September 2019 - mathfun@mailman.xmission.com

Re: [math-fun] Random nonempty intersection of cube and plane, revisited
by Keith F. Lynch 28 Sep '19

28 Sep '19

Sorry it took me a while to finish my program that finds the average perimeters and areas of the intersection of a plane with a cube. On July 28th, I wrote: Okay, I think I figured it out. Where n is the order of the intersection polygon, 3 through 6, there are n intersection points between the cube and the plane, hence n! possible sequences of intersection points. Since the polygon is convex, I'm pretty sure that the correct order to visit the n intersection points, returning to my starting point (i.e. a Hamiltonian circuit) will simply be the one with the shortest total distance. Someone please politely tell me if I'm wrong. This got me the correct perimeter length. Correct as supported by the program's agreeing with Fred Lunnon's and my ideas of what the smallest and largest perimeters should be for each n-gon. Since it doesn't matter which intersection point I start on, I only have to try (n-1)! sequences. Make that (n-1)!-1, since there are two correct solutions (the same solution in opposite directions) and they can't both be in the last place. So for the largest possible n, 6, 119 tries. Make that (n-1)!/2 sequences. I precomputed all permutations of 1-3, 1-4, 1-5, and 1-6, and discarded the ones that were reversals of others already in the table. So for the largest possible n, 6, 60 tries. On July 29th I wrote: Once I know the order of the intersection points, all I have to do is run a line from one intersection point to each of the other intersection points, to divide the polygon into triangles, then Heron each of the triangles and add their areas together. This should have worked. I spent most of my time trying, without success, to figure out why it didn't work. I knew it wasn't working because I was getting impossibly large areas. (Except for triangles, where it appeared to be correct, given the maximum area it found.) I was hampered in debugging by the fact that I have no way of generating graphics. All I have to look at are lists of numbers. And I'm not a visual thinker, anyway. Eventually, I figured it must be that I was somehow mistaken that the sequence of vertices with the smallest sum of successive distances was the actual perimeter. I couldn't visualize how that could possibly be, but visualization isn't my strong suit. Yes, the perimeter maxima were correct, but that's no surprise given that a hypothetical incorrect vertex order with a smaller distance sum wouldn't affect that maximum. I looked forward to finding an exception, then drawing it by hand. So I implemented a variant of James Propp's suggestion that I rotate the polygon into the x,y plane, then convert from rectangular to polar coordinates, with the centroid as the origin, and sort by theta. Since at that time I only cared about the order of vertices, I instead simply discarded the z coordinate rather than doing a 3D rotation. (I am perhaps irrationally prejudiced against unnecessary trigonometry, as I'm under the perhaps outdated impression that trig makes programs very slow.) Much to my surprise, I discovered that the orders were the same. (At least for all of ten million random planes.) The flaw in my area algorithm was elsewhere. I continued looking for it until I started running out of time. (James Propp's talk is on August 7th, and today is the 6th.) Then I tried getting the area from my projection into the x,y plane, using the shoelace formula (which was suggested by both James Propp and Tom Duff). The projection is obviously going to be foreshortened, so I tried multiplying it by the slope of the original polygon to fix it. Apparently I computed the slope wrong, as the areas of triangles (hence presumably also all other figures) were wrong. Fortunately, Tom Duff also said, "In 3 dimensions you can either rotate everything into the plane or compute the projected areas in the xy, yz and zx planes and take the square root of the sum of the squares." Since it was easy to adapt my code which projected into the x,y plane and worked there into doing the same with the other two orthogonal planes, so I did it that way. Then I had a tenacious bug that was due to the modulo function in C being broken. -1 mod 3 should be 2, not -1. Once I changed at[l] += fabs(v2[pe[g][im][j]][(l+1)%3] * (v2[pe[g][im][(j+1)%g]][(l+2)%3] - v2[pe[g][im][(j-1)%g]][(l+2)%3])); to j1 = (j-1)%g; if (j1 < 0) j1 += g; at[l] += fabs(v2[pe[g][im][j]][(l+1)%3] * (v2[pe[g][im][(j+1)%g]][(l+2)%3] - v2[pe[g][im][j1]][(l+2)%3])); it started working perfectly. (That's the shoelace formula.) In retrospect I should have documented better, used more meaningful variable names, and unrolled those nested subscripts. But the program started two years ago as a quick and dirty hack to measure the average number of edges of the intersection polygon, so I used "alpine programming," by analogy with "alpine style mountain climbing," which means not bothering with those tedious base camps and pre-positioning supplies, but instead just racing for the summit. In my defense, mountains, unlike programming tasks, seldom suddenly get much taller. Anyhow, without further ado, here are the results, based on the average of two runs of a billion random planes. Again, the cube's side length is 2. Smallest, average, and largest perimeters: 3 sides: 0.0000 3.0613 8.4736 4 sides: 4.0001 7.1700 9.6550 5 sides: 6.8314 8.2032 9.6521 6 sides: 8.4853 8.6827 9.6511 All over average perimeter: 6.2841 Smallest, average, and largest areas: 3 sides: 0.0000 0.5496 3.4546 4 sides: 0.0001 3.2519 6.9252 5 sides: 3.0088 4.6743 6.9218 6 sides: 5.1975 5.8310 6.9240 All over average area: 2.8819 Proportion of each polygon type: 3 sides: 27.97% 4 sides: 48.68% 5 sides: 18.70% 6 sides: 4.65% I'm CCing James Propp so that he'll get this in time for tomorrow's talk even if he's on the digest option and no digest goes out before that talk. Tomorrow I'll return to trying to figure out why my original area formula isn't working.

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[math-fun] Number Theorist Fears All Published Math Is Wrong
by Ray Tayek 28 Sep '19

28 Sep '19

https://science.slashdot.org/story/19/09/26/223218/number-theorist-fears-al… Posted by BeauHD <https://twitter.com/BeauHD> on Thursday September 26, 2019 @11:30PM from the future-of-mathematics dept. An anonymous reader quotes a report from Motherboard: /Kevin Buzzard, a number theorist and professor of pure mathematics at Imperial College London, believes that it is time to create a new area of mathematics dedicated to the computerization of proofs <https://www.vice.com/en_us/article/8xwm54/number-theorist-fears-all-publish…>. The greatest proofs have become so complex that practically no human on earth can understand all of their details, let alone verify them. He fears that many proofs widely considered to be true are wrong. Help is needed. What is a proof? A proof is a demonstration of the truth of a mathematical statement. By proving things and learning new techniques of proof, people gain an understanding of math, which then filters out into other fields. To create a proof, begin with some definitions. For example, define a set of numbers such as the integers, all the whole numbers from minus infinity to positive infinity. Write this set as: ... , -2, -1, 0, 1, 2, ... Next, state a theorem, for example, that there is no largest integer. The proof then consists in the logical reasoning that shows the theorem to be true or false, in this case, true. The logical steps in the proof rely on other, prior truths, which have already been accepted and proven. For example, that the number 1 is less than 2. New proofs by professional mathematicians tend to rely on a whole host of prior results that have already been published and understood. But Buzzard says there are many cases where the prior proofs used to build new proofs are clearly not understood. For example, there are notable papers that openly cite unpublished work. This worries Buzzard./ "I'm suddenly concerned that all of published math is wrong because mathematicians are not checking the details, and I've seen them wrong before," Buzzard told Motherboard while he was attending the 10th Interactive Theorem Proving conference <https://itp19.cecs.pdx.edu/> in Portland, Oregon, where he gave the opening talk. "I think there is a non-zero chance that some of our great castles are built on sand," Buzzard wrote in a slide presentation <http://wwwf.imperial.ac.uk/~buzzard/one_off_lectures/msr.pdf>. "But I think it's small." -- Honesty is a very expensive gift. So, don't expect it from cheap people - Warren Buffett http://tayek.com/

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[math-fun] "Prodigal"s are mainstream.
by Bill Gosper 28 Sep '19

28 Sep '19

What some of us have long called "prodigal" seems to have existed all along under the name https://en.wikipedia.org/wiki/Product_integral . Prodigal : Integral :: Geometric mean : Arithmetic mean. E.g., productIntegral[x!, {x, 0, t}] == E^Integrate[Log[x!], {x, 0, t}] == (2 π/E^t)^((1 + t)/2) Hyperfactorial[t] In[1176]:= Product[k^k, {k, n}] Out[1176]= Hyperfactorial[n] —rwg

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[math-fun] EndoBob StarPants
by Bill Gosper 27 Sep '19

27 Sep '19

Unfortunately, In[1142]:= PolyhedronData["Endododecahedron", "Volume"] // N During evaluation of In[1142]:= PolyhedronData::notdef: PolyhedronData has no value associated with the specified argument(s). The faces of the endododecahedron <http://gosper.org/endogram.png> are the equilateral trousers <http://gosper.org/starpants.png> (waistband = inseam = outseam) of a regular pentagram. —rwg The volume of the endodec inscribed in the unit cube (its convex hull) is tiny: 2 - PolyhedronData["Dodecahedron", "Volume"]/ GoldenRatio^3 // FullSimplify Out[1128]= 1/(3 + Sqrt[5]) = ϕ⁻²/2 ~ 0.190983005625053. in disagreement with (at least the first edition of) The Symmetries of Things. Has it been corrected? https://en.wikipedia.org/wiki/Dodecahedron#/media/File:Endo-dodecahedron_ho… https://www.reddit.com/r/mathpics/comments/572fpu/endododecahedron/

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[math-fun] Hilbert's ≠ Peano's
by Bill Gosper 26 Sep '19

26 Sep '19

On 2019-09-22 06:15, Joerg Arndt wrote: > Even "Peano's curve" isn't unique: > Walter Wunderlich: {\"{U}ber Peano-Kurven}, > Elemente der Mathematik, vol.~28, no.~1, pp.~1-10, (1973). > http://sodwana.uni-ak.ac.at/geom/mitarbeiter/wallner/wunderlich/ > Don't let the German scare you, just check the images. Wow, in http://sodwana.uni-ak.ac.at/geom/mitarbeiter/wallner/wunderlich/pdf/125.pdf the 3⨉3 edge-traverser (Figur 5) is a perfect(ly confusing) hybrid of Peano's with Hilbert's! Gotta "Julianize" these. —rwg > For the n-dimensional version of (one of) the Peano curve(s): > https://jjj.de/fxt/demo/comb/index.html#peano-ndim > This is following > A.\ J.\ Cole: {A note on space filling curves}, > Software Practice and Experience, vol.~13, no.~12, (1983) > and > A.\ J.\ Cole: {A Note on Peano Polygons and Gray Codes}, > International Journal of Computer Mathematics, vol.~18, no.~1, > pp.~3-13, (1985). > These papers win both my "horrible notation" and "most useless example" award. > > Best regards, jj > > P.S. regarding Peanistic curves: > https://jjj.de/tmp-math-fun/all-R29-curves.pdf > https://jjj.de/tmp-math-fun/all-R29-tiles.pdf > View both files side-by-side. > > * Bill Gosper <billgosper(a)gmail.com> [Sep 20. 2019 10:10]: >> For decades I misapprehended Hilbert's spacefill as Peano's. Today at >> last, I corrected my web >> page: http://www.tweedledum.com/rwg/samhilbert.htm . The impetus was >> discovering that my >> blunder has been copied onto other websites. Unfortunately, with correct >> attribution.-} —rwg >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > _______________________________________________ > 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] cubic polynomials & symmetric functions
by Henry Baker 26 Sep '19

26 Sep '19

One of the standard parts of any algebra course shows that every "symmetric function" (actually, a symmetric polynomial) can be expressed as polynomial in various different basis symmetric polynomials, including the "elementary symmetric polynomials" [coefficients of product(z-ri)], the "power basis" polynomials [sum(ri^n)], etc. But this doesn't satisfy me, because I want to compute other "symmetric functions" of the roots of a polynomial. For example, if p(z)=(z-r1)*(z-r2)*(z-r3) is a cubic polynomial with *complex* roots, I'd like to be able to compute the area(r1,r2,r3)^2 of the triangle whose vertices are r1,r2,r3 in the complex plane. I don't seem to be able to do this, because the triangle area seems to require the *conjugate* operation in addition to the usual +,-,* operations. One elegant approach to the *power basis* comes from powers of the *companion matrix* for the polynomial. If C is the companion matrix for the polynomial p(z) above, then C is diagonalized by the *Vandermonde* matrix of the roots of p(z), and the eigenvalues of C are the roots r1,r2,r3 of the polynomial p(z). V^(-1).C.V = diag(r1,r2,r3) But (V^(-1).C.V)^n = diag(r1,r2,r3)^n, so C^n = V.diag(r1^n,r2^n,r3^n).V^(-1). Note that C^n can trivially be computed by rational operations -- i.e., without solving for r1,r2, r3. This means that the characteristic polynomial of (C^n) will provide the expansion of the powers of the roots in terms of the elementary symmetric polynomials of r1,r2,r3. In fact, for any polynomial q(z) (or indeed any convergent power series q[z]), we can compute q(C) (or q[C]), and produce another cubic polynomial whose roots are q(r1), q(r2),q(r3) (q[r1],q[r2],q[r3]). In particular, we could compute approximations to the matrix sin(C) whose characteristic polynomial roots are sin(r1), sin(r2), sin(r3), or exp(C) whose characteristic polynomial roots are exp(r1), exp(r2),exp(r3), or even log(C), whose characteristic polynomial roots are log(r1),log(r2),log(r3). I suspect that Newton himself would have found this insight to be very cool! We can also rationally compute a new polynomial ps(z) whose roots are (r1+r2),(r2+r3),(r3+r1), and also a new polynomial pp(z) whose roots are r1*r2, r2*r3, r3*r1. Since (r1-r2) is not symmetric, we can square it and rationally produce a polynomial pdiff(z) whose roots are (r1-r2)^2, (r2-r3)^2, and (r3-r1)^2. But none of these methods enables the calculation of area(r1,r2,r3)^2, which is a positive real number and is a "symmetric function" of r1,r2,r3. Triangle areas seem to require the introduction of *conjugates*, so we have symmetric functions such as abs(r1)^2+abs(r2)^2+abs(r3)^2 = r1r^2+r1i^2+ r2r^2+r2i^2+r3r^2+r3i^2. For the life of me, I haven't been able to find symmetric functions using +,-,*,conjugate that compute symmetric functions such as this length, or the areas of triangles. But neither have I been able to prove that it can't be done. I've searched the literature online, but haven't been able to find anything which treats this question. One thought is to compute the *polar decomposition* of the companion matrix C = U.P, where U is unitary and P is positive semidefinite. Then (C*).C = (P*).U*.U.P = (P*).P. But the problem is that the eigenvalues of C=U.P aren't the same as the eigenvalues of P, so we would need some way to relate the eigenvalues of P to those of C. In general, I'd like to see as many of the questions about polynomial roots and coefficients be put into *matrix form* whenever possible, even though that doesn't follow the history of polynomial roots very accurately. Similarly, the SVD of C doesn't help, either, because the unitary matrices for conjugate(C) aren't the same as those for C itself. Any suggestions?

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Re: [math-fun] Hilbert's ≠ Peano's
by Bill Gosper 25 Sep '19

25 Sep '19

On Tue, Sep 24, 2019 at 5:38 PM rwg <rwg(a)ma.sdf.org> wrote: > > > -------- Original Message -------- > Subject: Re: [math-fun] Hilbert's ≠ Peano's > Date: 2019-09-24 08:24 > From: Brad Klee <bradklee(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Reply-To: math-fun <math-fun(a)mailman.xmission.com> > > Subj:Re: [math-fun] Hilbert's ≠ Peano's > Isn't the valency of a multi-point limited by the valency of vertices > in the unit > cell tiling? Square unit cells seem to imply that quadruple points have > the > most possible degeneracy. Or am I wrong? That sounds right. It's true of Julian's sextuple pointer: That sounds right. It's true of Julian's *http://howwords.com/triangles/?a_=sexttri&radiusPct=90&nRotors=6251&dtPct=3&rotorSort=1&cmn_=32&cmt_=highcontrast&cmabs_=false&stepsPerCycle_=20&delayMS_=0 <http://howwords.com/triangles/?a_=sexttri&radiusPct=90&nRotors=6251&dtPct=3…>* (In a fleeting moment of mental clarity, I managed to "invert" this fractal. Sure enough, piecewiserecursivefractal returned preimage sextuples.) But when he sent it he said he can go higher than 6. Maybe with a wild tiling? —rwg > > Considering only tiling topology, it would also be possible to have > triple points, > but it looks like the symmetry of the unit pattern forbids this. The > curve snakes > across the unit cell between opposite corners, so divides any Z^2/3^n > lattice > into subsets 2*Z^2/3^n and 2*Z^2/3^n + 1. One associates to quadruple > points > and the other to double points. This is already visible in the nice > "tetraskelion" pic. > > > From the perspective of multipoints, Wunderlich "Figur 5" is the most > interesting. It looks to have double, triple, and quadruple points. > > --Brad > > > On Tue, Sep 24, 2019 at 7:38 AM Bill Gosper <billgosper(a)gmail.com> > wrote: > > > > Here's a raw sandbox with Peano and Wunderlich samplings: > > https://www.wolframcloud.com/obj/1fa6d29d-7451-4e68-ae7b-dded77390040 > > The "tetraskelion"s are dense with quadruple points. I need to figure > > out > > how, > > with Julian's tools, to make inversepeano and see if there are > > quintuple or > > sextuple points (which would be surprising). > > —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] Is superseeker broken? Am I?
by Bill Gosper 24 Sep '19

24 Sep '19

I submitted the following trivalued sequence, based on the Dragon Curve. I don't see a connection with any of its suggestions. ? —rwg ---------- Forwarded message --------- From: <superseeker-reply(a)oeis.org> Date: Wed, Sep 11, 2019 at 7:14 PM Subject: Re: To: <billgosper(a)gmail.com> Greetings from The On-Line Encyclopedia of Integer Sequences! https://oeis.org/ > lookup 2 1 3 2 1 2 3 2 2 1 2 3 1 2 3 2 2 1 3 2 1 2 2 3 2 1 2 3 1 a(n) = 2, 1, 3, 2, 1, 2, 3, 2, 2, 1, 2, 3, 1, 2, 3, 2, 2, 1, 3, 2, 1, 2, 2, 3, 2, 1, 2, 3, 1 Note: Your sequence does not directly appear in the OEIS. If it is of general interest, please submit it at https://oeis.org/Submit.html. # Transformations These sequences match transformations of the original query. T099 abs(a(n)), sorted with duplicates removed = 1, 2, 3 oeis.org/A000045 Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1. 0, 1, <1, 2, 3>, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155 oeis.org/A000041 a(n) is the number of partitions of n (the partition numbers). 1, <1, 2, 3>, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 31185, 37338, 44583, 53174, 63261, 75175, 89134, 105558, 124754, 147273, 173525 oeis.org/A000027 The positive integers. Also called the natural numbers, the whole numbers or the counting numbers, but these terms are ambiguous. <1, 2, 3>, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77 oeis.org/A001222 Number of prime divisors of n counted with multiplicity (also called bigomega(n) or Omega(n)). 0, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 3, 1, 2, 2, 4, 1, 3, 1, 3, 2, 2, 1, 4, 2, 2, 3, 3, 1, 3, 1, 5, 2, 2, 2, 4, 1, 2, 2, 4, 1, 3, 1, 3, 3, 2, 1, 5, 2, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 4, <1, 2, 3>, 6, 2, 3, 1, 3, 2, 3, 1, 5, <1, 2, 3>, 3, 2, 3, 1, 5, 4, 2, 1, 4, 2, 2, 2, 4, 1, 4, 2, 3, 2, 2, 2, 6, 1, 3, 3, 4, 1, 3, 1, 4, 3, 2, 1, 5, 1, 3, 2 oeis.org/A001221 Number of distinct primes dividing n (also called omega(n)). 0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 1, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 3, 1, 2, 2, <1, 2, 3>, 1, 2, 2, 3, 1, 2, 1, 2, 2, 2, 2, 3, 1, 2, 1, 2, 1, 3, 2, 2, 2, 2, 1, 3, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, <1, 2, 3>, 2, 1, 2, 1, 3, 2 ... 25666 total # Transformations as Deltas The deltas of these sequences match transformations of the original query. T030 a(n+2) - a(n) = 1, 1, 2, 0, 2, 0, 1, 1, 0, 2, 1, 1, 2, 0, 1, 1, 1, 1, 2, 0, 1, 1, 0, 2, 0, 2, 1 (as deltas) oeis.org/A301849 The Pagoda sequence: a sequence with isolated zeros in number-wall over finite fields. -1, 0, <1, 0, -1, 1, 1, -1, -1, 0, 1, 1, -1, 0, 1, -1, -1, 0, 1, 0, -1, 1, 1, 0, -1, -1, 1, 1, -1, 0>, 1, -1, -1, 0, 1, 0, -1, 1, 1, -1, -1, 0, 1, 1, -1, 0, 1, 0, -1, -1, 1, 0, -1, 1, 1, 0, -1, -1, 1, 1, -1, 0, 1, -1, -1, 0, <1, 0, -1, 1, 1, -1, -1, 0, 1, 1, -1, 0, 1, -1, -1, 0, 1, 0, -1, 1, 1, 0, -1, -1, 1, 1, -1, 0>, 1, 0, -1, -1, 1, 0 oeis.org/A253414 G.f. satisfies (1+x^2)*g(x) = 1 + x*g(x^2). 1, 1, <-1, 0, 1, -1, -1, 1, 1, 0, -1, -1, 1, 0, -1, 1, 1, 0, -1, 0, 1, -1, -1, 0, 1, 1, -1, -1, 1, 0>, -1, 1, 1, 0, -1, 0, 1, -1, -1, 1, 1, 0, -1, -1, 1, 0, -1, 0, 1, 1, -1, 0, 1, -1, -1, 0, 1, 1, -1, -1, 1, 0, -1, 1, 1, 0, <-1, 0, 1, -1, -1, 1, 1, 0, -1, -1, 1, 0, -1, 1, 1, 0, -1, 0, 1, -1, -1, 0, 1, 1, -1, -1, 1, 0>, -1, 0, 1, 1, -1, 0, 1 T035 a(n) + 2 = 4, 3, 5, 4, 3, 4, 5, 4, 4, 3, 4, 5, 3, 4, 5, 4, 4, 3, 5, 4, 3, 4, 4, 5, 4, 3, 4, 5, 3 (as deltas) oeis.org/A317331 Indices m for which A058304(m) = 1. <4, 8, 11, 16, 20, 23, 27, 32, 36, 40, 43, 47, 52, 55, 59, 64, 68, 72, 75, 80, 84, 87, 91, 95, 100, 104, 107, 111, 116, 119>, 123, 128, 132, 136, 139, 144, 148, 151, 155, 160, 164, 168, 171, 175, 180, 183, 187, 191, 196, 200, 203, 208, 212, 215, 219, 223, 228, 232, 235, 239, 244, 247, 251 T038 a(n) - 2 = 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1 (as deltas) oeis.org/A014577 The regular paper-folding sequence (or dragon curve sequence). <1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0>, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, <1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0>, 0, 0, 1, 1, 0 oeis.org/A038189 Bit to left of least significant 1-bit in binary expansion of n. 0, <0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1>, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, <0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1>, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1 oeis.org/A014707 a(4n)=0, a(4n+2)=1, a(2n+1)=a(n). <0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1>, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0 oeis.org/A082410 a(1)=0. Thereafter, the sequence is constructed using the rule: for any k >= 0, if a(1), a(2), ..., a(2^k+1) are known, the next 2^k terms are given as follows: a(2^k+1+i) = 1 - a(2^k+1-i) for 1 <= i <= 2^k. 0, <1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0>, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, <1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0>, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1 oeis.org/A014710 The regular paper-folding (or dragon curve) sequence. <2, 2, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1>, 1, 2, 2, 2, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 2 ... 9 total T099 abs(a(n)), sorted with duplicates removed = 1, 2, 3 (as deltas) oeis.org/A000045 Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1. 0, 1, 1, <2, 3, 5, 8>, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155 oeis.org/A000217 Triangular numbers: a(n) = binomial(n+1,2) = n(n+1)/2 = 0 + 1 + 2 + ... + n. <0, 1, 3, 6>, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431 oeis.org/A001222 Number of prime divisors of n counted with multiplicity (also called bigomega(n) or Omega(n)). 0, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 3, 1, 2, 2, 4, 1, 3, 1, 3, 2, 2, 1, 4, 2, 2, 3, 3, 1, 3, 1, 5, 2, 2, 2, 4, 1, 2, 2, 4, 1, 3, 1, 3, 3, 2, 1, 5, 2, 3, <2, 3, 1, 4>, 2, 4, 2, 2, 1, 4, 1, 2, 3, 6, 2, 3, 1, 3, 2, 3, 1, 5, 1, 2, 3, 3, 2, 3, 1, 5, 4, 2, 1, 4, 2, 2, 2, 4, 1, 4, 2, 3, 2, 2, 2, 6, 1, 3, 3, 4, 1, 3, 1, 4, 3, 2, 1, 5, 1, 3, 2 oeis.org/A000700 Expansion of Product_{k>=0} (1 + x^(2k+1)); number of partitions of n into distinct odd parts; number of self-conjugate partitions; number of symmetric Ferrers graphs with n nodes. 1, 1, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 5, 5, 5, 6, 7, 8, 8, 9, 11, 12, 12, 14, 16, <17, 18, 20, 23>, 25, 26, 29, 33, 35, 37, 41, 46, 49, 52, 57, 63, 68, 72, 78, 87, 93, 98, 107, 117, 125, 133, 144, 157, 168, 178, 192, 209, 223, 236, 255, 276, 294, 312, 335, 361, 385 oeis.org/A112798 Table where n-th row is factorization of n, with each prime p_i replaced by i. 1, 2, 1, 1, 3, <1, 2, 4, 1>, 1, 1, 2, 2, 1, 3, 5, 1, 1, 2, 6, 1, 4, 2, 3, 1, 1, 1, 1, 7, 1, 2, 2, 8, 1, 1, <3, 2, 4, 1>, 5, 9, 1, 1, 1, 2, 3, 3, 1, 6, 2, 2, 2, 1, 1, 4, 10, 1, 2, 3, 11, 1, 1, 1, 1, 1, 2, 5, 1, 7, 3, 4, 1, 1, 2, 2, 12, 1, 8, 2, 6, 1, 1, 1, 3, 13, 1, 2, 4, 14, 1, 1, 5, 2, 2, 3, 1, 9, 15, 1, 1, 1, 1 ... 17444 total In transformation descriptions, Sn(z) denotes the ordinary generating function with coefficients a(n), and En(z) denotes the exponential generating function with coefficients a(n). ___________________________________________________________________________ o You can look up sequences online at the OEIS web site https://oeis.org/ o For an explanation of the format used in the OEIS, see https://oeis.org/wiki/Style_Sheet o If your sequence was not in the OEIS and is of general interest, please submit it using the submission form https://oeis.org/Submit.html o The email address <sequences(a)oeis.org> does a simple lookup in the On-Line Encyclopedia of Integer Sequences, a limited form of the search available on the web site. o If you send an empty message, you will be sent the help file. Sequentially yours, The On-Line Encyclopedia of Integer Sequences. Maintained by The OEIS Foundation Inc., https://oeisf.org. Please donate! P.S. This content is made available under the terms of The OEIS End-User License Agreement: https://oeis.org/LICENSE

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[math-fun] How many 5 digit #s are (US) zip codes?
by Bill Gosper 23 Sep '19

23 Sep '19

G4G once had an entertainer named "Mr. Zipcode"(?). You tell him yours and he tells you where, and then goes on to describe places to eat, etc. (He struggled with Ray Solomonoff's obscure town in upstate New York. Or maybe some offstage confederate struggled with https://m.usps.com/m/ZipLookupAction?search=zip .) Impressive, but maybe not 10^-5 impressive. I just mistakenly tried 44007, which seems to be Nantes, the site of a French Revolution atrocity. Does Mr. Zipcode mention atrocities? —rwg

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[math-fun] Hilbert's ≠ Peano's
by Bill Gosper 22 Sep '19

22 Sep '19

For decades I misapprehended Hilbert's spacefill as Peano's. Today at last, I corrected my web page: http://www.tweedledum.com/rwg/samhilbert.htm . The impetus was discovering that my blunder has been copied onto other websites. Unfortunately, with correct attribution.-} —rwg

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