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This doesn't mean there's anything interesting about 37. Rather, the only interesting fact here is that a finite median here exists at all. Once you've establi
by dataflow 3y ago
This doesn't mean there's anything interesting about 37.
Rather, the only interesting fact here is that a finite median here exists at all. Once you've established that, it's guaranteed to be some prime number, because we're defining the median of a list to be an element of the list. It just happens to be 37 for this list, but it may as well have been anything else.
What could make 37 interesting is if we relaxed the definition of the median to be outside the set itself (which is entirely possible), and yet the limit still, and it still converged to 37 somehow. That would be wild.
- 2-718-281-828 3y agois it trivial that the series of medians for the second prime factor of numbers up to N is a specific number? isn't it just as plausible that this median would increase?
- drexlspivey 3y agoThe median of the first prime factor is 2
- codetrotter 3y agoBut intuitively it’s weird that the median of the second prime factor is not 3, right?
- drexlspivey 3y agoOnly multiples of 6 (1 in 6 numbers) have 3 as their second prime factor so I wouldn’t expect the median to be 3
- deleted 3y ago[deleted]
- eru 3y agoI guess it depends on whether you count multiples of 4 to have 2 as both their second and first prime factor?
- deleted 3y ago[deleted]
- NikkiA 3y agoI'd expect the mode of the second prime factor to be 3, but not necessarily the median.
- deleted 3y ago[deleted]
- cmiller1 3y agoThe sequence of the medians of successive prime factors would be interesting
- deleted 3y ago[deleted]
- LanceH 3y ago> isn't it just as plausible that this median would increase? Every sixth number will have a factors 2 and 3. (so 3 is the second prime for 16.666% of numbers) Every 10th number will have factors 2 and 5. Every 15th number will have factors 3 and 5. Of course then you have to subtract out every 30th number which contains a 2, 3 and 5. They ignored duplicates and it would have been nice if they had defined "first" and "second" properly. Also note that a prime counts the 2nd factor as "infinite", adding one to the "above 37" bucket. Anyhow, you add those up and it's easy to at least see that in any range of numbers it very quickly approaches 50%. Keep going to write out the Venn diagram and it happens to come out at 37. It's low enough, I assume it could even be done by hand.
- LegionMammal978 3y agoAs we take the limit as N → ∞, the list continually grows larger. Thus, the number of instances of 37 continually increases, so that they occupy an asymptotically constant proportion (about 0.963%) of the values in the list; only ~49.061% of the values are less than 37, and only ~49.975% of the values are greater than 37. After a certain point, for even N, there will always be two instances of 37 surrounding the 50% mark in the list, so the median will still be exactly 37, and not some other value. I wrote up a longer explanation in another comment [0]. [0] https://news.ycombinator.com/item?id=38245162 https://news.ycombinator.com/item?id=38245162
- tromp 3y agoOut of the integers 1..2N+1, exactly N have smallest prime factor 2, while the remaining N+1 have smallest prime factor > 2. Thus it is fair to say that the median smallest prime factor is 3 rather than 2.
- tromp 3y agoOops, that's not correct, since 1 has no prime factor. We should instead look at the odd-sized range 2..2N, in which N elements have smallest prime factor 2, making 2 the unique median. Only in even sized ranges 2..2N+1 is the median shared between 2 and 3.
- LegionMammal978 3y agoIn their original paper, De Koninck & Tenenbaum take the convention that the smallest prime factor of 1 is ∞, presumably on the grounds that the minimum value of the empty set is ∞. Under this convention, the range 1..N does make sense, with the median smallest prime factor being 3 for all N ≥ 3 odd, and 2.5 for all N ≥ 4 even. So formally, the limit as N → ∞ wouldn't exist. (They further take the convention that the second-smallest factor of a prime number is ∞, but the choice is ultimately irrelevant for the median of the second-smallest factor, since the primes have natural density 0.)
- 3y ago
- sriku 3y agoI'm curious about the first number to have absolutely nothing interesting to be said about it ;)
- xigoi 3y agohttps://en.wikipedia.org/wiki/Interesting_number_paradox https://en.wikipedia.org/wiki/Interesting_number_paradox
- vouaobrasil 3y agoIn short, it's not really a paradox because in the original joke, the assumption is that the first number to have nothing interesting about it is itself an interesting property. That assumption isn't true when you actually think about it...it's actually rather boring.
- antonvs 3y ago> It just happens to be 37 for this list What do you mean by “this list” here? It seems to be misunderstanding the result.
- jameshart 3y agoWell sure - the median is just one particular percentile - not a magically special one. The existence of a median also implies that there are probably limits for every other percentile - or, indeed, that for any number, there is some in-the-limit proportion of numbers for which the second prime factor is greater than that number (and for 37 that happens to be .5).
- madcaptenor 3y agoYes, and in small cases you can work out the distribution explicitly. For example: - 1/6 of all have their second smallest prime factor equal to 3 (that is, they have a factor of 2 and a factor of 3) - 1/5 of integers have a factor of 5; of those, 1/2 have a factor of 2 or a factor of 3, but not both (that is, they're congruent to 2, 3, or 4 mod 6); so 1/10 of integers have second smallest prime factor equal to 5. So the proportion of integers with second smallest prime factor equal to 5 is 1/6 + 1/10 = 8/30. Thus the first quartile of second smallest primes is 5, and similarly for any quantile between 1/6 and 8/30. Don't ask me to work out the third quartile - this distribution has a really long tail. In particular I suspect the limiting distribution has no mean.