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[math-fun] The substring is our difference (no!)(product?)
by Éric Angelini 02 Apr '20

02 Apr '20

The answer is no, as it is impossible to find a term of k digits that comes after a term of k-1 digits. If we ask to a(n), a(n+1) and their _product_ to share at least one identical substring we have a lexico-infinite seq that starts, I guess, with: 1, 10, 11, 12, 2, 21, 15, 5, 25, ... and the successive shared substrings are: 1 1 1 2 2 1 5 5 etc. Best, É. > Hello Math-Fun ... > 1199, 1200, 1220, 1222, 1224, 1248,... > If we compute a(n) - a(n+1) we notice that the result is a substring of both a(n) and a(n+1). The above example gives: 1199, 1200, 1220, 1222, 1224, 1248,... 1 20 2 2 24 >Question: is an infinite such seq possible?

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[math-fun] The substring is our difference
by Éric Angelini 02 Apr '20

02 Apr '20

Hello Math-Fun ... 1199, 1200, 1220, 1222, 1224, 1248,... If we compute a(n) - a(n+1) we notice that the result is a substring of both a(n) and a(n+1). The above example gives: 1199, 1200, 1220, 1222, 1224, 1248,... 1 20 2 2 24 Question: is an infinite such seq possible? If yes, what would a(1) be of the lexico- first one (of distinct terms)? Note that it is always possible to build a seq W as long as wanted: just start W with 1 followed by a zillion zeroes -- and add 1 at each step. This might be old hat, sorry, but I don't know how to search the OEIS for seqs where I cannot compute at least a(1)... Best, É.

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[math-fun] Exponentially many needles in a double-exponential haystack
by James Propp 02 Apr '20

02 Apr '20

Dear math-funsters, It's been conjectured that for all n above some small bound that I forget (maybe 24?), you can arrange the numbers from 1 to n in a circle so that every two adjoining numbers sum to a perfect square. I believe this conjecture is due to Richard Guy. Is this attribution correct? Back in August 2018, I asked Richard about the problem; he replied "I can't give the BEST reference. Indeed, I can't give ANY (published) reference. Am copying to Elwyn Berlekamp, who has given talks on this and may have published. Let me see if I have any files that I can attach. R." Berlekamp sent me a PowerPoint file that touched on a related problem: arranging the numbers from 1 to n along a *path* so that every two adjoining numbers sum to a perfect square. (Of course every solution to the circle problem is a solution to the path problem as well.) Has anything been written about these two problems? A curious feature of both problems (I've been told) is that it's not too hard to find solutions for *individual* values of n, but devilishly hard to find systematic solutions that apply to *lots* of values of n. If I recall correctly, Guy described the problem as one of "looking for exponentially many needles in a double-exponential haystack". One of the Mathematical Enchantments essays I was thinking of writing (unless someone's already written it!) is a collection of problems with a similar flavor; specifically, problems of the form "Show that for all n, the set A_n is nonempty" with the property that for each specific n within computational range it's easy to construct elements of A_n, but not in a systematic way that would permit us to prove that A_n is nonempty for all n. The situation would be especially poignant if we couldn't even prove that A_n is nonempty for infinitely many n. I'm thinking it might be nice to write this essay soon, in Richard's honor. Another example of a "hay in a haystack" problem is the the Bricklayer's Challenge (which IIRC Richard named the Barrycades problem in honor of its inventor Barry Cipra). I'll check with Barry to see what he knows about the status of the problem. Do you have a favorite "hay in a haystack" problem? Thanks, Jim Propp

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[math-fun] New COVID-19 policy
by Keith F. Lynch 01 Apr '20

01 Apr '20

(Warning: This post contains some physics and chemistry, and perhaps some humor, but relatively little math. It may be offensive to Republicans and Democrats.) The US government is implementing two approaches to the pandemic, and has rejected a third: 1) Release several trillion tons of Freon, a clean, odorless, non-toxic, non-combustible refrigerant, into the atmosphere. This will eliminate the highly toxic ozone gas from the stratosphere, and will allow the sun's life-giving radiance to sterilize all coronavirus on all outdoor surfaces. It will also get rid of all other kinds of viruses, bacteria, and fungi. Also mold, mildew, roaches, rodents, and all other vermin. As an added bonus, this will discourage people from leaving home, since people who don't mind the risk of a viral illness may be deterred by the risk of a severe sunburn. Another bonus is that it will decrease the incidence of both smoking and obesity, by reducing crop yields. Also, it will revitalize America's chemical industry. Very little Freon is produced anywhere in the world today, so the US can easily become the world's leader in this field. The only downside is that it won't eliminate coronavirus indoors or inside people. It can be made to eliminate coronavirus indoors, at least in sunny rooms, if windows are left wide open, or their glass replaced with UV-transparent silica glass. An earlier proposal suggested using Halon rather then Freon. It was decided against when it was realized that, although Halon wouldn't interfere with respiration, it would suppress all combustion. It would mean no more home furnaces, gas stoves, blast furnaces, or internal combustion engines. It would also mean no smoking of tobacco, marijuana, or crack cocaine. 2) All experts agree that the pandemic will last until 2021. There will be no vaccine until then, nor will it be practical to reopen businesses, schools, churches, and entertainment venues. Inspired by the 1753 (in Britain and its then-colonies, earlier or later elsewhere) calendar shift, in which 11 days were skipped, and by our annual ritual of getting more sunlight by setting clocks ahead by one hour, the Department of Calendar has ruled that the day after April 1, 2020 will be April 2, 2021. Also, there will be a long-overdue change to the Gregorian leap-year rule. Instead of adding a day if the year is divisible by 4, subtracting a day if the year is divisible by 100, then adding a day if the year is divisible by 400, the new rule will be to add a day for every odd number that the year is divisible by and subtracting a day for every even number that the year is divisible by. In addition to bringing a quick end to the pandemic, this has the advantage of ensuring full employment to computer programmers (to update all software that deals with calendars), and making every American eligible for Medicare and Social Security a year sooner than they otherwise would have been. 3) A proposed Coronavirus Manhattan Project was vetoed by President Trump. He said that although nuclear weapons would certainly kill viruses, their use has undesirable side effects. Specifically, they harm vital infrastructure, such as hotels, casinos, and golf courses, and roads and bridges that lead to those facilities. When Fauci told him that it's only an analogy, the president replied that he has known for years that it's only an analogy, and that he's the world's greatest expert on analogies, and on Manhattan. They only people who know anywhere near as much about analogies as he does are his trusted friends Vladimir Putin, Xi Jinping, Kim_Jong-un, and Mohammed bin Salman. "Especially the last, as he knows just how to deal with journalists from the fake news Washington Post." Ms. Clinton responded that if only Obamacare were still mandatory, every American would get unlimited lifesaving treatment for covid-19, hence nobody would die of it. (My past April 1st messages to this list had a lot more math in them. Those who are relatively new to the list, please email me directly if you want me to send them to you. Thanks.)

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Re: [math-fun] Pandigital Trivia Spoiler
by Keith F. Lynch 01 Apr '20

01 Apr '20

Allan Wechsler <acwacw(a)gmail.com> wrote: > Andy Latto <andy.latto(a)pobox.com> wrote: >> I find the pattern in the first row of your table very interesting. Thank you. >> Is there any intuitive reason that pandigital numbers in odd bases >> only have very small powers of two as divisors? > I'm making a very big free-associative leap, but this topic seems > vaguely related to something I've thought of as "reverse long > division". I'd like to know whether it's been studied; I think the > problem under discussion is in the same conceptual area. .... Interesting. I had not thought of that. However, there is a different reason why every other odd-numbered base has 0 for its highest power of 2. I finished calculating the 14th term, i.e. the highest power of 2 that divides any 14-digit base-14 pandigital number. It's 36. This took more than a full day. Calculating it for 15 would take two weeks, and for 16 nearly a year. But I didn't need to, as I solved those two powers in my head. For 15 it's 0, and for 16 it's 7. And there are more than 87 billion solutions for 7. (I won't list them all.) Can you see how I was able to solve those in my head? Warning: My next post to this list should not be taken seriously. It will contain some physics and chemistry, and perhaps some humor, but relatively little math. It may be offensive to Republicans and Democrats.

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Re: [math-fun] Pandigital Trivia Spoiler
by Keith F. Lynch 31 Mar '20

31 Mar '20

Hans Havermann <gladhobo(a)bell.net> wrote: > "Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: >> What do 3076521984, 3718250496, and 6398410752, and no other >> numbers, have in common?" > They are obviously pandigital numbers and thus share the divisor > 3^2. The GCD of the three numbers divided by 9 is a large power of > 2. A quick check shows that no other pandigital has that large a > power-of-two divisor (the next largest being for 9805234176). Correct. My three numbers are all divisible by 2^21. No 10-digit base-10 pandigital numbers are divisible by 2^22. I've looked into the highest powers of small primes among the n-digit base-n pandigital numbers for small n. Here's a table. As always with tables, it should be viewed in a fixed-width font. base: 2 3 4 5 6 7 8 9 10 11 12 13 2: 1 0 3 1 11 0 5 2 21 0 29 1 3: 0 1 3 5 7 9 10 3 15 17 18 21 5: 0 1 2 1 5 5 6 9 9 11 12 13 7: 0 1 2 2 3 1 7 6 8 9 9 13 11: 0 1 1 2 2 3 5 7 5 1 8 9 13: 0 0 1 2 2 3 3 5 5 6 8 1 I can easily push it to higher primes, but not to higher bases. Computing the entries for base n will take roughly n times longer than computing them for all smaller bases put together. I may extend this table to powers of composite numbers. It should be obvious why the entry equals 1 whenever the row and column numbers are equal. The "opposite" problem, finding n-digit base-n pandigital primes, isn't very interesting. There are only three of them. Can you find them? (I haven't looked into non-standard bases, such as negabinary, Fibonacci, or balanced ternary.) Which of these rows and columns would be of interest to OEIS? None of them appear to be there yet.

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[math-fun] An array with all the negative and positive integers, plus zero
by Éric Angelini 31 Mar '20

31 Mar '20

Hello Math-Fun, have a look here, on my personal page: https://bit.ly/2wGyzpv You will see how to build this array; it starts (fixed font): > A new array for the OEIS? 0 +1 +3 +7 +11 +5 +14 ... -1 +2 -4 +4 -6 +9 ... -3 +6 -8 -10 -15 ... -9 -14 -2 -25 ... -5 -12 -23 ... -7 -35 ... -28 ... ... The idea is to insert in an infinite “triangular” array, all integers from -∞ to +∞, with no duplicates. We start, as usual, in the upper left corner with 0. The array now is filled with those two rules: >>Rule 1) we always try to extend the upper row with the smallest positive integer a(n) not present so far in A (A is the array), integer that doesn’t lead to a contradiction immediately, or later in the rows below it; >>Rule 2) a(n) and a(n-1), the term at the left of a(n), produce another integer in the row below them; this term is the difference (smallest term – biggest term). If this leads to a contradiction (immediately or later in the antidiagonal), we try (biggest term – smallest term). If both operations lead to a contradiction, we try a(n)+1 instead of a(n) and repeat the whole process, filling A by antidiagonals. Example: After 0, we write 1 because this 1 doesn’t lead to a contradiction: we will have -1 below the pair (as 0 – 1 = -1): (...) Best, É.

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Re: [math-fun] math-fun Digest, Vol 205, Issue 35
by Steve Witham 31 Mar '20

31 Mar '20

From: Mike Stay <metaweta(a)gmail.com> I do get the "not secure" complaint, but the server sends an EarthLink 404 page, not a page on hyperbolic pairing. On Mon, Mar 30, 2020 at 12:39 AM Steve Witham <sw(a)tiac.net> wrote: > > http://www.rong.address Oops, habit. http://www.mac-guyver.com/switham/2020/03/HyperbolicPairing/index.html --Steve

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[math-fun] Occam's Razor and COVID analysis.
by Brad Klee 31 Mar '20

31 Mar '20

Part of the problem using the COVID-19 calculator is that it has too many parameters relative to available data, especially with the outbreak only 99% ended in China (for now). If we take only one data set, for example Chinese deaths/day, then we are better off fitting to a one or two parameter model. Quite amazingly, COVID data looks enough like the Popillia japonica seasonality data from iNaturalist. It is also well-fit by a logistic map (thanks Cris!). For reference: https://mathworld.wolfram.com/LogisticMapR=2.html With China data, we need to bin data 6 days into one bin, and then find 98.9% accuracy by overlap dot-product. From peak outbreak to 99.98% quell is six intervals, or 36 days. Depending on peak-flattening the extracted parameter 6 (days / bin) should be expected to change by region. This is a good way to compare distinct outbreaks, only when stated next to a distribution accuracy statistic. Another interesting data set is the iNaturalist seasonality for the Chaser and King Skimmer dragonflies of genus Libellula. Good news, there should be an outbreak soon! --Brad

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[math-fun] PDF link Re: Hyperbolic pairing
by Steve Witham 30 Mar '20

30 Mar '20

I've put up the "Hyperbolic Pairing" draft so one can see the math. http://www.tiac.net/switham/2020/03/HyperbolicPairing/index.html V0.4, slightly improved. There is a link to the PDF in the middle of the (short) page. Your browser should complain that it is not secure. I need to get and install a certificate. --Steve

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