- mathfun@mailman.xmission.com - mailman.xmission.com

[math-fun] Another Lex. Earliest Sequence: Scott Shannon's "Enots Wolley sequence A336957
by Neil Sloane 21 Aug '20

21 Aug '20

Obviously this is an inverted version of the Yellowstone sequence A098550 ! The name Enots Wolley is for personal use only, it must not be mentioned in the OEIS! We frown on such made-up names. Definition: Lexicographically earliest sequence {a(n)} of distinct positive numbers such that, for n>2, a(n) has a common factor with a(n-1) but not with a(n-2). 1, 2, 6, 15, 35, 14, 12, 33, 55, 10, 18, 21, 77, 22, 20, 45, 39, 26, 28, 63, ... The original idea was due to Scott, with a different sequence, but this is my (canonical!) version. Could someone please prove the conjecture that this is a permutation of the set {1, all numbers with at least two distinct prime factors} ? I can't even prove that every number 2*p (p prime) appears, or that there are infinitely many even terms (although I've found a dozen false proofs). It's a slippery problem. Neil

9 13

Re: [math-fun] FactorInteger
by Bill Gosper 20 Aug '20

20 Aug '20

Slight improvement: Mathematica's front end by default displays a centerdot for implicit multiplication. Strangely, this doesn't work in old the FI function below, which displays blanks. Blanks also (by default) separate digits in blocks of 3. So the display is sometimes ambiguous. But Wolfram CE Harry Calkins showed me this workaround: FI⨉[n : _Rational | _Integer] := Defer @@ {FactorInteger@n} /. {{a_Integer, 1} -> a, {a_Integer, b_} -> a^b} /. {a_} -> a /. List -> Times E.g. In[144]:= FI⨉@BernoulliB@50]@// tim During evaluation of In[144]:= 0.006522,1 (seconds) Out[144]= (5^2 417202699 47464429777438199)/(2 3 11) (The first Bernoulli number with a square factor!) Dirty trick. I used a tiny ⨉ symbol instead of x, so you need Front End completion to input FI⨉. You may want to edit the definition. Instead of little centerdots between the factors, there's a Preferences variable Appearances|Numbers|Multiplication|Automatically display multiplication symbol between multiplied numbers Use symbol: which you can bash to be a mini ⨉. Sorry, I don't know how to bash it under program control, as in an init file. Unfactor is way nicer than with FactorInteger: UnFI⨉@Defer[x_] := x In[153]:= UnFI⨉@%144 Out[153]= 495057205241079648212477525/66 —rwg On Wed, Jan 4, 2017 at 12:44 AM Bill Gosper <billgosper(a)gmail.com> wrote: > There are > > In[14]:= 52! > Out[14]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > ways to shuffle a deck. > That's just a "little" over a googol ... > > In[366]:= Log[10.^100, %] > Out[366]= 0.679066483922048 > > to the 2/3 power. The square of the cube root. And vice versa. So if > anybody asks how many > ways to shuffle a deck, "Oh, about the cube root of a googol squared." > But this is actually low > by a fairly substantial factor of > In[372]:= (10.^100)^(.679 - 2/3) > > Out[372]= 17.1132830416181 > > The actual number of shufflings is enormously composite: > > In[15]:= FactorInteger@%14 > Out[15]= > {{2,49},{3,23},{5,12},{7,8},{11,4},{13,4},{17,3},{19,2},{23,2},{29,1},{31,1},{37,1},{41,1},{43,1},{47,1}} > > But this is ugly and hard to read. The doc suggests > > In[16]:= CenterDot@@(Superscript@@@%) > Out[16]= > 2^49\[CenterDot]3^23\[CenterDot]5^12\[CenterDot]7^8\[CenterDot]11^4\[CenterDot]13^4\[CenterDot]17^3\[CenterDot]19^2\[CenterDot]23^2\[CenterDot]29^1\[CenterDot]31^1\[CenterDot]37^1\[CenterDot]41^1\[CenterDot]43^1\[CenterDot]47^1 > > This actually looks neat. Except for the ^1s. Try it. But it's unclear > (actually, a nice little Mathematica exercise) how to get your original > number back. > So I wrote something called FI > > In[364]:= FI[52!] > Out[364]= 2^49 3^23 5^12 7^8 11^4 13^4 17^3 19^2 23^2 29 31 37 41 43 47 > > In[365]:= ReleaseHold@% > Out[365]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > Here is the definition: > FI[n : _Rational | _Integer] := > Times @@ ((HoldForm @@ {#1[[1]]})^#1[[2]] &) /@ FactorInteger[n] > > And here's a spoiler for how to invert FactorInteger: > > In[367]:= Times @@ (Power @@@ FactorInteger[73 137]) > Out[367]= 10001 > > which also works on that CenterDot expression: > > In[368]:= Times @@ (Power @@@ %16) > > Out[368]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > Finally, you might enjoy finding a fairly nice formula for how many 0s are > on the end of n!, and, > more generally, what are exponents of the various prime factors. > --Bill >

1 1

[math-fun] (> 40)? yr old Life pattern for Roth's constant.
by Bill Gosper 19 Aug '20

19 Aug '20

(Which I incorrectly labeled Thue's. It has a fulminating continued fraction.) This (admittedly trivial) pattern has been waiting years to get included as Golly's Patterns>Hashlife>roth.mc. It spells out the bit sequence with blinkers. Formulas: https://www.jjj.de/fxt/fxtbook.pdf, 38.2 Screenshot <http://gosper.org/rothshot.png>: Golly file <http://gosper.org/roth.mc>: —rwg

1 0

[math-fun] Sage on iMac: on no account read the instructions!
by Fred Lunnon 16 Aug '20

16 Aug '20

The SAGE/SageMath CAS is free software, maintained by volunteer enthusiasts. This has pros and cons, of which I have experienced a handsome dose in the last couple of days. Starting with the cons: every hotshot programmer-mathematician who feels a ferocious urge to maintain an up-to-date step-by-step installation guide for dummies, please raise your hand now ... Uh-huh, thought so. Besides, none of those guys could do that job anyway, even if they wanted to: by the time anyone starts to take such things seriously, familiarity has already blinded them to many of the difficulties faced by a newcomer. Thus the download file directory comprises large number of alternative file identifiers, with no clue how to distinguish (say) sage-8.9-OSX_10.14.6-x86_64.app.dmg [for my operating system] sage-9.1-OSX_10.11.6-x86_64.app.dmg [for latest CAS release]. On randomly downloading some not-obviously unsuitable selection, one might (eventually) succeed in finding a way past Apple's security to click hopefully on the nice blue icon, calling up a browser panel labelled "Jupyter" which looks very business-like, but cannot apparently be persuaded to actually DO anything. At which point your newbie rolls up a sleeve and falls straight into a novice trap: he opens the installation README file, and attempts to execute the procedure described therein in deceptively precise and comforting detail. Then rolls up the other sleeve and tackles the Q&A support site, which has lots of Q's and A's and a search option, but (naturally) no procedure apparent for actually submitting an enquiry. Eventually all these obstacles prove to be minor, given a modicum of dogged perseverance, which proves rewarded by a couple of major pros. The level of support is astounding: once broken into the frankly weird bug-report system, my query was responded to within minutes by a fellow user, who pursued it through the weekend. And when one does succeed in invoking it, the software appears robust, efficient, cleanly designed, fast, (relatively) compact. For the benefit of anyone else contemplating installing SageMath on an Apple iMac, there is now a step-by-step description in the author's answer at https://ask.sagemath.org/question/53036/installation-fails-at-readme-step-5… which tries to carry on from where https://doc.sagemath.org/html/en/installation/binary.html http://www.mirrorservice.org/sites/www.sagemath.org/osx/intel/index.html leave the novice dangling. Fred Lunnon

1 0

Re: [math-fun] 'in-band' signaling in voting systems
by Keith F. Lynch 16 Aug '20

16 Aug '20

Brent Meeker <meekerdb(a)verizon.net> wrote: > Andy Latto wrote: >> There are two downsides to that. One downside is shared with >> absentee voting -- the voter could have been coerced or rewarded to >> vote in a specific way. The advantage of in-person voting is that >> that's not possible, at least not unless the coercer or rewarder >> hooks the voter up to a reliable lie detector. I, not Brent, wrote the above. > Or asks for a cel phone video of the ballot being marked. Not possible here in Virginia. You mark the ballot however your coercer or rewarder wishes, take a photo of it, then take the ballot to the poll worker and tell her that you made a mistake. She will discard the spoiled ballot and hand you a new one, which you then mark however you like. I assume the same would be true in most US states. And that the coercer or rewarder would know this, hence wouldn't bother.

1 0

Re: [math-fun] 'in-band' signaling in voting systems
by Keith F. Lynch 16 Aug '20

16 Aug '20

Henry Baker <hbaker1(a)pipeline.com> wrote: > I.e., in order to properly test the voting network, it may be > necessary to inject a *known* number of 'false' votes into the > network, and subsequently subtract these false votes from the > totals. How would that help confirm that the votes are getting through, unless the total number of votes for candidate X were less than the number of test votes for candidate X? Perhaps test votes should be for a non-existent candidate. Hypothetical hacked software can't accept those votes while rejecting votes for a real candidate, unless the name of one or both candidates appears somewhere in the code, which it shouldn't. Of course an encrypted version of one or both names could be hidden in the code. A better approach is to use the live votes as tracers. Every voter is encouraged to make a note of their ballot serial number. The whole database of who everyone voted for is made publicly available online, except for the link between the voter's name and the ballot serial number. Each voter could look up his own ballot and see that it's correct. There are two downsides to that. One downside is shared with absentee voting -- the voter could have been coerced or rewarded to vote in a specific way. The advantage of in-person voting is that that's not possible, at least not unless the coercer or rewarder hooks the voter up to a reliable lie detector. The other downside is that the voter could falsely claim that his vote was missing or wrong. Also, this doesn't address the opposite problem, ballot-box stuffing. > Clearly, there are great dangers in a poorly implemented voting > system with 'in-band' signaling. However, there are also great > dangers in *not testing* a voting system for link failures and > active hacking attacks. Re hacking, who says voting has to rely on computers? See https://xkcd.com/2030/ Of course non-computerized voting systems are also subject to hacking. But it's harder to automate, and harder to hide, just as before computers spies could steal secret information, but couldn't steal millions of pages of it at once. > Has anyone else considered these kinds of problems? Of course. For instance it's a perennial topic on the Risks Digest, which has been running for 35 years, making it perhaps the oldest still extant email list. Note that even if there's an easy and foolproof solution, that doesn't mean that it will be implemented if those in charge prefer the flaws in the current system. For instance they've been simultaneously making government-issued ID harder to get, and required for voting. Here in Virginia, the ID requirement was recently abolished, but the online portal for requesting an absentee ballot won't let you submit the request unless you provide the serial number on your current ID. Analogously, there's an obvious solution to the many forensic lab scandals: Double-blind testing. Never tell the techs what samples the prosecutors want to match what other samples. But since prosecutors run the system, that will never happen. Some labs are actually paid per conviction, which is an enormous conflict of interest. And some are overworked to the point where it's physically impossible to actually run all the tests.

3 2

[math-fun] Geometry puzzle
by Dan Asimov 16 Aug '20

16 Aug '20

Consider the cubic 3-dimensional torus T^3 = R^3/Z^3, a unit cube with opposite faces identified. Now connect each of 4 alternate corners of the cube, say {(1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1)}, to the center (1/2,1/2,1/2) of the cube by a line segment. Now we want to shorten this union of 4 segments with these moves: 1) Any vertex (endpoint of a segment) may be moved continuously, while keeping the same connectivity (segments that connect 2 vertices will still connect them even if they are moved); and 2) Any vertex v may split in two to become 2 new vertices v', v''. v' and v'' are assumed to be connected by a line segment of length 0. When that happens, one or more segments that had v as an endpoint (but not all of them) are assigned to have v' as endpoint, and the remaining segments are assigned to have v'' as endpoint. 3) The reverse of 2): any 2 vertices may (by 1) coalesce to form a new vertex. Puzzle: Using only these moves, what is the shortest possible network that the original network of 4 segments can be transformed to? —Dan

1 0

[math-fun] Sage on iMac ?
by Fred Lunnon 16 Aug '20

16 Aug '20

I'm currently attempting to install SageMath on my iMac, and would appreciate hearing (directly please) from anyone who has done so: what version of OSX, what version of Sage, and what outcome? WFL

1 0

Re: [math-fun] nth digit * nth term
by Éric Angelini 16 Aug '20

16 Aug '20

Many thanks Hans — and beautiful illustration! I will submit this sequence together with Kris Katterjohn in a couple of hours and link it to your blue spikes, if you don’t mind. Best, É. Catapulté de mon aPhone > Le 16 août 2020 à 14:39, Hans Havermann <gladhobo(a)bell.net> a écrit : > > EA: "We start S with a(1) = 2 and always extend S with the product « nth digit x nth term of S ». (When the product is = 0, we don’t extend S with 0 but with the smallest integer not yet present in S.)" > > See my graph of S to 10^5 here: > > http://chesswanks.com/num/nthDigit*nthTerm.png > > Large values in S are created by lengthy non-zero stretches in the digit list, appearing in the graph as spikes that culminate in a peak when the digit is again zero. Based only on the digit list for my calculated S, the peak at a(28385) can be predicted to be superseded by a slightly larger peak at a(350165). > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

1 0

[math-fun] nth digit * nth term
by Éric Angelini 16 Aug '20

16 Aug '20

Hello Math-fun, We start S with a(1) = 2 and always extend S with the product « nth digit x nth term of S ». (When the product is = 0, we don’t extend S with 0 but with the smallest integer not yet present in S.) S = 2, 4, 16, 16, 96, 96, 576, 5184, 31104, 279 936,... We see that 2 x 2 = 4 = a(2) 4 x 4 = 16 = a(3) 1 x 16 = 16 = a(4) 6 x 16 = 96 = a(5) 1 x 96 = 96 = a(6) 6 x 96 = 576 = a(7) 9 x 576 = 5184 = a(8) 6 x 5184 = 31104 = a(9) 9 x 31104 = 279 936 = a(10) 6 x 279 936 = 1 679 616 = a(11) 5 x 1679 616 = ... etc. The first column of the above array is the succession of the digits of S; the second column is the succession of the terms of S. Question: what is the biggest of the first 100 terms of S? Best, É. à+ É. Catapulté de mon aPhone

3 2