December 2011 - mathfun@mailman.xmission.com

[math-fun] The Count of Heron Triangles with Diameter<N grows like N^(1+o(1)) when N-->infinity
by Warren Smith 12 Dec '11

12 Dec '11

The Count of Heron Triangles with Diameter<N grows like N^(1+o(1)) when N-->infinity ==================Warren D. Smith====December 2011================================== I believe I finally proved this. But in order to do it, I had to call in artillery support from one of the heaviest guns in mathematics, the "Green-Tao theorem." Just standing near that big gun has practically deafened me. And I suspect that the way I did that -- the "Counting Lemma" below -- will have many other uses too -- I don't see why counting Heron triangles should be the only use. A "Heron triangle" is a triangle with integer sidelengths and integer area. It is "primitive" if gcd(sidelengths)=1. Its "diameter" is its longest side. It was shown by Wm. F. Cheney: Heronian triangles, Amer. Math'l. Monthly 36,1 (1929) 22-28 that triangles with integer sides and rational area always have integer area which always is divisible by 6. (Fred Lunnon, myself, and Jan Fricke all rediscovered Cheney's theorem during the last 3 years... and quite likely Cheney himself rediscovered it too, since I find it hard to believe Brahmagupta did not know that.) THE MAIN THEOREM: The count of primitive (and also of nonprimitive, for that matter) Heron triangles with Diameter<N (more precisely the count of their sidelength triples a,b,c with a>=b>=c) when N-->infinity, is bounded between two functions of N which both grow like N^(1+o(1)). REMARK: Whether the triangles are primitive or not, does not matter because it is easily seen that the primitive and nonprimitive counts can at most differ by a factor bounded by a bounded power of logN at most, which is N^o(1). PROOF OF MAIN THEOREM LOWER BOUND (EASY): It has been known apparently since the ancient Indian mathematician Brahmagupta (598-668 AD) that every primitive Heron triangle (with sides a,b,c) arises from a = (n*n + k*k)*m; b = (m*m + k*k)*n; c = (m + n)*(m*n - k*k); and then divide out gcd(a,b,c) to primitivize it. Furthermore, this procedure for(n=1; (2*n+1)*(n+1)*n<=U; n++){ for(m=n+1; m*n*(m+n)<=U; m++){ for(k=1; k*k*(m+2*n)<=m*n*n ; k++){ a = (n*n + k*k)*m; b = (m*m + k*k)*n; c = (m + n)*(m*n - k*k); G = gcd(a,b,c); a /= G; b /= G; c /= G; output(a,b,c); }}} Will generate every primitive Heron triangle exactly once in the limit U-->infinity. Hence... CountOfHeronTrianglesOfDiameterUpToN > PositiveConstant * N. QED COUNTING LEMMA (DIFFICULT BUT HIGHLY USEFUL): Let R be a random integer [i.e. selected uniform in {1,2,3...,X} in the limit X large] or a random square [i.e. R=y*y where y selected uniform in {1,2,3...X} in same limit] or indeed a random Jth power for any fixed integer J>0. Let K be a fixed integer>=1, for example K=4. Let R=F1*F2*F3*...*FK be a random factorization of R into K parts, each a positive integer (different orderings counted multiply; "random" means uniform selection from all possible such factorizations). Let E be a fixed integer K-tuple, for example (-1, 1, 1, 1). THEN: The probability that the vector dot product F.E is zero, is asymptotically X^(o(1) - J/K) or less, in the limit X-->infinity. PROOF will be the bulk of this post, see below, but we defer it temporarily. PROOF OF MAIN THEOREM UPPER BOUND FROM THAT LEMMA: >From Heron's formula (Hero of Alexandria, lived about 10-70 AD) the area of triangle with sides a,b,c obeys 16*area^2 = (a+b+c) * (a+b-c) * (a+c-b) * (b+c-a). Choose a random square R as above as a putative value of 16*area^2. Plainly R<N^4 hence there are at most N^2 possible values of R. Choose a random factorization N=F1*F2*F3*F4. We can solve for a,b,c from F2,F3,F4 via linear equation solve. (This may yield rational a,b,c but the denominators if so are bounded so that does not affect any counts by more than a constant factor.) The probability that F1 really equals a+b+c=F2+F3+F4 is then seen via the LEMMA to be at most N^(o(1)-4/4) = N^(o(1)-1). We only get a Heron triangle if this probability comes true. Finally, use the fact proven by Carl Severin Wigert (life 1871–1941; see http://en.wikipedia.org/wiki/Divisor_function) that the number of divisors of N is <N^o(1), more precisely 2^(logN/loglogN + o(1)), [and hence the number of 4-factor factorizations of N is also N^o(1)]. Hence, the total number of Heron triangles of diameter<N is N^2 * N^(o(1)-1) * N^o(1) = N^(1+o(1)) at most. QED Finally, to prove the difficult lemma. The proof will have 3 main ingredients. INGREDIENT 1: Linear equations in primes (or supersets thereof) =============================================================== THEOREM. Let S denote the set of primes, or any superset of them, within the positive integers. Let A be a kXn integer matrix, and let B be an integer k-vector. We ask the following basic question: are there integer-valued vectors x so all coordinates of Ax+B lie in S, and if so, how often? Answer: If no two rows of A are linearly dependent, and if (A,B) is not trivially obstructed -- that is, if for each prime p there exists an integer vector x causing all coordinates of Ax+B to be nonzero mod p and there are an infinite set of integer vectors x such that all coordinates of Ax+B are positive -- then: there exist an infinite set of x so that all coordinates of Ax+B are simultaneously is S, and further the multidimensional density of such x is just what you'd naively expect up to a constant factor. Specifically, for S=primes the number of vectors x that work with N/2<MinCoord(x)<=MaxCoord(x)<N is asymptotically (N^k)/(logN)^n * PositiveConstant where the constant depends on A and B (and is calculable explicitly); and for S a superset of the primes this bound increases by a factor at most of order (logN)^n [obviously]. REMARK: The linear dependence clause means you cannot use this to prove an infinity of twin primes exist. (Roughly speaking, you need at least 2 more unknowns than equations to solve linear equations in primes.) But you can use it to prove an infinity of all-prime K-term arithmetic progressions exist for any fixed K. ABOUT THE PROOF. This theorem was proven by Benjamin J. Green & Terence Tao: Linear equations in primes, Annals of Maths. (2) 171,3 (2010) 1753-1850 (98 pages!), http://arxiv.org/abs/math/0606088 depending on two conjectures they called MS(s) and GI(s). They proved conjecture MN(s) in Benjamin J. Green & Terence Tao: The Moebius function is strongly orthogonal to nilsequences, to appear, Annals of Maths, (23 pages) http://annals.math.princeton.edu/articles/3163 http://arxiv.org/pdf/0807.1736 And the conjecture GI(s) was proved in 2010 by B.J.Green+T.Tao+Tamar Ziegler: An inverse theorem for the Gowers $U^{s+1}[N]$-norm, submitted Annals of Maths http://arxiv.org/abs/1006.0205 http://arxiv.org/abs/1009.3998 (116 pages!) http://arxiv.org/abs/0911.5681 to appear in Glasgow J. Math. 49 pages. I warn you I have essentially no idea how Green, Tao, & Ziegler did it. This is over 200 pages of absolutely brutal mathematics involving super-advanced combinatorics, harmonic analysis, and number theory, at a fields medal level. Fortunately for the purpose of proving the Heron triangle result, I think this earlier weaker result Antal Balog: Linear equations in primes, Mathematika 39,2 (1992) 367-378 MR1203292 (93m:11103) suffices, which is only 12 pages. An earlier even weaker result of the same ilk (which is not good enough, far as I can see) is P.G.L. Dirichlet's prime density theorem (1837), see http://en.wikipedia.org/wiki/Dirichlet's_theorem_on_arithmetic_progressions Here is a brief description of the Green+Tao result, plagiarised+altered from math'l reviews: The paper under review is a landmark contribution to analytic number theory. The authors establish a vast multilinear generalization of the classical Dirichlet theorem on primes in arithmetic progressions. Consider the following problem: Let P denote the set of primes. Let A be a $k\times n$ integer matrix, and let B be an integer vector. We ask the following basic question: are there integer-valued vectors x so all coordinates of $Ax+B$ are prime, and if so, how often? Answer: If no two rows of A are linearly dependent, and if (A,B) is not trivially losing -- that is, if for each prime p there exists an integer vector x causing all coordinates of Ax+B to be nonzero mod p and there are an infinite set of integer vectors x such that all coordinates of Ax+B are positive -- then there exist an infinite set of x so that all coordinates of Ax+B are simultaneously prime, and further the multidimensional density of such x is just what you'd naively expect up to a constant factor (and the constant factor always is calculable explicitly, with more work). INGREDIENT 2: Kalai's algorithm for generating random numbers in factored form: =============================================================================== Adam Kalai: Generating Random Factored Numbers Easily, Journal of Cryptology 16,4 (2003) 179-193 http://ttic.uchicago.edu/~kalai/papers/old_papers/factorcryptology.pdf Kalai's algorithm is incredibly simple. Here it is: Input: Integer n>0. Output: Exactly-uniformly random number 1<=r<=n with factorization into primes. Step 1: Generate a sequence n>=s1>=s2>=...>=sL by choosing s1 in {1,2,3,...,n} uniformly at random, then and s[i+1] uniformly at random from {1,2,3,...,s[i]}, until reaching 1. Step 2: Let r be the product of the PRIME s[i]'s. Step 3: If r<=n, output r with probability r/n. Otherwise, RESTART. COMMENT ON ALGORITHM: Runs in expected number O(log(n)^2) of primality tests and O(log(n)) restarts. Note the "triangular density rejection step" in step 3. This triangular is not uniform, but within the interval [n/2, n] it is uniform to within a factor of sqrt(2). INGREDIENT 3: Known facts about max number of divisors an integer N can have ============================================================================ We already mentioned Wigert's bound #divisors(N) <= N^o(1). Actually, if you want to try to get the strongest possible form of the thing I am sloppily calling "o(1)" then look into Erdos-Kac theorem of 1939-1941 and related work about how many prime factors random integers have and its asymptotic distribution. http://en.wikipedia.org/wiki/Erdos–Kac_theorem Peter DTA Elliott: Probabilistic Number Theory vols 1 & 2, Springer 1980, (GMW 239&240); QA241.7.E55. This seems to be a stronger result that would allow us to safely ignore numbers N with more than about sqrt(logN)*loglogN prime factors as so rare that they cannot affect the count much. Anyhow something like this ought to exist and be stronger than the Wigert bound, but I do not need it. FINALE: HOW TO MAKE A PROOF OUT OF THOSE 3 INGREDIENTS: ======================================================== We use Kalai's algorithm to generate (the Jth powers of) uniform random numbers. Then we argue using Green-Tao-Ziegler plus Kalai's analysis that the probability that the 4th factor is the sum of the other 3 factors, is at most about what you'd expect it to be (up to distortion by constant and log^power factors at most, asymptotically... anyhow clearly N^o(1) factors at most). Note, in Kalai's algorithm, certain things get distorted away from uniform density. The distortion in the interval [n/2, n] is at most by a factor of sqrt(2). If T things are multiplied the density distortion factor could be as large as log(X) * 2^(T/2). There might be considerably larger distortion on the decrease side -- this bound is for increase-factor -- but that is ok since we are only asking for an upper bound on an integrated probability density. However, T for us is going to be at most T=O(logX/loglogX) by the Wigert bound on #prime factors of X. Also the Wigert bound on #divisors and hence #factorizations of X into K parts (K and J are fixed, remember) will be X^o(1). These distortions are going to alter the Green-Tao-Ziegler count estimates (which were based on uniform density) by at most a factor X^o(1). QED. REMARK: I think the "o(1)" more precisely should be PositiveConstant / loglogN + lower order terms which explains why our computers haven't clearly realized that the Heron Triangle Count is N^(1+o(1)). The computer counts think it is behaving more like N^1.4. The reason is loglogN goes to infinity too slowly for computers to realize that o(1) goes to 0. [Using Erdos-Kac or something like that, my expression for the "o(1)" could probably be improved to something better, but I think it'd still slow enough to blur the vision of computers.] (End of proof). -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Another counting problem
by Dan Asimov 11 Dec '11

11 Dec '11

>From this Sunday's New York Times Magazine section: << WHO MADE THAT? The Stop Sign Wasn’t Always Red By HILARY GREENBAUM and DANA RUBINSTEIN How a 2-by-2-feet sheet of metal with black lettering on a white background evolved into a red hexagonal icon. >> < http://www.nytimes.com/pages/magazine/index.html > --Dan ________________________________________________________________________________________ It goes without saying that .

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[math-fun] Diophantine equation a^4+b^4=d*c^2 and the specialness of 17, 68, 82...
by Warren Smith 10 Dec '11

10 Dec '11

If d is fixed, then any solution (a,b,c) yields a boring infinite set of solutions (a*k, b*k, c*k^2). Here are non-boring solutions with d=17: a=1, b=2, d=17, c=1 a=2, b=13, d=17, c=41 a=38, b=43, d=17, c=569 a=314, b=863, d=17, c=182209 a=859, b=1186, d=17, c=385241 a=2297, b=12134, d=17, c=35732401 If both b and d are fixed, then a^4+b^4=d*c^2 is an "elliptic curve" enabling you to generate an infinite set of rational solutions (a,c) from just a few (indeed usually only one) starter solutions by using the elliptic curve group. This observation is much more powerful than Desbove's method of getting new solutions from old. Then by multiplying by appropriate denominators we get an infinite set of integer solutions now with b not fixed. I do not know of any parameterized family of non-boring solutions.

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Re: [math-fun] math-fun Digest, Vol 106, Issue 11
by Tom Duff 10 Dec '11

10 Dec '11

On Tue, 6 Dec 2011, math-fun-request(a)mailman.xmission.com wrote: > Date: Tue, 6 Dec 2011 16:17:33 -0800 (GMT-08:00) > From: Dan Asimov <dasimov(a)earthlink.net> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Counting problem > Message-ID: > <10736092.1323217054054.JavaMail.root(a)elwamui-norfolk.atl.sa.earthlink.net> > > Content-Type: text/plain; charset=UTF-8 > > Ha -- and in my spare moments today I counted 87 (by hand). > > So, can we split the difference and agree on 86 ? > > --Dan > > > Allan wrote: > > << > As luck would have it, this afternoon I had a boring staff meeting in which > I enumerated all of them ... I think. And the answer is 85 ... I think. I found 88. Here they are. Where is my mistake? (Each string has 3 for a 30 degree angle, 6 for 60, 9 for 90. These are the lexicographically smallest representatives of their equivalence classes.) 333333333333 33333333336 3333333339 3333333366 3333333636 333333369 3333336336 333333639 333333666 33333399 3333363336 333336339 333336366 33333669 33333696 33333939 3333633336 333363339 333363366 333363636 33336369 33336396 33336639 33336666 3333699 33339339 3333969 333633366 333633636 33363369 33363396 33363639 33363666 33363936 3336399 33366339 33366366 3336669 3336696 3336939 33393339 3339369 333999 336336336 33633639 33633666 3363399 33636339 33636366 33636636 3363669 3363696 3363939 3363966 33663366 3366369 3366396 3366639 3366666 336699 3369339 336969 336996 3393639 339399 339669 36363636 3636369 3636639 3636666 363699 363969 3663666 366399 366669 366696 366939 366966 369369 369396 369639 36999 393939 39699 666666 66699 66969 9999 -- Tom Duff. Without the Daemons, there ARE no Penguins.

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Re: [math-fun] math-fun Digest, Vol 106, Issue 11
by Dan Asimov 10 Dec '11

10 Dec '11

Another math-fun member, using a brief Mathematica program, also found 88. It looks as though that's the correct answer. (Though, I haven't yet compared the two sets of output, the other of which is in terms of 1's, 2's, and 3's.) --Dan Tom Duff wrote: << On Tue, 6 Dec 2011, math-fun-request(a)mailman.xmission.com wrote: > Date: Tue, 6 Dec 2011 16:17:33 -0800 (GMT-08:00) > From: Dan Asimov <dasimov(a)earthlink.net> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] Counting problem > Message-ID: > <10736092.1323217054054.JavaMail.root(a)elwamui-norfolk.atl.sa.earthlink.net> > > Content-Type: text/plain; charset=UTF-8 > > Ha -- and in my spare moments today I counted 87 (by hand). > > So, can we split the difference and agree on 86 ? > > --Dan > > > Allan wrote: > > << > As luck would have it, this afternoon I had a boring staff meeting in which > I enumerated all of them ... I think. And the answer is 85 ... I think. I found 88. Here they are. Where is my mistake? (Each string has 3 for a 30 degree angle, 6 for 60, 9 for 90. These are the lexicographically smallest representatives of their equivalence classes.) >> ________________________________________________________________________________________ It goes without saying that .

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[math-fun] The count of Heronian tetrahedra of diameter<N is bounded between two power laws
by Warren Smith 10 Dec '11

10 Dec '11

Oops! I had made a MAPLE script to illustrate the proof in action & verify it. The script worked, and the proof was verified... but... when MAPLE fully simplified the stuff in the script, it turned out 2 of my 4 parameters in my "4-parameter family" of Heron tetrahedra... vanish! They cancel themselves out! So it is really only a 2-parameter family and my power law like N^0.4999 is really going to be a considerably weaker power law, I must retract and correct that: REVISED CONCLUSION: We have found a class of Heron tetrahedra (actually several classes, but just the first will suffice) with sides s01=a, s23=b, s02=s03=s12=s13=c such that the number of primitive tetrahedra in this class with max(a,b,c)<N grows with N at least as the same order as N^(1/14). ----- #Here is a MAPLE script which computes members of my #infinite 4-param family of Heron tetrahedra and verifies their Heronitude: #Warren D. Smith, Dec 2011 aa := (n,m) -> 16*m*n*(n^2+m^2)*(n^4+m^4)*(n^4+14*m^2*n^2+m^4)* (n^4+6*m^2*n^2+m^4)*(n^2-4*m*n-m^2)* (n^2+4*m*n-m^2)*(17*n^8+28*n^6*m^2+166*n^4*m^4+28*n^2*m^6+17*m^8); bb := (n,m) -> 6*(n^2+m^2)*(n^8-4*n^6*m^2+134*n^4*m^4-4*n^2*m^6+m^8)* (5*n^4+6*m^2*n^2+5*m^4)*(n^8+60*n^6*m^2+134*n^4*m^4+60*n^2*m^6+m^8)*(m-n)^3*(m+n)^3; cc := (n,m) -> (n^8-4*n^6*m^2+134*n^4*m^4-4*n^2*m^6+m^8)*(m^4+n^4)* (n^8+60*n^6*m^2+134*n^4*m^4+60*n^2*m^6+m^8)*(17*n^8+28*n^6*m^2+166*n^4*m^4+28*n^2*m^6+17*m^8); print("the following (a,b,c,area1,area2,vol) tuples should be all-integer:"); for n from 1 to 9 do for m from n to 9 do a := aa(n,m): b := bb(n,m): c := cc(n,m): g := gcd(a, gcd(b,c)): a := abs(a/g): b := abs(b/g): c := abs(c/g): if(a<>0 and b<>0 and c<>0) then print( "(", a, b, c, #sides sqrt(a^2*(2*c+a)*(2*c-a)/16), #triarea sqrt(b^2*(2*c+b)*(2*c-b)/16), #triarea sqrt( a^2*b^2*(4*c^2-a^2-b^2)/144 ), #tetvol ")" ); fi;od;od;

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[math-fun] The count of Heronian tetrahedra of diameter<N is bounded between two power laws
by Warren Smith 09 Dec '11

09 Dec '11

THE COUNT OF HERONIAN TETRAHEDRA IS BOUNDED BETWEEN TWO POWER LAWS ================Warren D. Smith Dec 2011================================= In this post, I will prove that the number of primitive Heronian tetrahedra ["Heronian" means tetrahedra with all sides, face-areas, and volume all integer, and "primitive" means gcd(sidelengths)=1] with diameter<N grows with N at least proportionally to N^0.4999. (Fred Lunnon helped somewhat, but I'd say this proof is 99% due to me.) It had previously been proven that the number of Heronian tetrahedra was infinite. However, I think all the previous proofs employed integer points on elliptic curves and therefore only yield a count-growth law of order loglog(N) roughly. While we know that loglogN goes to infinity, this has never actually been observed. The present power-law count-growth is the correct kind of growth, since a power-law upper bound can be shown trivially. What is the correct power? -- that is now the question. The PROOF is divided into parts I, II, II, IV: I. Solutions of "the 4.2.2 diophantine equation" x^4+y^4=u^4+w^4: ================================================================= This is discussed here by Tito Piezas: http://sites.google.com/site/tpiezas/013 A.Gerardin (about 1910?) 1-parameter family: (a+3*a^2-2*a^3+a^5+a^7)^4 + (1+a^2-2*a^4-3*a^5+a^6)^4 = (a-3*a^2-2*a^3+a^5+a^7)^4 + (1+a^2-2*a^4+3*a^5+a^6)^4. This shows that the number of primitive solutions (x,y,u,w) with max(x,y,u,w)<N grows at least as fast as order N^(1/7). The following 2-parameter solution is mentioned by Hardy & Wright: Intro to the theory of numbers (I think it is supposed to be a cleaned up version of Euler, but not sure where it ultimately came from). [E.g. see GH Hardy, EM Wright, JH Silverman: Intro to the theory of numbers, EQ13.7.10 and thereabouts] as "the simplest... but not in any sense complete": (a+b)^4+(c-d)^4 = (a-b)^4+(c+d)^4 where a = n*(m^2+n^2)*(18*m^2*n^2-m^4-n^4), b = 2*m*(m^6+10*m^4*n^2+m^2*n^4+4*n^6), c = 2*n*(4*m^6+m^4*n^2+10*m^2*n^4+n^6), d = m*(m^2+n^2)*(18*m^2*n^2-m^4-n^4). It shows that the number of primitive solutions (x,y,u,w) with max(x,y,u,w)<N grows at least as fast as order N^(2/7). J.Stegall 2-parameter family (sort of): (a^3+b)^4 + (1+a*d)^4 = (a^3+d)^4 + (1+a*b)^4, where d=2*a*c-b and {a,b,c} must satisfy a quadratic in b, namely b^2*c - 2*a*b*c^2 - a^4+3*a*2*c + 2*a^2*c^3 - 1 = 0. Solving this in the rationals entails making its discriminant a square, hence involves the elliptic curve z^2 = (1+a^4)*c+3*a^2*c^2-a^2*c^4 If we do the substitution a = (n+1)/(n-1), then one solution is c = 9*n^4/(2*n^8-2*n^6+n^4-2*n^2+2) from which we can compute other rational points. Then we can solve the quadratic for b trivially using quadratic formula. Just using this c-from-n formula without trying to generate other rational points on the elliptic curve(s), we would obtain a 1-parameter family of solutions also showing power-law growth, but even slower than Gerardin. Euler: found several solutions. Dickson Introduction 60-62. II. Certain kinds of "imperfect pythagorean boxes": =================================================== If x=p^2*q^2-r^2*s^2, y=2*p*q*r*s, z=p^2*r^2-q^2*s^2 then x^2+y^2 = (p^2*q^2+r^2*s^2)^2 and y^2+z^2 = (p^2*r^2+q^2*s^2)^2 are squares. [But x^2+z^2, in general, is nonsquare.] If, further, p^4+s^4=q^4+r^4 (see solutions in part I), then x^2+y^2+z^2 = (p^4+s^4)*(q^4+r^4) is a square too. Indeed, more generally, this is true if (p^4+s^4)*A^2=(q^4+r^4)*B^2 which means we can take any solution of p^4+s^4=q^4+r^4 and then multiply p and s by A and q and r by B, then the above x,y,z will still be such that x^2+y^2, y^2+z^2, and x^2+y^2+z^2 all will simultaneously be squares. Using the (I)solution from Hardy & Wright we see that this shows that the number of primitive (x,y,z) with max(x,y,z)<N satisfying these three simultaneous squareness conditions, grows with N at least as fast as order N^(0.4999). Proof: from (I) the number of (p,q,r,s) with max<M grows like M^(2/7) at least; the number of (A,B) with max<J grows like J^2, and then for (x,y,z) we have N of order M^4*J^4. Pick J=N^0.24999 and M=N^0.00001 to get M^4*J^4=N and J^2*M^(2/7)=N^0.499982857... QED. III. Certain kinds of Heron tetrahedra: ======================================= A tetrahedron with these sidelengths s01=a, s23=b, s02=s03=s12=s13=c has exactly two kinds of triangle faces (both isoceles) (a,c,c) and (b,c,c). Its two face areas arise from 16*TriArea^2 = (a+2*c) * (a)^2 * (2*c-a) 16*TriArea^2 = (b+2*c) * (b)^2 * (2*c-b) and these both are integer iff T^2 = (2*c+a)*(2*c-a) = 4*c^2 - a^2 U^2 = (2*c+b)*(2*c-b) = 4*c^2 - b^2 both are squares. Its volume is given by 144*TetVol^2 = a^2*b^2*(4*c^2-b^2-a^2) Thus for the tetrahedron to be Heronian we also need that S^2 = 4*c^2 - b^2 - a^2 also be a square. Note that multiplying by 4 does not affect squaritude. Comment: Jan Fricke discovered that this kind of tetrahedron always arises from gluing together 4 congruent copies of a tetrahedron with all faces being right triangles. IV. Certain Heron tetrahedra from certain kinds of imperfect pythagorean boxes. =============================================================================== Now suppose we have an imperfect pythagorean box as in (II) having diagonal c where c^2 = x^2+y^2+z^2. We can now make a tetrahedron with sides s01=2*z, s23=2*x, s02=s03=s12=s13=c. This will be Heron of the type in (III) since T^2 = 4*c^2-4*z^2 = 4*(x^2 + y^2) = square U^2 = 4*c^2-4*x^2 = 4*(y^2 + z^2) = square S^2 = 4*c^2 - 4*x^2 - 4*y^2 = 4*z^2 = square. I have not checked the following as carefully: Related Heron tetrahedron (i): sides s01=s02=c, s13=s23=squareroot(x^2+y^2)=integer, s03=z, s12=2*x; Related Heron tetrahedron (ii): sides s01=c, s02=squareroot(y^2+z^2)=integer, s13=squareroot(x^2+y^2)=integer, s03=z, s12=x, s23=y; Related Heron tetrahedron (iii): sides s01=2*z, s02=s12=c, s03=s13=squareroot(y^2+z^2), s23=x. CONCLUSION: We have found a class of Heron tetrahedra (actually several classes, but just the first will suffice) with sides s01=a, s23=b, s02=s03=s12=s13=c such that the number of primitive tetrahedra in this class with max(a,b,c)<N grows with N at least as the same order as N^0.4999. --end. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] Convolution and continued fractions
by Bill Gosper 09 Dec '11

09 Dec '11

MikeS> Great! I'd love to see the rest of the slides. They're pretty cryptic: http://gosper.org/mitslides.pdf MikeS> I take it the denominator in each sum is the generalized q-Pochhammer symbol described at the bottom of the page here:http://mathworld.wolfram.com/q-PochhammerSymbol.html Precisely. (eqn 18) Here's the lhs written out: (c104) PRODCONTRACT(INTOPROD(MAKEPROD(D94[1]))); 2 n n - 1 2 -- - n/2 /===\ n n 2 | | 1 (- 1) %i q | | -------------------------------------- | | i i inf i = 0 %i (- 1) q i i + 1 ==== (------------ + 1) ((- 1) q + 1) \ c (c + %i) > ------------------------------------------------------------------ / n ==== c n = 0 (d104) --------------------------------------------------------------------------------- 2 n n - 1 2 -- + n/2 /===\ n 2 | | 1 %i q | | ------------------------------------------ | | i i + 1 inf i = 0 i i + 1 %i (- 1) q ==== ((- 1) q + 1) (1 - ----------------) \ c > --------------------------------------------------------------- / n ==== c n = 0 Numerically, for c=1, q=1/2, (c107) CF(RECTFORM(APPLY_NOUNS(POCHSIMP(APPLY_NOUNS(SUBST([Q = 5.0b-1,C = 1,INF = 15],D94[1])))))); (d107) [1, 2, 4, 8, 16, 32, 64, 128, 260] (This container was filled at the manufacturing facility, but some settling may have occurred during shipping.) An interesting partial fractions identity in those slides is Sum[(x + a)^-k/k*x^(k - n)/(n - k), {k, 1, n - 1}]== Sum[a^(k - n)/(n - k)/ k*((-1)^(n - k)/Binomial[n, k] - k/n)*(1/(x + a)^k + (-1)^(n - k)/x^k), {k, 1, n - 1}] but Mma turns it into a useless pile of garbage before I can feed it to << HolonomicFunctions.m (whose authors have since kindly shown me how to smuggle into their subfunctions before Mma finds out it has Sums.). On Thu, Dec 8, 2011 at 5:46 AM, Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote:> I just stumbled upon slides from some forgotten talk containing a matrix> sketch of> an identity which specializes to> 2> n> inf 2 -- - n/2> ==== n n 2> \ (- 1) %i q> (c + %i) > ---------------------> / n %i> ==== c (- --, - q; - q)> n = 0 c n> ------------------------------------> 2> n> inf 2 -- + n/2> ==== n 2> \ %i q> > --------------------> / n %i q> ==== c (- q, ----; - q)> n = 0 c n>> => 1> c + -----------------> c 1> - + -------------> q c 1> -- + --------> 2 c 1> q -- + ---> 3 .> q .> .>>> (Sorry, no Mma. It was being unusably, inconceivably stupid on this.)> --rwg>> On Thu, Dec 1, 2011 at 10:59 AM, Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote:>>> On Wed, Nov 30, 2011 at 2:51 PM, Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com> <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote:> This promised to be quite tedious, except that it was possible to take>> "unconscionable> shortcuts" (I think I called them). I think I can dig up some results, if you want.>>>>>> Yes, please; I don't get any hits on searching my archive for>> "unconscionable" except this email.>> -->> Mike Stay - metaweta(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=metaweta%40gmail.com> <http://gosper.org/webmail/src/compose.php?send_to=metaweta%40gmail.com>http://www.cs.auckland.ac.nz/~mikehttp://reperiendi.wordpress.com>>>> You can pretty much find everything searching for ContinuedFractionK. But>> I see a big garble in a msg containing>> [...] And constant/linear gives 1F0>>>>>> ContinuedFractionK[e, a n + b, {n, 1, Infinity}] ==>>>> b e>> e Hypergeometric0F1[2 + -, --]>> a 2>> a>> ------------------------------------,>> b e>> (a + b) Hypergeometric0F1[1 + -, --]>> a 2>> a>> and a pattern emerges. But not quite.[...]>> ------------->> The pattern in question predicted the values of p and q in the pFqs based>> on>> the degrees of the two polynomials in the ContinuedFractionK, to which I'd>> found a>> strange exception. Unfortunately, I seem to have spazzed the editing and>> stated>> the same example twice in succession.>> --rwg>> I think the aforementioned shortcuts somehow combined the determination of>> the>> ODE integration constants with the subsequent limit-taking.>>

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Re: [math-fun] Convolution and continued fractions
by Bill Gosper 08 Dec '11

08 Dec '11

On Wed, Nov 30, 2011 at 2:51 PM, Bill Gosper <billgosper(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=billgosper%40gmail.com>> wrote:> This promised to be quite tedious, except that it was possible to take "unconscionable> shortcuts" (I think I called them). I think I can dig up some results, if you want. Yes, please; I don't get any hits on searching my archive for "unconscionable" except this email. -- Mike Stay - metaweta(a)gmail.com <http://gosper.org/webmail/src/compose.php?send_to=metaweta%40gmail.com>http://www.cs.auckland.ac.nz/~mikehttp://reperiendi.wordpress.com You can pretty much find everything searching for ContinuedFractionK. But I see a big garble in a msg containing [...] And constant/linear gives 1F0 ContinuedFractionK[e, a n + b, {n, 1, Infinity}] == b e e Hypergeometric0F1[2 + -, --] a 2 a ------------------------------------, b e (a + b) Hypergeometric0F1[1 + -, --] a 2 a and a pattern emerges. But not quite.[...] ------------- The pattern in question predicted the values of p and q in the pFqs based on the degrees of the two polynomials in the ContinuedFractionK, to which I'd found a strange exception. Unfortunately, I seem to have spazzed the editing and stated the same example twice in succession. --rwg I think the aforementioned shortcuts somehow combined the determination of the ODE integration constants with the subsequent limit-taking.

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Re: [math-fun] Counting problem
by Dan Asimov 07 Dec '11

07 Dec '11

Ha -- and in my spare moments today I counted 87 (by hand). So, can we split the difference and agree on 86 ? --Dan Allan wrote: << As luck would have it, this afternoon I had a boring staff meeting in which I enumerated all of them ... I think. And the answer is 85 ... I think. On Tue, Dec 6, 2011 at 12:36 PM, Richard Guy <rkg(a)cpsc.ucalgary.ca> wrote: > This seems to be a particular case of the necklace problem > which was solved by Hazel Perfect in Math Gaz, when? > (more than half a century ago -- not in MR) R. > > > On Tue, 6 Dec 2011, Dan Asimov wrote: > > What are all the ways that 30-, 60-, and 90-degree angles can be arranged >> about the origin in the plane? >> ________________________________________________________________________________________ It goes without saying that .

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