April 2013 - mathfun@mailman.xmission.com

[math-fun] point sets with integer [or sqrt(integer)] distances
by Warren D Smith 04 Apr '13

04 Apr '13

http://ginger.indstate.edu/ge/Graphs/GEOMETRY/fig5.png http://ginger.indstate.edu/ge/Graphs/GEOMETRY/fig6.png there is more work of this kind...

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Re: [math-fun] Tan[Sin[x]]-Sin[Tan[x]] puzzles
by Bill Gosper 04 Apr '13

04 Apr '13

Mike Hirschhorn (using Maple) solved both puzzles off list. Rich>asin(tanh(x))-atan(sinh(x)) is even flatter! --- Quoting Bill Gosper <billgosper(a)gmail.com>: >From Neil Bickford (under protest): Plot[Tan[Sin[x]]-Sin[Tan[x]], {x, 0, π/2}] is ultraflat near 0 and goes berserk at π/2. Ultraflat: x^7/30+(29 x^9)/756+(1913 x^11)/75600+O[x]^12 Berserk: Local minima and maxima cluster at π/2, vibrating between roughly .5 and 2.5. What limits, if any, do they approach? Note that by analytic continuation, an arbitrarily short segment of the flat part near zero completely predicts the berserk part west of π/2. Are all the terms of the series approximated above positive? Plot[ArcSinh[Tan[y]] - ArcTanh[Sin[y]], {y, -2, 2}, PlotRange -> All] looks that flat for awhile. Then, a Big Bang violates analytic continuation. So THAT's what caused it! --rwg

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[math-fun] Combinatorial line of 10 primes
by Keith F. Lynch 04 Apr '13

04 Apr '13

On March 18th in http://cp4space.wordpress.com/2013/03/18/generalising-erdos-conjecture/ Adam P. Goucher writes: > If you can find a combinatorial line of 10 primes, I'll be impressed. 39402*9*7*3 where * takes the values 0 through 9 (same value for each *), i.e. 39402090703, 39402191713, ..., 39402999793, are all primes. The increment had to be divisible by 210, since otherwise at least one of the 10 numbers would have been divisible by 2, 3, 5, or 7. And it had to be all 1s and 0s when expressed in base 10 for the 10 numbers to constitute a combinatorial line by your definition. 101010 was the smallest number which met those criteria. And the above is the smallest set of primes with that increment. That doesn't necessarily mean it's the smallest such set of primes, as there are several other allowable increments less than those primes. Fortuitously, 101010 is also divisible by 13. Otherwise chances would be 10/13 that one of any set of 10 candidate prime numbers is divisible by 13. As it is, chances are 10/11 that one of any set of 10 candidate prime numbers is divisible by 11. I did confirm that no smaller sequence of 10 primes exist with an increment of of 1111110, the smallest increment that's divisible by 2,3,5,7, and 11. (It too is also divisible by 13. Unsurprisingly, given that it's simply 11 times the previous increment.) That makes me wonder, is it always possible to find a number which is all 1s and 0s in base 10 that is divisible by whatever numbers you choose? It's not possible to pick and choose which numbers you want it to be divisible by, given that no number in that form can be divisible by 2 but not by 5 or vice versa. But other than that, is it always possible to pick and choose which primes it is and isn't divisible by? For instance to find a number in that form which is divisible by only the primes (other than 2 and 5, if applicable) that divide your home phone number? That reminds me, some years ago I came up with a proof that you can always find a power of 2 that begins with any desired digit string. In fact, you an always find any number of them. Powers of 2 that begin with your home phone number, powers of 2 that begin, not just with a few billion 1s and 0s, but that begin with the specific pattern of 1s and 0s that comprise your favorite movie DVD. I'm also reminded that on my wall I have a newspaper ad that says, "10101, TENTENONE, The Prime Number." It's an ad for a luxury condo with that street address. Of course 10101 isn't a prime number in base 10 or in base 2. I soon proved it wasn't a prime in any base whatsoever (if the base is an integer and is greater than 1). Getting back to combinatorial lines, the idea seems to be to regard successive digits of an N-digit number as coordinates on successive axes in an N-dimensional space. If several such numbers fall on a line in that space, they constitute a combinatorial line. However, the definition implies that (when the numbers are put in order) the digits in each place must always remain the same or increase in sync. Why can't digits decrease? Why isn't 14 23 32 41 a valid combinatorial line? Why can't some digits increase faster than others? Why isn't 11 23 35 47 a valid combinatorial line?

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[math-fun] Futility Closet
by Jason 04 Apr '13

04 Apr '13

This blog is thoroughly delightful, with brain teasers and historical oddities. In particular, I thought the solution to this tiling problem was quite lovely, and generalizes to other shapes: http://www.futilitycloset.com/2012/09/30/tiling-task/ Upon further reading, it appears to be related to the mutilated chessboard problem.

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[math-fun] chromatic number kicks proof theorist butt
by Warren D Smith 04 Apr '13

04 Apr '13

>Perhaps a proof theorist can give a truly horrific bound? >Charles Greathouse --It sounds like what you are asking for, is this question: Is there a function F(N), such that every N-chromatic graph, contains an N-chromatic subgraph with at most F(N) vertices? I believe the answer is "no." In particular F(3) does not exist because any odd-cycle is 3-chromatic but no subgraph is.

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[math-fun] How often does every bit matter?
by Warren D Smith 03 Apr '13

03 Apr '13

OK, I may have the correct conjecture now. Let C2=0.660161815846869573927812110014... be the Hardy Littlewood twin prime constant described here http://en.wikipedia.org/wiki/Twin_prime#First_Hardy.E2.80.93Littlewood_conj… Then let B = exp(-2*C2 / ln2) = 0.14884878474999065378100135978. Then my conjecture is, that the proportion of primes which are "every bit matters" primes, (i.e. if you alter any bit in their binary expansion, you get a composite; here we do not regard leading 0s as "bits") is B. Not hugely convincingly, I computed that among the 50847534 primes below 10^9, 7179982, which is 14.121%, are every-bit-matters primes. Perhaps going to 10^12 or something would be more convincing. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] How often does every bit matter?
by Warren D Smith 03 Apr '13

03 Apr '13

Letting C be the fraction of primes with the property that altering any bit destroys primality, it occurs to me I can prove limsup C <= 3/8 by arguing that every prime p=1 mod 60 has p+2=composite, p=7(60) ==> p-2 composite, p=11(60) ==> p-2 compo, p=13(60) ==> p+2 compo, p=23(60) ==> p-2 compo, and similarly for -1,-7,-11,-13,-23. One could of course get much stronger upper bounds by considering congruence classes to appropriate higher moduli by computer; the best moduli for this purpose perhaps are of the form 2^j * 3*5*7*11*13*... I still lack a proof that lim C exists, merely have limsupC<=3/8 and liminfC>10^(-29). But I suspect an appropriate subsequence of the upper bounds alluded to above, converges to limC from above.

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[math-fun] Minimal squares in a rectangle
by ed pegg 03 Apr '13

03 Apr '13

B. Felgenhauer has found minimal solutions for dividing a rectangle into squares, up to size 300x300. http://int-e.eu/~bf3/squares/ http://int-e.eu/~bf3/squares/view.html#13,11http://demonstrations.wolfram.c… An n×m rectangle can be tiled with a minimum k squares. Can rectangle of size 2(n×m) be tiled with k-1 or fewer squares? Conjecture-- No. That's an item at http://math.stackexchange.com/questions/240277/ So far, no counterexample is known. --Ed Pegg Jr

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[math-fun] quintic AGM
by Joerg Arndt 03 Apr '13

03 Apr '13

A message from the department of mindless q-staring. Here is what looks to me like the paper I have been waiting for a long time: Tim Huber: A theory of theta functions to the quintic base http://arxiv.org/abs/1304.0684 The AGM relations are given on p.11. Two more fine papers: http://arxiv.org/find/math/1/au:+Huber_T/0/1/0/all/0/1 Best, jj

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[math-fun] direction of fastest convergence
by Bill Gosper 03 Apr '13

03 Apr '13

Using path-invariance, we can, in polynomial vs exponential time, exhaustively search a taxicab-metric "sphere" for the direction of fastest convergence. Starting with the matrices i: {{(i + k) (i + n)/((1 + i) (i + j + k + n)), -j k n}, {0, 1}}, j: {{(j + k) (j + n)/((1 + j) (i + j + k + n)), -i k n}, {0, 1}}, k: {((i + k) (j + k)/((1 + k) (i + j + k + n)), i j n}, {0, 1}}, n: {{(i + n) (j + n)/((1 + n) (i + j + k + n)), i j k}, {0, 1}} which in the {n,k} plane give Gauß's identity 2F1(a,b,c;1)=Gamma Gamma/Gamma/Gamma, and adding the h dimension via Sunday's "cheap trick", h: {{(h (h + i - j))/((h + k) (h + i + n)), -1 + j + k + n}, {0, 1}}}, i: {{((h + i - j) (i + n))/((1 + i - j - k) (h + i + n)), ((-1 + h + k) (-1 + j + k + n))/(-1 - i + j + k)}, {0, 1}}}, j: {{((i - j - k) (j + n))/((-1 + h + i - j) (j + k + n)), ((-1 + h + k) (-1 + h + i + n))/(-1 + h + i - j)}, {0, 1}}}, k: {{(k (-i + j + k))/((h + k) (j + k + n)), -1 + h + i + n}, {0, 1}}}, n: {{((i + n) (j + n))/((h + i + n) (j + k + n)), -1 + h + k}, {0, 1}}} then searching the neighborhood of starting point n=1,i=0,j=0,h=1,k=1 at a radius of 18 says that "best" is to take 6 n steps, 6 h steps, and 6 k steps, yielding the 6 bits/term identity (combining just one step of n, h, and k) π^2/6==Sum[(13 + 21*k)*π^(3/2)*k!^3/(64^(k + 1)*(1/2 + k)!^3), {k, 0, ∞}] which Mathematica converts to a useless pair of 4F3s. Notice how notationally and computationally inferior are both Sum and pFq notations vs this teensy matrix product In[893]:= {{(2/j + 4)^-3, 21 j/8 - 1}, {0, 1}} In[894]:= Dot @@ Table[%, {j, 3}] Out[894]= {{1/2744000, 23687/14400}, {0, 1}} In[895]:= 23687/14400 - π^2/6. Out[895]= -3.51129*10^-6 (~3 ppm from 3 terms). The Sum identity appears to express π^2 in terms of π^(3/2) ! But of course you don't need π to evaluate the summand. I quoted "best" above because, e.g., we could have combined the n, h, and k steps with a coordinate transformation, yielding fewer steps with messier matrices. Subjectivity persists. --rwg Soon: the "best" matrices (from this system) for csc πx, sin πx, π, and 1/π .

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