January 2021 - mathfun@mailman.xmission.com

Re: [math-fun] September 10000, 1993
by Keith F. Lynch 20 Jan '21

20 Jan '21

"Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: > My full table, which shows just the most recent solution for each > day of a full Gregorian period (400 years = 146097 days), can be > downloaded from http://KeithLynch.net/146097.txt. The code that > generated it can be downloaded from http://KeithLynch.net/146097.c. If those URLs don't work for you, try leaving out the periods at the ends of the sentences. I wrongly thought that was obvious, but apparently not.

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Re: [math-fun] September 10000, 1993
by Keith F. Lynch 20 Jan '21

20 Jan '21

"Keith F. Lynch" <kfl(a)KeithLynch.net> wrote: > The solution to a related but more difficult calendar puzzle *does* > depend on the details of our calendar. What about power-of-ten dates > of *all* months? Is there any day which is *not* such a day? (Using > the proleptic Gregorian calendar.) The 1st and the 10th of each month > are of course the 0th and 1st powers of 10 of that month. But what > about the other 26, 27, 28, or 29 days of that month? . . . > Also, since the calendar does repeat, if we find solutions for > 146097 consecutive days, we'll have found a solution for all days. I have done exactly that, and confirmed that *every* day can indeed be expressed as day 10^N of some prior month in infinitely many ways, and that the most recent such N never exceeds 306. Here are the last few and next few days: W 1/13/2021 4/100000/1747 R 1/14/2021 12/10^55/1795 F 1/15/2021 6/10^64/1971 S 1/16/2021 9/10000/1993 N 1/17/2021 6/10^18/1913 M 1/18/2021 4/10^35/1954 T 1/19/2021 8/10^183/1833 W 1/20/2021 9/10^28/1963 R 1/21/2021 11/10^52/1790 F 1/22/2021 6/10^176/1713 S 1/23/2021 4/10^23/2087 N 1/24/2021 5/1000/2018 M 1/25/2021 3/1000000/2083 T 1/26/2021 2/10^12/1814 W 1/27/2021 1/10^16/1736 Note that the year on the right is usually wrong, sometimes by an immense number of years. But always an exact multple of 400 years. For instance the 24th of this month is of course not actually day million of March 2083, it's day million of March 718 BCE, exactly 2800 years earlier. > Since that number (400/146097) is rational, the decimal expansion > repeats, so there are only a finite number of distinct fractional > numbers of years ago, no matter how many powers of 10 you try. I > don't know the period of repetition of 400/146097, but since 146097 > is 3^3 * 7 * 773, I think it's at most 6*772=4632 decimal digits. It is exactly 1/4 of that. 1158 digits. Hence January 16 wasn't just September 10^4, 1993, it was also September 10^1162 of an enormously earlier year, September 10^2320 of an even earlier year, ad infinitum. Phrased another way, 10^1159 = 1 (mod 146097). My full table, which shows just the most recent solution for each day of a full Gregorian period (400 years = 146097 days), can be downloaded from http://KeithLynch.net/146097.txt. The code that generated it can be downloaded from http://KeithLynch.net/146097.c.

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[math-fun] "smallest" degree n irreducible polynomial over GF(p)
by Mike Speciner 19 Jan '21

19 Jan '21

Let $n = 2^{e_0} q_1^{e_1} \cdots q_i^{e_i}$ be the prime factorization of $n$. Let $p$ be an odd prime. Is it true that $\exists k$ such that $x^n+k$ is irreducible over GF($p$) $\iff$ $(2^{min(e_0,2)} q_1 \cdots q_i) | (p-1)$ ? Are there similar criteria for "larger" polynomials such as $x^n + k_1 x + k_0$ ?

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[math-fun] Tictactoid game with arithmetic sequences.
by Allan Wechsler 18 Jan '21

18 Jan '21

I am sure this two-parameter family of games has been analyzed up the wazoo, so I am hoping somebody can just tell me what's known. X and O take turns claiming an integer from [1,n]. The winner is the first to have claimed an arithmetic sequence of length k. The classic strategy-stealing argument shows that this game is always either a win for the first player or a draw. What values of (n,k) have known outcomes? If k = 3, it's a draw for k < 6 and a win for k > 6; I haven't quite analyzed k = 6. Higher k's are more interesting. Is there a complete theory out there?

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[math-fun] Simpsonian conjecture
by Bill Gosper 18 Jan '21

18 Jan '21

If a plane P through the center of an ellipsoid E intersects it in area A, the volume of E is 2Ah/3, where h is the distance between the two planes parallel to P and tangent to E. —rwg

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[math-fun] Conjecture about the distribution of nontrivial powers.
by Allan Wechsler 17 Jan '21

17 Jan '21

A number b^e, where b is a positive integer, and e an integer at least 2, we will call a "nontrivial power". This term is from Neil Sloane, and I think it's superior to the term "perfect power" which seems to have some currency. The sequence of perfect powers is collected as A001597 at OEIS. We naturally turn our attention to the b's and e's of the entries in this sequence, but there is some ambiguity here, exemplified by 16 = 4^2 = 2^4. Let us, for the moment, resolve the ambiguity by preferring, for this purpose, the representation with the smallest exponent. With this choice, the sequence of b's is A072813, while the e's are at A072814. The exponent sequence A072814 is dominated by 2's, because cubes and other higher powers are much rarer than squares. Our attention is attracted by the rare cases when two higher powers are sandwiched between consecutive squares. The smallest example is the pair 3^3 = 27 and 2^5 = 32, sandwiched between the squares of 5 and 6. The next example occurs between the squares of 11 and 12, 5^3 = 125 and 2^7 = 128. The third example has 3^7 = 2187 and 13^3 = 2197, sandwiched between the squares of 46 and 47. Are there an infinite number of such examples? Are there any examples of *three* higher powers between consecutive squares? Hugo Pfoertner and I have expressed cautious skepticism; if I read Hugo's tone right, he and I both suspect there aren't any. Perhaps it is too ambitious to call this a conjecture, as I did in the subject of this message. Similar questions about the existence of primes in intervals of various sizes are notoriously hard, but I would expect this problem to be much easier.

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[math-fun] September 10000, 1993
by Keith F. Lynch 17 Jan '21

17 Jan '21

There's a bit of Internet folklore that says September 1993 never ended. (Look up "Eternal September" for details.) September 100, 1993 was otherwise known as Thursday, December 9th, 1993. September 1000, 1993 was otherwise known as Monday, May 27, 1996. September 10000, 1993 is today, otherwise known as Saturday, January 16, 2021. I thought about the pattern of the days of the week of the successive rollover days, and concluded that there's just one day of the week that they will never land on. Can you see which one, and why?

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Re: [math-fun] September 10000, 1993
by Keith F. Lynch 17 Jan '21

17 Jan '21

Allan Wechsler <acwacw(a)gmail.com> wrote: > I'm guessing that anybody that has dived into calendar math for any > time at all can figure this one out. > I don't want to give spoilers, so I will digress a bit, with > apologies to Keith. A vaguely related curiosity is that given dates > of the year fall on the seven days of the week with *almost* equal > probability -- almost but not exactly equal. Right. Most notoriously, the 13th is slightly more likely to land on a Friday than to be any one of the other six days of the week. > This is because the central postulate of the Gregorian calendar is > that 400 years equals exactly 146,097 days. Unfortunately, this > period in days is exactly divisible by seven, so the day-of-the-week > pattern in our calendar repeats exactly every four centuries. Right. Isaac Asimov got this wrong, saying that the calendar repeats every 2800 years. As you said, it repeats every 400. (Except for the dates of Easter, which repeat every 5,700,000 years.) But the calendar repetition period has nothing to do with the solution to my problem, which would be the same even if the calendar never repeated. (An example of a non-repeating calendar is one in which a day is added to the year for every odd number the year number is divisible by and subtracted from the year for every even number the year number is divisible by. The average year length would then be a transcendental number, hence the calendar would never repeat. (There would also be zero-length and negative-length years, but those would be extremely rare.)) > Calendar math is full of weirdness like this. I *still* haven't > wrapped my head around the Hebrew calendar. The Islamic calendar > is far simpler in principle, and (intentionally) full of mystery > in practice. I thought the Islamic calendar was unpredictable, as it depends on when, after each new moon, the moon is first observed by someone in authority, which in turn depends on the weather. The solution to a related but more difficult calendar puzzle *does* depend on the details of our calendar. What about power-of-ten dates of *all* months? Is there any day which is *not* such a day? (Using the proleptic Gregorian calendar.) The 1st and the 10th of each month are of course the 0th and 1st powers of 10 of that month. But what about the other 26, 27, 28, or 29 days of that month? About once a month is the 100th day of some previous month, and about once a month is the 1000th day of some other previous month, and about once a month is the 10,000th day of some other previous month, etc. So most likely yes. But with an average of about 30.4 days per month, we'd have to go to at least 30th powers of 10, meaning most such previous months would be long before the Big Bang. Probably a lot more than 30th powers, since (presumably) nothing keeps such days from landing on the same dates, leaving other dates empty. Since there's no shortage of negative-numbered years (given that we've abandoned the constraint that the Gregorian calendar, the Earth, or even the universe, needs to have existed at the time) is it inevitable that every day has this property? Not necessarily, since the calendar does repeat. Successive powers of 10 number of days are 2.7379070069885... 27.379070069885..., 273.79070069885... years ago, etc. Since that number (400/146097) is rational, the decimal expansion repeats, so there are only a finite number of distinct fractional numbers of years ago, no matter how many powers of 10 you try. I don't know the period of repetition of 400/146097, but since 146097 is 3^3 * 7 * 773, I think it's at most 6*772=4632 decimal digits. Also, since the calendar does repeat, if we find solutions for 146097 consecutive days, we'll have found a solution for all days. A simpler version is what if every month had the same length? If so, I'm pretty sure that there would be no solutions for some days unless that month length was relatively prime to 10.

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Re: [math-fun] Conjecture about the distribution of nontrivial powers.
by Keith F. Lynch 17 Jan '21

17 Jan '21

Mike Speciner <ms(a)alum.mit.edu> wrote: > Some of the cubes and fifth powers are also squares. And some of > the cubes and fifth powers are the same, as they are fifteenth > powers. In the text you quoted in that very message, Allan Wechsler said: >> ... but there is some ambiguity here, exemplified by 16 = 4^2 = >> 2^4. Let us, for the moment, resolve the ambiguity by preferring, >> for this purpose, the representation with the smallest exponent.

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[math-fun] Regular (until they're flexed) dodecahedra are slightly flexible?
by Bill Gosper 16 Jan '21

16 Jan '21

But only this much <http://gosper.org/dodecasquish.gif>? (No stretch; some squish? I don't trust this.) —rwg

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