube
by hartmut.holzwart＠allianz.de 26 Dec '05

26 Dec '05

Ich werde ab 24.12.2005 nicht im Büro sein. Ich kehre zurück am 30.12.2005. Ich werde Ihre Nachricht nach meiner Rückkehr beantworten. Ursprüngliches Thema: [math-fun] Squares on a Cube

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Re: [math-fun] Squares on a Cube
by Daniel Asimov 26 Dec '05

26 Dec '05

Christian wrote: << Frank Rubin asks this interesting problem: --------------------------------- "It is possible to write a square on each of the 6 faces of a cube (such as a die) so that the 3 faces surrounding each of the 8 vertices sum to a square. For example, write 1 on two opposite faces, and 4 on the other 4 faces. The sum of the 3 faces surrounding each vertex is then 1+4+4=9, a square. Can this be done using the squares of 6 distinct whole numbers? If so, what set of 6 such squares has the smallest sum?" . . . >> On the principle that it's always easier to ask more questions than to answer one, how about the other 4 regular polyhedra? How about the 13 Archimedean solids? How about the regular and Archimedean tilings of the plane? And last (and least), how about regular polygons? (I.e., for a given n >= 3, do there exist distinct nonnegative integers {X_k | 0 <= k < n} such that for each index k we have (X_k)^2 + (X_(k+1))^2 is a square number?) -------------------------------------------------------------------------------- And there's no reason not to extend these questions to other regular tilings of surfaces or even higher-dimensional things: E.g., given the tiling of a torus by a rosette of 7 hexagons, same question for the 14 vertices. Or given a 4D regular solid -- say the 24-cell (composed of 24 octahedra, 6 per vertex), can we assign a square number to each of the 24 octahedra such that the sum of the 6 squares in the octahedra containing any vertex v add up to a square number? (Or why stick with only square numbers; why not higher powers?) --Dan

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[math-fun] Squares on a Cube
by Christian Boyer 26 Dec '05

26 Dec '05

Frank Rubin asks this interesting problem: --------------------------------- "It is possible to write a square on each of the 6 faces of a cube (such as a die) so that the 3 faces surrounding each of the 8 vertices sum to a square. For example, write 1 on two opposite faces, and 4 on the other 4 faces. The sum of the 3 faces surrounding each vertex is then 1+4+4=9, a square. Can this be done using the squares of 6 distinct whole numbers? If so, what set of 6 such squares has the smallest sum?" --------------------------------- As for the "3x3 magic squares of squares" unsolved problem, it is a system of 8 equations summing 3 squares. If the die is composed by the 6 integers: a² b², c², d², e² f² (couples a²/f², b²/d² and c²/e² being on opposite faces) the 8 equations are: a²+b²+c² = s² a²+c²+d² = t² a²+d²+e² = u² a²+e²+b² = v² f²+b²+c² = w² f²+c²+d² = x² f²+d²+e² = y² f²+e²+b² = z² I am able to find numerous solutions of 7 equations on 8, for example: 0²+132²+176² = 220² 0²+176²+468² = 500² 0²+468²+351² = 585² 0²+351²+132² = 375² 1200²+132²+176² = 1220² 1200²+176²+468² = 1300² 1200²+468²+351² = 1335² 1200²+351²+132² = 1580625 = 75² x 281 = (1257.229...)² But I am unable to find a complete solution of the 8 equations. Do you see any mathematical impossibility? Christian.

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[math-fun] Casey's theorem
by Fred lunnon 22 Dec '05

22 Dec '05

Around July 1997 under the subject line "geometry puzzle", there was a discussion in math-fun about Ptolemy's and Purser's theorems, which I did my best to derail by whacking Ptolemy into n-space using determinants. I recently came across this thread again, recalled failing to discover what Purser's theorem said, and while repairing the omission encountered Casey's theorem: [See http://mathworld.wolfram.com/CaseysTheorem.html; PursersTheorem.html] Four (oriented) circles in the Euclidean plane are tangent to a fifth (circle or line) just when t_12 t_34 +/- t_13 t_24 +/- t_14 t_23 = 0, where t_ij denotes the length of tangent common to i-th and j-th circles. For this to be interpreted correctly circles and lines must be oriented: think of them as being hairy on one side or the other, the hairyness matching at points of tangency. [Synthetic proofs not being my forte, I leave them to those better equipped.] Casey's theorem follows in a way similar to Ptolemy's, from the case n = 2 of another beautiful determinant which I didn't mention previously: Given n+4 oriented spheres in Euclidean n-space, the determinant of their common tangent squared lengths has value zero: | (t_ij)^2 | = 0. There is a puzzling feature of Casey's theorem which seems to have gone unremarked. There are two distinct ways in which a fifth (oriented) circle may be tangent to four others: e.g. in the simplest case where all four are external to each other and oriented alike, a tangent circle may be either surrounded by them or surrounding them; but the Casey criterion fails to distinguish these alternatives. How may the two cases be distinguished? More interestingly, what is an analogous criterion for four circles to be tangent to two others? Fred Lunnon 23/12/05

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[math-fun] Normality of sqrt(N)
by Schroeppel, Richard 21 Dec '05

21 Dec '05

A shocking abstract: http://arxiv.org/abs/math/0512404 On the simple normality to base 2 of the square root of s, for s not a perfect square. Richard Isaac Lehman College and Graduate Center, CUNY 13 pages We show that each number of the form, the square root of s for s not a perfect square, is simply normal to the base 2. The argument uses some elementary ideas from the calculus of finite differences. ---- My comments [Rich]: "Simply Normal" means only individual bit statistics are considered, not groups of bits. This comes down to 0 and 1 each being 50% in the limit. The author says at the end of the paper he thinks he can get "Normal". The method seems to be restricted to base 2, and depends on "there are only two different digits to consider". It also depends on the algebraic degree being 2 for the number being examined, although there might be some wiggle room here, with pseudo- Fibonacci-ratio numbers like x in x^3 = x+1 being accessible. The only previous result I'm aware of is trivial remark that the first N bits of sqrt2 must have at least O(sqrtN) 1 bits, simply to get enough cross product terms in the square to make the 1s in 1.11111111... . I looked over the paper, but can't figure out the hard part. It seems to have some good ideas, but I'm skeptical. For S=2, i.e. normality of sqrt2, his argument looks at the relationship between u = sqrt2 - 1 = .4142135... and v = u^2 = 3 - 2 sqrt2 = 1 - 2u = .1715728... Since v = 1-2u, the bit patterns for u and v are complementary and shifted one place. He somehow relates this to the relationship between (the bit pattern of) a number and its square. He looks at the temporary, diminishing, effect of a bit position in v, where a bit has only the effect 1/N in the average bit value of the first N bits of v, while the complementary shifted bit in u has a more permanent effect on the average bit value of u^2. The overall argument puzzles me, since I would expect to see more complicated machinery that looked at carries between bit positions, and at dot products of bit patterns. The exception for s a perfect square enters into the proof as simply that non-square s have non-terminating binary expansions. Rich

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[math-fun] Codes - note from Berlekamp
by Schroeppel, Richard 21 Dec '05

21 Dec '05

I received the note below from Berlekamp, about recent developments in Error Correcting Codes. Posted with his permission. --Rich -------------- (from Elwyn Berlekamp, Nov 22 2005) Coding theory remains an active and dynamic subject, as it has been for the past 58 years. Although I haven't published any papers in this area since 1996, I remain an interested spectator. Low Density Parity Check Codes were invented and studied by Bob Gallager in his PhD thesis in the early 1960s. Even now, most of what is known about them was published by him then. He became an assistant professor at MIT soon after he finished his PhD. Coding theory was then very active at MIT. I was Gallager's second PhD student. Shannon, Elias, and Wozencraft were also on my committee. No one favored further work on low density parity check codes at that time, because they were felt to be relatively well understood but impractical. Convolutional codes with sequential decoding enjoyed popularity in the early and mid 1960s. In the 1970s, Viterbi decoding became more popular, especially in military voice communication systems. Reed-Solomon codes, mostly using decoding algorithms with which I was personally involved, also became popular, especially in computer memory devices such as magnetic discs, optical discs, and magnetic tapes. They are now enjoy very widespread use. Turbo codes appeared in the late 1990s. A revival of interest in low density parity check codes began about the same time. It is true that, for certain channels including the white Gaussian noise channel, both turbo codes and low density parity check codes can achieve performance much closer to the Shannon capacity than any of their competitors. Each is now the topic of many papers. Many theorists study turbo codes in an effort to better understand their good performance, which was initially demonstrated almost entirely empirically. There are some system environments in which one or the other of these two types of coding systems is the clear winner. It is also true that most communication and computer memory systems, even new ones, continue to use older coding techniques such as Reed-Solomon. That's because real systems require tradeoffs among many parameters. In 1980 I published an overview entitled "The Technology of Error-Correcting Codes" in the Proceedings of the IEEE. Most of the considerations discussed there, including code rate, block length, latency, and interface issues, remain highly relevant. The advantages of low density parity check codes become decisive only at VERY long block lengths. A decade or two ago, memory was viewed as too expensive. Latency has now become a bigger concern. Turbo codes require considerably more decoding computation than Viterbi codes or RS codes. The relative performance advantage/disadvantage is strongly dependant on the code rate, and for various reasons not directly related to the coding, most memory systems use very high-rate codes, an environment which is especially favorable to Reed-Solomon. So I would say that the performance claims of turbo codes and low density parity check codes are true, but their impact is often exaggerated. Another major development of the 1990s is the work of Madhu Sudhan and others on list decoding of low-rate Reed-Solomon codes. This buys improved RS performance at the cost of more computation. It's primary use, to date, has been in computer science programs doing probabilistic search rather than in any of the more traditional applications of coding theory.

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[math-fun] minimising leaf-count
by wouter meeussen 21 Dec '05

21 Dec '05

It is possible to write any integer as sums of expressions containing only digits 1 or 2 or powers of 2, so that the exponents are also such expressions. Cfr Bill Dubuque's funny transliterations a few years back. But is it feasible to minimise the 'leaf count' of the result by expressing {n} as {2^k}-{n-2^k} if the count of 1-bits exceeds the count of 0-bits in the binary representation of n? So, 14 becomes -2+2^(2^2) and not just 2+2^2+2^(1+2) What I aim for is an over-all minimised leaf-count, not a minimisation at each step. a trial on n= 30197521 could give: 1 + 2^2^2 + 2^2^(1 + 2) + 2^(-1 + 2^2^2) + 2^(2 + 2^2^2) + 2^(1 + 2 + 2^2^2) + 2^(2 + 2^2 + 2^2^2) + 2^(1 + 2^(1 + 2)) + 2^(2 + 2^(1 + 2)) + 2^(2 + 2^2 + 2^(1 + 2)) + 2^(2^2^2 + 2^(1 + 2)) + 2^(-1 - 2^(1 + 2) + 2^(1 + 2^2)) but I doubt if it is optimal. W.

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[math-fun] yet an other Mersenne prime?
by wouter meeussen 18 Dec '05

18 Dec '05

keep an eye on http://mersenne.org/primenet/ a (as yet unconfirmed) candidate M43 between 25,000,000 and 33,000,000 (or abouts) seems to be in the verification stage. (found on http://www.mersenneforum.org/) ------- Mersenne Exponent Test State ------- Assigned in Tests Cleared Since Last Synchronization Factoring only : 9620 Factored composite : 860 Lucas-Lehmer testing : 56487 Lucas-Lehmer composite: 4577 Double-checking LL : 20150 Double-checked LL : 2671 Prime, UNVERIFIED : 1 ---------------------- ------- ---------------------- ------- TOTAL : 86257 TOTAL : 8109 'recent' finds: February 18, 2005: Prime M(25964951) May 15, 2004: Prime M(24036583) November 17, 2003: Prime M(20996011) W.

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