November 2017 - mathfun@mailman.xmission.com

Re: [math-fun] sockrilege
by Bill Gosper 05 Nov '17

05 Nov '17

On 2017-11-03 18:09, Neil Sloane wrote: > Quicker to look it up! https://oeis.org/A051288. > Duh, a month in the laboratory can often save an afternoon in the library. > I added the socksual interpretation. > Wow, I had never connected socksuality with foot-fetishism. But (sniff), all my socks are achiral, in more ways than one! "Think socks (those weird "chiral" kind.)" --rwg > 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 > > > On Fri, Nov 3, 2017 at 8:38 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > >> Our old friend STAN.K bought n pairs of white "chiral" (podal!) socks >> clearly stamped L and R, >> which his housekeeper nevertheless paired willy-nilly. Stan wondered, for >> n pairs, how likely >> were 0, 2, 4, ... mismatches. >> I couldn't remember how to map a washer-dryer onto an urn, so I just >> tabulated >> >> 2 >> >> 4 2 >> >> 8 12 >> >> 16 48 6 >> >> 32 160 60 >> >> 64 480 360 20 >> >> 128 1344 1680 280 >> >> 256 3584 6720 2240 70 >> >> 512 9216 24192 13440 1260 >> >> and fit a formula with Mathematica. The nth row represents n pairs, >> row sum = binomial(2n,2). Thus the 2nd row, 4 2, represents 2 pairs, >> LRLR,LRRL,RLLR,RLRL = 4 perfectos, and >> LLRR,RRLL = 2 mismatches. >> (The left column is how many perfectos for n pairs.) >> The empirical formula I got was >> StringReverse@"]j2-n,j,j[laimonitluM)j2-n(^2" >> which then Julian promptly derived in a single line: >> >> StringReverse@".LR dna RL neewteb esoohc ot syaw )j2-n(^2 eht yb >> deilpitlum >> neht si hcihw ,)!)j2-n(*2^!j(/!n si LR ro RL rehtie eb ot j2-n dna ,RR eb >> ot j ,LL eb ot sriap j esoohc ot syaw fo rebmun eht" >> >> I guess that makes it fairly clear why the left (perfecto) column is 2^n. >> And, why the right diagonal is 2n choose n. --rwg

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Re: [math-fun] Are you a Gammaturge?
by Dan Asimov 05 Nov '17

05 Nov '17

That sure is a lot of digits of accuracy for a Boolean variable. —Dan -----Original Message----- >From: Bill Gosper <billgosper(a)gmail.com> >Sent: Nov 5, 2017 4:12 AM >To: math-fun(a)mailman.xmission.com >Subject: [math-fun] Are you a Gammaturge? > >Out[344]= (2^(2/5) Sqrt[5] Gamma[1/5]^4)/(Gamma[1/10]^2 Gamma[3/10]^2) == >GoldenRatio > >In[345]:= N[%, 11111] > >Out[345]= True >--rwg >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] sockrilege
by Bill Gosper 04 Nov '17

04 Nov '17

Our old friend STAN.K bought n pairs of white "chiral" (podal!) socks clearly stamped L and R, which his housekeeper nevertheless paired willy-nilly. Stan wondered, for n pairs, how likely were 0, 2, 4, ... mismatches. I couldn't remember how to map a washer-dryer onto an urn, so I just tabulated 2 4 2 8 12 16 48 6 32 160 60 64 480 360 20 128 1344 1680 280 256 3584 6720 2240 70 512 9216 24192 13440 1260 and fit a formula with Mathematica. The nth row represents n pairs, row sum = binomial(2n,2). Thus the 2nd row, 4 2, represents 2 pairs, LRLR,LRRL,RLLR,RLRL = 4 perfectos, and LLRR,RRLL = 2 mismatches. (The left column is how many perfectos for n pairs.) The empirical formula I got was StringReverse@"]j2-n,j,j[laimonitluM)j2-n(^2" which then Julian promptly derived in a single line: StringReverse@".LR dna RL neewteb esoohc ot syaw )j2-n(^2 eht yb deilpitlum neht si hcihw ,)!)j2-n(*2^!j(/!n si LR ro RL rehtie eb ot j2-n dna ,RR eb ot j ,LL eb ot sriap j esoohc ot syaw fo rebmun eht" I guess that makes it fairly clear why the left (perfecto) column is 2^n. And, why the right diagonal is 2n choose n. --rwg

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[math-fun] Primitive roots
by James Propp 04 Nov '17

04 Nov '17

Anyone have a favorite recreational math tidbit that involves primitive roots? Three that I know about are card shuffling, radix expansions of fractions, and the Ducci process. (Shuffle a deck of 2n cards with successive perfect inshuffles. It might take 2n-1 iterations before the original order is restored; for what n does it happen? For what n does the binary expansion of 1/n repeat with period n-1? The Ducci process is described at https://en.m.wikipedia.org/wiki/Ducci_sequence ; for what n is it the case that the length of the eventual repeating cycle depends only on n and not on the specific choice of not-all-equal integers written at the vertices of the original n-gon? The answers to all three questions involve the issue of which primes p have 2 as a primitive root.) Jim Propp

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[math-fun] peculiar product constant
by Bill Gosper 03 Nov '17

03 Nov '17

In[103]:= Product[Sin[n π/(2 n + 1)], {n, ∞}]==0.7435726169400367135862 During evaluation of In[103]:= Product::div: Product does not converge. And, as usual, I am the Virgin Mary. No joy from ries (now with Lambert-W) nor ISC nor Finch. The (log of the) slows-to-a-crawl product converts to a ~1 digit/term sum: Sum[(4^k - 1 )*BernoulliB[2*k]*((1 - 4^k)*π^(4*k)*BernoulliB[2*k] - 2*(-1)^k*π^(2*k)*(2*k)!)/(4*k*(2*k)!^2), {k,∞}] =-0.2962888487994070997 (How often do you see squared Bernoulli numbers?) FullSimplify splits this into two divergent series, one of which it formally recognizes as a simple constant, while the Bernoulli² sum just sits there smirking while approaching a harmonic series. --rwg

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[math-fun] a small formula
by François Mendzina Essomba 03 Nov '17

03 Nov '17

Hello, This morning, I noticed this formula: ln(x)=Limit(k^(n)*((2*x^(1/k^n)-1)^(1/2)-1),n=infinity); we consider k integer, This formula seems all the more effective as k is large. Best regards.

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[math-fun] some formulas of pi in infinte product (2)
by François Mendzina Essomba 01 Nov '17

01 Nov '17

Hello, The first writing of my previous formulas was not very clear .. Their writing is more appropriate in introducing infinite limits. We will have: Pi=Limit(2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product((2-1/(cos(Pi*2^(-i-2)))^2)^(2^(i-1)),i=1..n))^(1/2^n))),n=infinity); Pi=Limit(2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product((1-tan(Pi*2^(-i-2))^2)^(2^(i-1)),i=1..n))^(1/2^n))),n=infinity); Pi=2^(n+5/2)*sqrt(1-1/(2^(2^(-n-1))*(product(2^(2^(i-1))/(1/cos(Pi*2^(-i-1))+1)^(2^(i-1)),i=1..n))^(1/2^n))); Pi/4=Limit(sqrt(3*2^(2*n+1)-sqrt(3)*2^(4*n+2)*sqrt(2^(-2^(-n-1)-4*n-1)/(product((1-tan(Pi*2^(-i-2))^2)^(2^(i-1)),i=1..n))^(1/2^n)+2^(-4*n-2))),n=infinity); Pi/4=Limit(sqrt(3*2^(2*n+1)-sqrt(3)*2^(4*n+2)*sqrt(2^(-2^(-n-1)-4*n-1)/(product((2-1/(cos(Pi*2^(-i-2)))^2)^(2^(i-1)),i=1..n))^(1/2^n)+2^(-4*n-2))),n=infinity); Best regards.

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[math-fun] Random sections of a cube
by James Propp 01 Nov '17

01 Nov '17

In "On random sections of the cube" (https://arxiv.org/pdf/math/9812123.pdf) Yossi Lonke considers the intersection of an n-cube with a randomly-chosen k-plane that goes through the origin (chosen from the unique rotationally-invariant probability measure on k-planes). Lonke shows (among other things) that the average number of vertices of a random 2-dimensional central section of the 3-cube is exactly (24/Pi) arctan (1/Sqrt(2)), or about 4.7. Our "Wall of Fire problem" is different: we average over all nonempty sections, not just the central ones. It can be shown that when k is n-1, the expected number of j-dimensional faces of a random cross-section is the same as the number of j-dimensional faces of the k-cube; moreover, we can restrict to just the k-planes of any particular orientation, and the result still holds. To what extent does this generalize to the case where the codimension is bigger than 1? Here's the first interesting case (I think): Fix a 2-plane in 4-space. Look at all the translates of that 2-plane that intersect a fixed 4-cube. The intersections are polygons. What can be said about the average number of sides? It is not obvious to me that the answer is independent of the original choice of a 2-plane. (Thanks to Neil Sloane for bringing the Lonke article to my attention.) Jim Propp

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