July 2020 - mathfun@mailman.xmission.com

[math-fun] Fwd: diGamma sums
by Bill Gosper 10 Jul '20

10 Jul '20

---------- Forwarded message --------- From: Bill Gosper <billgosper(a)gmail.com> Date: Fri, Jul 10, 2020 at 7:32 AM Subject: diGamma sums To: Wolfram Tech Support I don't know if there's a general theory, but FindIntegerNullVector seems to crank these out reliably. Sum[PolyGamma[0, k]^2/k^2, {k, Infinity}] == (EulerGamma^2*Pi^2)/6 + (11*Pi^4)/360 - 2*EulerGamma*Zeta[3] Sum[PolyGamma[0, k]^3/k^2, {k, Infinity}] == (-(1/6))*EulerGamma^3*Pi^2 - (11*EulerGamma*Pi^4)/120 + 3*EulerGamma^2*Zeta[3] + (1/6)*Pi^2*Zeta[3] + (15*Zeta[5])/2 Sum[PolyGamma[0, k]^2/k^3, {k, Infinity}] == -((EulerGamma*Pi^4)/180) + EulerGamma^2*Zeta[3] + (1/6)*Pi^2*Zeta[3] - (3*Zeta[5])/2 Similarly for HarmonicNumber vs PolyGamma. Perhaps you should offer such results via the function ConjectureExpand.-) —Bill

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[math-fun] One of Ron Graham’s neat papers
by Victor Miller 10 Jul '20

10 Jul '20

The largest hexagon is not regular! https://pdf.sciencedirectassets.com/272565/1-s2.0-S0097316500X01720/1-s2.0-…

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[math-fun] Ron Graham
by rcs＠xmission.com 10 Jul '20

10 Jul '20

Neil Sloane reports that Ron Graham has died. --Rich

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[math-fun] Dragon[Thue] empirical results
by Bill Gosper 07 Jul '20

07 Jul '20

D:=Dragon[Thue] d:=Dragon[1-Thue] Then Im[D] + Re[D] == 7/15 + Abs[D]^2, Im[d] + Re[d] == 7/15 + Abs[d]^2, D + I d == I, 113/15 + 15 Abs[d]^4 + 15 Abs[D]^4 == 16 Abs[d]^2 + 16 Abs[D]^2 . Test: D ~ (552141176700154938708639716987262633 + 449568033846806255846939851654337899 I)/ 1329227995784915872903807060280344576 d ~ (879659961938109617056867208626006677 + 552141176700154938708639716987262633 I)/ 1329227995784915872903807060280344576 In[258]:= {Im[D] + Re[D] | 7/15 + Abs[D]^2, Im[d] + Re[d] | 7/15 + Abs[d]^2, D + I d | I, 113/15 + 15 Abs[d]^4 + 15 Abs[D]^4 | 16 Abs[d]^2 + 16 Abs[D]^2} /. {D -> (552141176700154938708639716987262633 + 449568033846806255846939851654337899 I)/ 1329227995784915872903807060280344576., d -> (879659961938109617056867208626006677 + 552141176700154938708639716987262633 I)/ 1329227995784915872903807060280344576.} Out[258]= {0.753602251625348009164368437081203650 | 0.753602251625348009164368437081203650, 1.077167455980927276355373650781728282 | 1.077167455980927276355373650781728282, 0.*10^-37 + 1.000000000000000000000000000000000000 I | I, 14.3589819883670712349825400724735776 | 14.3589819883670712349825400724735776} It's a little disappointing that we don't get valuations of Dragon[Thue] and Dragon[1-Thue] in terms of Thue, the way we do with Hilbert[Thue] = Thue + I and Hilbert[1-Thue] = 1 - Thue + I, which follow from treating Hilbert[t] as a four(?) state automaton that produces one bit of x and one bit of y for every two bits of t. —rwg Unlike the superfluous constant Glaisher, Thue (or a similar name) deserves a definition in Mathematica.

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[math-fun] Incredible property of HeighwayDragon[Thue's Constant]
by Bill Gosper 06 Jul '20

06 Jul '20

Thue's Constant: In[186]:= 1/2 - Product[1 - 2^-2^k, {k, 0, ∞]}]/4 In[189]:= N[%186, 69] Out[189]= 0.412454033640107597783361368258455283089478374455769557573379415348794 Check: In[190]:= RealDigits[%, 16] Out[190]= {{6, 9, 9, 6, 9, 6, 6, 9, 9, 6, 6, 9, 6, 9, 9, 6, 9, 6, 6, 9, 6, 9, 9, 6, 6, 9, 9, 6, 9, 6, 6, 9, 9, 6, 6, 9, 6, 9, 9, 6, 6, 9, 9, 6, 9, 6, 6, 9, 6, 9, 9, 6, 9, 6, 6, 9, 9}, 0} Offer Dragon tasty dyadic rational: In[212]:= Round[%189, 2^-239] Out[212]= 364371599346514549123253464270244268067660704514265010052337909024213813/ 883423532389192164791648750371459257913741948437809479060803100646309888 In[213]:= $RecursionLimit = 9999; dragun@%212 Out[213]= {552141176700154938708639716987262633/ 1329227995784915872903807060280344576 + (449568033846806255846939851654337899 I)/ 1329227995784915872903807060280344576} D:=dragun[thue]. Then Re@D+Im@D == 7/15 + |D|² !! In[223]:= N[List @@ (Im[D] + Re[D] == 7/15 + Abs[D]^2) /. D -> %213[[1]], 33] Out[223]= {0.753602251625348009164368437081204, 0.753602251625348009164368437081204} —rwg

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Re: [math-fun] Number of ways to pick 4 points from an m X n grid?
by Keith F. Lynch 06 Jul '20

06 Jul '20

"Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: > Dan Asimov <dasimov(a)earthlink.net> wrote: >> One method I did not see listed in that linked MathWorld article >> is to choose the points from a standard normal distribution in the >> plane, i.e., with density given by >> p(x,y) = (2?)^(-1) exp(-(x^2 + y^2)/2). >> This is of course circularly symmetric, but it's not at all clear to >> me that four points picked from this distribution will have the same >> chance of their convex hull being a quadrilateral as four points >> picked from the uniform distribution on a disk. > It seems to be different. I've only started, but the first 3 > billion give me 0.64905, as contrasted with 0.704480 for a circle > and 0.694444 for a square. I'll let it run for a couple days. With a quarter trillion sets, i.e. a trillion random points, I get 0.6490403. At the halfway point I had 0.6490408, so probably all but the last digit are correct. > To make sure I'm generating a standard normal distribution in the > plane correctly, please tell me what the means of the r values > (distances from the origin) should be. I'm measuring the > arithmetic, quadratic ("RMS"), and cubic means. The respective means were 1.253315, 1.414214, and 1.554989. I hope those are close to what they should be. (I didn't track the geometric or negative power means, since those would be dominated by how close the one point closest to the origin happened to be.) The only other test I made was to count how many of the trillion random points were in each quadrant: 249999663573, 250000431639, 249999574409, and 250000329923. Those sum to 999999999544. The remaining 456 points must have been on the borders of the quadrants, i.e. with an x or y value of 0. More specifically, given how theta was generated, x must have been positive and y must have been zero. And indeed 10^12 / 456 is close to the expected 2^31. As always, I checked for parallel lines and vertical lines. There were none. The only divisions by zero was when I took a log to get the r value. I prevented that by replacing any zero, before taking the log, with 10^-10. Again, I generated the pseudo-random numbers using Unix's random() function, which gives 31 bit integers. This time, after every billion sets (4 billion points) I switched to exclusive-oring the randoms with a different *pair* of distinct rows of the same order-32 Hadamard matrix as last time, again with its -1 entries replaced with 0s, and the leftmost bit (which was always 1) stripped off. This is to keep the random numbers from repeating. There are 496 (32 choose 2) such pairs of rows. Each such pair, when exclusive-ored together, has 16 1-bits and 15 0-bits. Each of the 122760 (496 choose 2) *pair* of pairs differs from every other pair of pairs in 12, 16, or 20 bit positions. If you want to take it further, or search for bugs, I can email you my C code. >> Also on Sunday, I posted: >> For the circular five-point case I calculated 0.3561808623 and >> posted 0.356181. That page doesn't give the solution. Assuming >> the solution is in the same form as the circular four-point case, >> 1 - (i / (j pi^2)), where i and j are integers, (35 and 12 in the >> 4-point case), if I try 1 - (305 / (48 pi^2)), I get 0.356188312... >> which differs from my answer by 7 in the sixth digit. Does anyone >> know if that's the correct formula? > That was with about 9 billion sets. With 59 billion (and > counting) I get 0.35618796, so it's looking like it really is > 1 - (305 / (48 pi^2)), as it differs from that formula by only > 3 in the *7th* digit. That's a hell of a way to find a formula. With 216 billion sets, I get 0.356188363, even closer to that formula. That's the probability of a convex pentagon given five random points in a circle. The probability of a quadrilateral with one point in it is 0.548822606, and the probability of a triangle with two points in it is 0.094989031. I got 1089 vertical lines and no parallel lines in that run.

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Re: [math-fun] OEIS-inspired research project for an undergrad
by Steve Witham 04 Jul '20

04 Jul '20

> From: Allan Wechsler <acwacw(a)gmail.com> > Date: 7/3/20, 5:02 PM > > Look at the logarithmic scatterplot of A001414. > > Explain the feathery diagonal bands at the bottom edge. [& a couple later posts] I want to start by saying, patterns in semi-random looking computer plots can always be artifacts of the plotting process. Sometimes not artifacts but real (but boring) phenomena aliased against pixels, bin sizes, etc. But I think the feathers are real. Hans Havermann had already done:https://oeis.org/A001414/a001414.png which is more spread out. My version... http://www.mac-guyver.com/switham/2020/07/OEIS_A001414/A001414_lin_log.png ...has white gaps in the upper black lines that I believe are artifacts. Havermann's doesn't have the same artifacts, but possibly has others. The clear upper lines are n (the primes), n/2, n/3, n/4... but there is a dark band at sqrt(n). This is a log-log plot instead of linear-log: http://www.mac-guyver.com/switham/2020/07/OEIS_A001414/A001414_log_log.png Differently interesting at the lower edge. Higher up, you can see sqrt(n), sqrt(n)/2, maybe sqrt(n)/3, but I can't convince myself there's an n^(1/3). Â --Steve

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[math-fun] OEIS-inspired research project for an undergrad
by Allan Wechsler 04 Jul '20

04 Jul '20

Look at the logarithmic scatterplot of A001414. Explain the feathery diagonal bands at the bottom edge.

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Re: [math-fun] math-fun Digest, Vol 209, Issue 9
by Stuart Anderson 03 Jul '20

03 Jul '20

"But how about the famous expression 1^2 + 2^2 + 3^2 + ... + L^2 = N^2 for L = 24 and N = 70, the unique nontrivial such equation. QUESTION: --------- Can a 70x70 square torus be tiled with one KxK square tile for each K in the range 1 ? K ? 24 ??? Possibly in a non-parallel fashion? ?Dan " Tiling or dissecting a given shape, with successive single squares of sides 1,2,3... n seems difficult - there are no known examples, however if we consider a 'patch' of tiles, a collection of squares, adjacent to each other, except at the boundary of the patch then it is possible. Adam Ponting has a method for tiling 1 to n^2 squares where n is an odd number. The key is to create a chessboard like nxn matrix with entries 1 to n^2 and then translate adjacent matrix entries into adjacent squares. see https://demonstrations.wolfram.com/PontingSquarePacking/ http://www.adamponting.com/squaring-the-plane/ http://www.adamponting.com/square-packing/ http://www.adamponting.com/update-ponting-packing/ http://www.adamponting.com/rectiling-i/ Stuart > >

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Re: [math-fun] A COMPLETELY different kind of "squared square"
by Dan Asimov 03 Jul '20

03 Jul '20

Ross Millikan posted this on Math StackExchange (July 26, 2018) in answer to a question about squares that are sums of three squares, giving families of such equations. Can either of these be turned into a periodic tiling of a square torus (not necessarily parallel to the edges of the square that the torus is based on)? ----- We know that (n+1)^2 − n^2 = 2n+1, so pick your favorite Pythagorean triple a^2 + b^2 = c^2 with c odd. Let c^2 = 2n+1, n = (c^2−1)/2, and (*) a^2 + b^2 + n^2 = (n+1)^2. If you pick a triple with c even, we can use (n+2)^2 − n^2 = 4n + 4, so we can let n = (c^2 − 4)/4 and have (**) a^2 + b^2 + n^2 = (n+2)^2. If c is even, c^2 is divisible by 4, so the division will come out even. ----- —Dan

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