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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 22 Dec '20

22 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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[math-fun] Is this entire function surjective?
by Dan Asimov 22 Dec '20

22 Dec '20

A complex-valued function f : C —> C on the complex plane C is called "entire" if it's analytic for all z ∊ C. Or in other words it can be defined by a power series f(z) = ∑ c_n z^n with an infinite radius of convergence. 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? —Dan

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[math-fun] Tapered peaks
by Éric Angelini 22 Dec '20

22 Dec '20

Hello Math-Fun, How high will you climb? Read one digit d at the time, starting from the first one, and extend S with the product d*(last term of S). When a zero is read, replace this product by the smallest integer not yet present in S and go on. S = 2,4,16,16,96,96,576,5184,31104,279936,1679616,8398080,58786560,352719360,1763596800,1763596800,14108774400,56435097600,169305292800,169305292800,1,4,8,56,... Integers with two or more consecutive zeros must be read carefully. I hope I did. P.-S. How I read and multiplied is visible here, on my blog: http://cinquantesignes.blogspot.com/2020/12/tapered-peaks.html Best, É.

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Re: [math-fun] The True Tau Day(s)
by Keith F. Lynch 20 Dec '20

20 Dec '20

Greg Huber <greg.huber(a)czbiohub.org> wrote: > c'mon people, seriously, grow up -- there's no need to throw away > another Greek letter, 'cause they ain't making any new ones!) There are lots of other alphabets. I suggest using the whole of Unicode for the names of mathematical constants. Some upgrades to Mathematica may be necessary. Last month Randall Munroe discovered that raising an emoji to an emoji power will crash Mathematica.

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[math-fun] Question about perfect matchings
by Dan Asimov 20 Dec '20

20 Dec '20

For a finite graph G (assume G is connected and without loops or multiple edges), a "perfect matching" is a set of disjoint edges all of whose endpoints form the vertex set V(G) of G. (Fun exercise: For each of the five Platonic solids, the graph of its edges has a perfect matching.) Obviously, a *necessary* condition for G to have a perfect matching is that |V(G)| be *even*. Question: --------- Is a simple sufficient condition for G to have a perfect matching known? (A vertex v contained in only one edge e of G would force e to be in any perfect matching, and this leads to easy counterexamples, like the Y-shaped graph on 4 vertices. (So let's also assume there are no vertices contained in only one edge.) —Dan

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[math-fun] Braced Heptagon
by ed pegg 20 Dec '20

20 Dec '20

In Mathematica GraphData[{"UnitDistance", {21, 2}}] Is that a rigid graph? If so, it uses just 42 unit rods, beating the record at https://erich-friedman.github.io/mathmagic/0100.html

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[math-fun] Super-weak Skolem-Langford numbers
by Éric Angelini 18 Dec '20

18 Dec '20

Hello Math-fun, find the first 5 consecutive numbers that are also Super-weak Skolem-Langford. (I guess 6 is impossible). http://oeis.org/A339803 à+ É. Catapulté de mon aPhone

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