December 2020 - mathfun@mailman.xmission.com

[math-fun] I wonder how old this is.
by Bill Gosper 28 Dec '20

28 Dec '20

The arclength of one period of sin x = the circumference of an ellipse with semiaxes 1 and √2. In[112]:= #1 == #2 == FunctionExpand@#1== N@# &[ArcLength[Circle[{0, 0}, {√2, 1}]], ArcLength[Sin@x, {x, 0, 2 π]}]] Out[112]= 4 EllipticE[-1] == 4 √2 EllipticE[1/2] == 4 √2 π^(3/2)/Gamma[1/4]^2 + Gamma[1/4]^2/√(2π) == 7.64039557805542 𝚪(¼) is the rightful value of the symbol 𝛕. And someone should write Beckmann II: A History of 𝛕. —rwg

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Re: [math-fun] The 2021 problem
by Keith F. Lynch 27 Dec '20

27 Dec '20

ed pegg <ed(a)mathpuzzle.com> wrote: > Two consecutive primes can have a product giving a concatenation of > consecutive integers. Two simple examples are43?47 = 20~21 > 891077215721081784886888257701070827?891077215721081784886888257701070829 =? > 794018604377235322848433897872605582~794018604377235322848433897872605583?? I'm curious how you found the latter, and whether you're claiming those are the first two examples, or whether there could be another example between them. (As I've mentioned before, non-ASCII characters appear on the digest option as question marks. It's obvious that two of the question marks were intended as multiplication signs, but I'm curious what they other three were intended to be. Thanks.)

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[math-fun] The 2021 sequence
by ed pegg 27 Dec '20

27 Dec '20

semiprimes of consecutive primes that are also formed by concatenating consecutive integers. 2021, 794018604377235322848433897872605582794018604377235322848433897872605583, 25187118108481597700189092548093595916723774714848814417444367033247162518711810848159770018909254809359591672377471484881441744436703324717 --Ed Pegg Jr

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[math-fun] Random domino tiling.
by Allan Wechsler 27 Dec '20

27 Dec '20

Today I learned from a Mathologer video (Burkard Polster, who apparently knows funster Jim Propp), that there is a fairly simple formula for the number of domino tilings on an arbitrary simply-connected polyomino. (Mathologer was ambiguous about whether the formula fails when the polyomino has narrow "necks", but my guess is that these cases are fine.) You checkerboard-color the thing. In order for there to be any tilings at all, there have to be an equal number n of black and white squares. Then, assigning columns to white and rows to black, you construct an adjacency matrix; horizontal connections are marked 1, vertical connections are marked i (!), with 0's elsewhere; and then finally you take the determinant. Magically, either the entire real part or the entire imaginary part cancels away, and the absolute coefficient of the result (either pure imaginary or pure real) is the number of tilings. Mathologer didn't say it, but this gives a reasonably efficient way to generate a random tiling of any tilable simply-connected polyomino P. Procedure: use the magic determinant to count how many tilings of P there are; call this N, and pick a random number K between 0 and N-1. This will be the "index" of the random tiling we will select. Now, prune off all the parts that have their domino coverings forced in some way or other. In the part that's left, pick the leftmost cell of the topmost row: it can be covered in exactly two ways, horizontally, and vertically. Imagine it's horizontal, and remove that domino to get a new polyomino Ph. Count how many tilings of Ph there are; call that N'. If K < N', then that upper-left cell will be covered horizontally; if K >= N', it will be covered vertically (and the remaining polyomino we will call Pv). Now repeat this procedure with index K and polyomino Ph, or with index K-N' and polyomino Pv, depending on the outcome of the comparison. Eventually you will have tiled all the cells. I would like to see an app that takes a given polyomino and picks random tilings for you. I would also like to see one that colors the cells of a given polyomino by the probability that they are covered horizontally or vertically by a random tiling.

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[math-fun] Rolling polyhedra
by James Propp 26 Dec '20

26 Dec '20

I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees. For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is). Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)? Jim Propp

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[math-fun] The 2021 problem
by ed pegg 26 Dec '20

26 Dec '20

Two consecutive primes can have a product giving a concatenation of consecutive integers. Two simple examples are43×47 = 20~21 891077215721081784886888257701070827×891077215721081784886888257701070829 = 794018604377235322848433897872605582~794018604377235322848433897872605583 Are there any others? --Ed Pegg Jr

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[math-fun] Reverse and add (my digits)
by Éric Angelini 25 Dec '20

25 Dec '20

Hello Math-fun, As 1,2,4,8,16,68 is not in the OEIS, I was wondering: do all such integers end on a palindrome? (a 1-digit integer is a palindrome, of course). Let’s see with 10: 10 —> 01 + 1 = 2 (palindrome) With 11: 11 —> 11+2=13 —> 31+4=35 —> 53+8=61 —> 16+7=23 —> 32+5=37 —> 73+10=83 —> 38+11=49 —> 94+13=107 —> 701+8=709 —> 907+16=923 —> 329+14=343 HIT! With 12 we get (if I’m not wrong): 12,24,48,96,84,60,12,... loop Ok we have our answer: no, some integers loop and don’t show any palindrome in the process. Merry Xmas to the list, Best, É.

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[math-fun] Sum of peculiar primes
by Éric Angelini 24 Dec '20

24 Dec '20

Hello Math-Fun, We look for integers that are the sum of certain primes: 32 = 3 + 29 117 = 19 + 19 + 79 119 = 11 + 11 + 97 127 = 19 + 29 + 79 139 = 11 + 31 + 97 ... If we keep the first digit of the summed terms and concatenate them, we form the sum again, in those equations. Any others? (yes, there is sometimes a trick: the last equation produces 169 = 11 + 61 + 97 and 179 = 11 + 71 + 97, for instance). Best, É.

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[math-fun] Platonic surface puzzle
by Dan Asimov 23 Dec '20

23 Dec '20

Let M be a compact surface and let N be the surface of one of the Platonic solids (a "Platonic surface"). A function f : M —> N is an "n-fold branched covering of N, branched over its vertices" if both 1) and 2) hold: ----- 1) For every point p of M such that f(p) is *not* a vertex of N, p has a neighborhood U that f takes homeomorphically onto the subset f(U) of N. Also, f^(-1)(f(p)) contains exactly n points of M. —and— 2) If p is a point of M such that f(p) *is* a vertex of N, then p has a neighborhood U that f takes to f(U) by wrapping around f(p) n times (just like the mapping z |—> z^n near the origin of the complex plane). In this case, f^(-1)(f(p)) contains only the point p of M. ----- Puzzle: ------- For each Platonic surface N and for each integer n ≥ 2, construct such a mapping f : M —> N for some compact surface M. —Dan

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Re: [math-fun] Is this entire function surjective?
by Dan Asimov 23 Dec '20

23 Dec '20

Thanks, Gareth! Nice proof (below), makes complete sense. (I had got as far as: If z exp(-z) omits the value c, then z exp(-z) - c = exp(g(z)) for some entire function g(z), which looked quite impossible. But I didn't see how to finish the proof.) —Dan I wrote: ----- The "little theorem" of Picard states that an entire function is surjective or omits just one complex value, or else it is a constant. Clearly the function f(z) = z exp(-z) is not a constant. Does it omit a value? In more generality, what about entire functions of the form g(z) = P(z) exp(Q(z)) where P and Q are polynomials? ----- Gareth McCaughan replied (slightly edited for my readability): ----- An entire function that's never zero can be written as exp(F(z)) where F is entire. So an entire function that omits the value u can be written as u + exp F(z) where F(z) is entire. If z exp(-z) = u + exp(F(z)) then exp(F(z)) = z exp(-z) - u equals -u for _exactly one_ choice of z, namely z=0. Hence there is only one z for which F(z) is _any_ logarithm of -u. But there are lots of logarithms of -u, so F(z) omits almost all of them, which is impossible for entire F(z). (What if u=0? Impossible, because z exp z definitely doesn't omit the value 0.) If we have P exp Q (P,Q polynomials) instead, we can still do more or less the same. ... ... ... -----

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