October 2015 - mathfun@mailman.xmission.com

[math-fun] Maximum size of a simplex in projective space
by Dan Asimov 23 Oct '15

23 Oct '15

I had heard some time ago of the tricky problem of finding the largest regular simplex in projective space P^(d-1). I.e., the largest number of lines through the origin of R^d that can all make the same angle with each other. For d = 2 the number is clearly 3. It's a bit surprising that for d = 3 the number is 6. (A fun exercise if you haven't seen it.) Also surprising that for d = 4 the number is still 6 (which I just learned today). (It's OEIS A002853 <http://oeis.org/A002853>OEIS A <http://oeis.org/A002853>002853.) It happens that P^3 is topologically the rotation group SO(3), so you'd think that maybe its symmetry would help, but apparently not. Here's a summary of the first few values (ranges for N(d) mean the exact value is not known): d 1 2 3 4 5 6 7<=d<=13 14 15 16 N(d) 1 3 6 6 10 16 28 28-30 36 40-42 d-1 0 1 2 3 4 5 6<=d-1<=12 13 14 15 The d-1 is in case you prefer to think of the problem as one of points in P^(d-1) instead of lines in R^d. Kind of surprising that you can't do any better than P^2 in P^3 !!! A little like maximizing the minimum distance for 5 points on S^2 does no better than 6 points (i.e., pi/2). —Dan

1 0

[math-fun] Terminology: "Characteristic Generating Function"?
by Bill Gosper 23 Oct '15

23 Oct '15

"Indicator Generating Function"? "Predicate Generating Function"? What is the name for the G.F. that encodes an integer sequence in the exponents with coefficients in {0,1} ? E.g., the OGF for the squares is Sum[n^2*x^n, {n, ∞}] == -((x*(1 + x))/(-1 + x)^3), but the thing I'm looking for is Sum[x^n^2,{n,0,∞}]==(1/2)*(1 + EllipticTheta[3, 0, x]). (Years ago I sent some of you the GF for expressions for k as the sum of n (optionally distinct) members of such a set, in terms of the GF predicating membership in the set. It turned out to be a superspecial case of a formula in Generatingfunctionology that was so general I didn't even recognize it as the same problem.) The reason I ask is that Neil just came up with 1/2 (-1+EllipticTheta[2,0,x]/x^(1/4)+EllipticTheta[3,0,x]) predicating the quartersquares<http://en.wikipedia.org/wiki/Multiplication_algorithm#Quarter_square_multip…> A002620 <https://oeis.org/A002620>, from which he got (1/(2 (-1 + x)^2))((-1 + x) x^(3/4) EllipticTheta[2, 0, x] + x (3 - x + (-1 + x) EllipticTheta[3, 0, x])) generating what we think is the number of volumes you can pump with n of the recently discussed rotors, arranged optimally. --rwg

4 3

Re: [math-fun] paper still not free after 136 years
by Axel Vogt 22 Oct '15

22 Oct '15

I can download it, https://goo.gl/3DZxi1, faking a location in Canada.

1 0

[math-fun] Martin Gardner testimonials
by James Propp 22 Oct '15

22 Oct '15

Who wants to post the lucky hundredth testimonial to Martin? It would be great if we could get the total up to 101 (or beyond) in time for the 101st anniversary of Martin's birth (tomorrow). Jim Propp

5 6

[math-fun] paper still not free after 136 years
by Neil Sloane 22 Oct '15

22 Oct '15

Dear Math Fun, This paper J. D. H. Dickson, <a href=" http://plms.oxfordjournals.org/content/s1-10/1/120.extract">Discussion of two double series arising from the number of terms in determinants of certain forms</a>, Proc. London Math. Soc., 10 (1879), 120-122. has some nice sequences in it. I HAVE a photocopy, which I made myself about 45 years ago, but I was hoping that by now it would be freely available on the web, so I could put a link to a free copy into sequence https://oeis.org/A047920 and many others that the paper contains. Fat chance - Oxford Univ. Press will rent it to me for one day for $39.00. No doubt I am naive, but I find this very annoying. It was published by the London Mathematical Society in 1879. Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane(a)gmail.com

6 7

Re: [math-fun] paper still not free after 136 years
by Dave Dyer 22 Oct '15

22 Oct '15

scan your copy and publish it yourself.

1 0

[math-fun] There's probably a simple derivation of this,
by Bill Gosper 22 Oct '15

22 Oct '15

but I sure missed it. Pi 1 4 2 E (-)! 1 + 2 n 4 Sum[------------------------------, {n, 0, oo}] == ----------- 2 n Pi -2 (1 + 2 n) Pi 3 E (1 - E ) Pi I got this as the q -> E^(-2 Pi) special case of Sum[DivisorSigma[1,2*n+1] q^n,{n,0,∞}]== Sum[(1 + 2 n) q^n/(1 - q^(1 + 2 n)), {n, 0, ∞}] == ( I (Derivative[1][DedekindEta][-((I Log[q])/(4 \[Pi]))]/ DedekindEta[-((I Log[q])/(4 \[Pi]))] - ( 3 Derivative[1][DedekindEta][-((I Log[q])/(2 \[Pi]))])/ DedekindEta[-((I Log[q])/(2 \[Pi]))] + ( 2 Derivative[1][DedekindEta][-((I Log[q])/\[Pi])])/ DedekindEta[-((I Log[q])/\[Pi])]))/(2 \[Pi] Sqrt[q]) As suggested by Sunday's "notational modularity" item, I have a bunch more special values of eta and eta' to crank into this. (I bet you can hardly wait.) Also, eta[q], eta[q^2], and eta[q^4] are polynomially related by Jacobi's aequatio identica satis abstrusa <https://www.google.com/search?client=safari&rls=en&q=aequatio+identica+sati…> , which should have an eta' analog. --rwg Vaguely related: ArcLength[{t, Sqrt[Cos[t]]}, {t, -π/2, π/2}] == Integrate[Sqrt[1 + (1/4)*Sin[t]*Tan[t]], {t, -Pi/2, Pi/2}] but the ISC says HUH? Really? (Connection: Area under that arc = π^(3/2)/(2 Sqrt[2] (1/4)!^2). )

3 2

[math-fun] Another infinitely differentiable egg (d/ds).
by Bill Gosper 21 Oct '15

21 Oct '15

A while ago we sought these, and it wasn't even Easter. Plot[{Sqrt[-ChebyshevT[4, x]], -Sqrt[-ChebyshevT[4, x]]}, {x, -(1/2) Sqrt[2 + Sqrt[2]], -(1/2) Sqrt[2 - Sqrt[2]]}, AspectRatio -> 3/4] Actually, this is how people remember and draw eggs. Real eggs are more nearly spheroids. The aspect ratio (spare us the obvious synonym) is known to vary, Spherical Egg Discovered in England - Gizmodo <https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uac…> , and the hen's egg: factors affecting shell splitting during boiling <https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uac…> fails to implicate it as a predictor. --rwg It takes a pretty smooth egg to earn a dentistry diploma.

1 0

[math-fun] Digit scanning question
by David Wilson 21 Oct '15

21 Oct '15

OEIS A036903(n) gives the number of digits of pi you must scan before seeing every possible n-digit substring. Over all infinite strings of digits, what is the expected number of digits you must scan before seeing every n-digit substring?

4 3

[math-fun] Wilson's sequence prime-cover problem
by Warren D Smith 21 Oct '15

21 Oct '15

0, 1, 3, 11, 39, 139, 495, 1763, 6279, 22363, 79647, 283667, 1010295, 3598219, 12815247, 45642179, 162557031, 578955451, 2061980415, 7343852147, 26155517271, 93154256107, 331773802863, 1181629920803, 4208437368135, 14988571946011,... > with a(n) = 3a(n-1) + 2a(n-2) for n >= 2. > for which primes p do all residues r appear in {a(n)} (mod p) > We'll shorten that to "a covers p". Formula: a(n) = ( [(3+sqrt(17))/2]^n - [(3-sqrt(17))/2]^n ) / sqrt(17) now, take a(n) mod p with p prime. Using quadratic reciprocity we see 17 is a square mod p if and only if p is a square mod 17: 0, 1, 4, 9, 16, 8, 2, 15, 13. (i) If so, then a(n) can be evaluated mod p using the above formula using plain old mod p arithmetic. Then by Fermat little theorem, a(n) mod p has period=p-1 (or a divisor of p-1), and hence cannot cover. (ii) If not (p mod 17=3,5,6,7,10,11,12,14) then we need to go to an extension field to do the arithmetic. In that case the period should be p^2-1=(p-1)(p+1) or a divisor. Empirically according to Wilson, case (ii) usually yields coverage mod p, but not always (exceptions p=683, 1217, 2731, 11299, 48817) while case (i) never yields coverage. Example of case (i) is p=19, a(n) = 0, 1, 3, 11, 1, 6, 1, 15, 9, 0, 18, 16, 8, 18, 13, 18, 4, 10, 0, ... with period=18 and not covering since missing 2,5,etc. Example of case (ii) is p=5 with a(n) = 0, 1, 3, 1, 4, 4, 0, 3, 4, 3, 2, 2, 0, 4, 2, 4, 1, 1, 0, 2, 1, 2, 3, 3, 0, ... with period=24 and covering. Obviously in cases with period >> p, it is highly likely that coverage will happen, but not logically forced. Further, period=(p-1)(p+1)/k will be unlikely to happen for large divisors k, and more likely to happen for small k, so coverage is always going to be likely, but not logically forced. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

1 1