Re: [math-fun] Weird leap-year rule
apgoucher@gmx.com (Adam P. Goucher) wrote:
Year 2^n has 29 - n days, and year 3^n has 29 + n days. Hence, February can be of any integer length ...
Correct. gareth.mccaughan@pobox.com (Gareth McCaughan) wrote:
... which I think equals 365 + log 2.
I didn't follow your reasoning, but that's the correct answer. The way I figured it: All numbers are divisible by 1, so add 1. Half the numbers are divisible by 2, so subtract 1/2. A third of the numbers are divisible by 3, so add 1/3. A fourth of the numbers are divisible by 4, so subtract 1/4. A fifth of the numbers are divisible by 5, so add 1/5. etc... And it's well known that 1 - 1/2 + 1/3 - 1/4 + ... converges to the natural log of 2. To make it really rigorous I'd probably have to deal with the fact that that series isn't absolutely convergent, i.e. by rearranging the terms you can make it converge to something else; in fact you can get it to converge to any real number whatsoever. Log 2 comes from the most natural (npi) ordering, and as year numbers ascend, the integers are introduced in their natural order, for whatever that's worth. Anyhow, like Hans Havermann, I added up the first few terms (for sufficiently large values of "few") and saw that it does indeed seem to converge to that value. "I have only proved it correct, not tried it." -- Donald Knuth "Trust, but verify." -- Ronald Reagan gladhobo@teksavvy.com (Hans Havermann) wrote:
The February days-added numbers are < https://oeis.org/A048272 >. Checking the 1 0000-term link that is provided there for numbers < -28 tells us that this first happens in the year 2880.
Correct, that is indeed the first year in which February has a negative length. I ran a program to test every number until it found it. A friend of mine to whom I gave the problem in person worked it out in front of me with pencil and paper.
What then will be the first year in which a whole year has a negative length?
... 17297280.
Correct. One curiosity about this calendar is that, like ours, it's on a four-year cycle, even though the number four isn't explicitly in the rule. Years are successively: * long (>365 days) * normal length (365 days) * long (>365 days) * short (<365 days) and this pattern repeats forever. I also came up with simple rules for calendars which gave pi in the average year length. Can anyone else find such a rule? Or rules that give e, Euler's constant, or other interesting numbers in the average year length? Has anyone else done this sort of thing before, or is it original with me?
On 10/02/2013 13:59, Keith F. Lynch wrote:
The way I figured it:
All numbers are divisible by 1, so add 1. Half the numbers are divisible by 2, so subtract 1/2. A third of the numbers are divisible by 3, so add 1/3. A fourth of the numbers are divisible by 4, so subtract 1/4. A fifth of the numbers are divisible by 5, so add 1/5. etc...
And it's well known that 1 - 1/2 + 1/3 - 1/4 + ... converges to the natural log of 2.
Oh, that's sooooo much simpler than what I did! And I bet it's possible to make it rigorous enough despite the fact that the series isn't absolutely convergent. -- g
Keith's February puzzle led me to wonder: Let pf(n) denote the number of prime factors of n, counting multiplicity. Now define Q: Z -> {1,-1} via Q(n) := (-1)^pf(n), n > 1, and Q(1) := 0. QUESTION: As N -> oo, does the sequence S_N := { s(k) = Q(N+k) | k in Z+ } approach being indistinguishable from the outcomes of a sequence of independent Bernoulli trials -- i.e. of repeated flips of a fair coin (if H = 1, T = -1) ??? Such questions are always philosophically tricky, since the outcomes of independent random trials will have probability 0 of conforming exactly to any number-theoretic function like Q. Maybe a good statistical test would be essentially the one used to define when a number is "normal", base 2, i.e.: ---------------------------------------------------------------------- REPHRASED QUESTION: For every L in Z+, does the distribution of the values in the multiset S_(N,L) := {{ Q(N+1),...,Q(N+L) }} approach the distribution of L independent Bernoulli trials as N -> oo, ??? ---------------------------------------------------------------------- --Dan
This paper by Churchouse and Good is relevant: http://www.ams.org/journals/mcom/1968-22-104/S0025-5718-1968-0240062-8/S0025... Victor Sent from my iPhone On Feb 10, 2013, at 20:24, Dan Asimov <dasimov@earthlink.net> wrote:
Keith's February puzzle led me to wonder:
Let pf(n) denote the number of prime factors of n, counting multiplicity.
Now define Q: Z -> {1,-1} via
Q(n) := (-1)^pf(n), n > 1,
and Q(1) := 0.
QUESTION: As N -> oo, does the sequence
S_N := { s(k) = Q(N+k) | k in Z+ }
approach being indistinguishable from the outcomes of a sequence of independent Bernoulli trials -- i.e. of repeated flips of a fair coin (if H = 1, T = -1) ???
Such questions are always philosophically tricky, since the outcomes of independent random trials will have probability 0 of conforming exactly to any number-theoretic function like Q.
Maybe a good statistical test would be essentially the one used to define when a number is "normal", base 2, i.e.:
---------------------------------------------------------------------- REPHRASED QUESTION: For every L in Z+, does the distribution of the values in the multiset
S_(N,L) := {{ Q(N+1),...,Q(N+L) }}
approach the distribution of L independent Bernoulli trials as N -> oo, ??? ----------------------------------------------------------------------
--Dan
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participants (4)
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Dan Asimov -
Gareth McCaughan -
Keith F. Lynch -
Victor S. Miller