He's really familiar with Numberphile lore! I love the idea that Brady's videos have entertained and inspired so many (including me) to build a love for mathematics.

@greatquux

2 жыл бұрын

Yeah it’s incredible how this channel has now influenced mathematics- kids have grown up on it!

@peterwolf8092

2 жыл бұрын

And I gave that comment the 123rd like. So pleasing... Thats what those videos do to me 😆👌

I love how you buried the lead that his Erdos number is 2. That is impressive!

@renerpho

2 жыл бұрын

It also hints, without mentioning it, at the fact that Lichtman has been working with Carl Pomerance for a number of years. In fact, some of the results mentioned in Lichtman's video here are due to Lichtman&Pomerance. Carl Pomerance, of course, shares his story of how he got Erdős number 1 in this Numberphile video: kzread.info/dash/bejne/k3elkpaHaczaaLg.html

@bsharpmajorscale

Жыл бұрын

I know someone who's Erdos number is 2, but maybe 1.5? He had driven Erdos around a couple times when he was in our neck of the woods (southeast US). I can't remember specifically how he said his was +0.5, I think it was something to do with talking maths with Erdos and writing a paper with the benefit of those insights.

2:04 I love how he is so hesitant to say the complement of 24.7%... Don't get me wrong, it is very relatable for a mathematician

@DrTacoPHD665

2 жыл бұрын

If I was someone with enough mathematical prowess to prove a conjecture, I would be deathly afraid to get a calculation wrong on Numberphile.

@kannix386

2 жыл бұрын

the measure of the complement set

@maxwellhosler4918

2 жыл бұрын

Never, never do arithmetic in public.

@bsharpmajorscale

Жыл бұрын

That's the beauty of math(s). You get so far into advanced stuff that you forget the easy stuff. Can't count the times I've done recurrence relations, only to find out I've messed up way back at the beginning because I forgot to carry or said 15-9 was 4 or something like that.

Love this guy, he's so into it! You should have let him explain why perfect numbers form a primitive set. It seems like Jared was eager to prove it.

@clanknfrends1

2 жыл бұрын

Honestly I was hoping he would

@GerSHAK

2 жыл бұрын

+

@thomasanderson9383

2 жыл бұрын

Yes! That would be a cool proof to see

@0:55 I find the gleam in his eyes so satisfying. Reminiscent of my kids when I express interest into whatever interests them.

@recklessroges

2 жыл бұрын

I'm loving seeing these young mathematicians thriving.

I love how he causally mentions a numberphile video from 2011.

The perfect numbers form a primitive set because any (nontrivial) multiple of a perfect number is an abundant number. Suppose n is a perfect number with proper divisors d_1, d_2 ... d_i, which add upto n. Then kn will have divisors kd_1, kd_2 ... kd_i which add upto kn, but also will have at least one other divisor, namely 1. Hence, kn is abundant.

Please have Jared Lichtman back, he’s the queen’s precious diamond

This guy was likely born after Erdos died in February 1996 (just guessing since he graduated Dartmouth in 2018). Very impressive with his whole career still ahead of him

@JCCyC

2 жыл бұрын

From the news articles on the proof, he's just a couple of years older than that. So his Erdos number can only be 1 if Erdos changed his diapers.

Hearing him talk about Erdos all the video long, I was wondering about his Erdos number too. I am so impressed that his number is 2 that said "WOW" out loud. ^^

@renerpho

2 жыл бұрын

You're in for a nice surprise then to learn WHY he has Erdős number 2. Lichtman has been working with Carl Pomerance for a number of years. In fact, some of the results mentioned in Lichtman's video here are due to Lichtman&Pomerance. Carl Pomerance, of course, shares his story of how he got Erdős number 1 in this Numberphile video: kzread.info/dash/bejne/k3elkpaHaczaaLg.html

@GuidoHaverkort

2 жыл бұрын

can you explain what it means that his number is 2? I'm sorry I'm a noob

@Ceelvain

2 жыл бұрын

​@@GuidoHaverkort Erdos number of Paul Erdos is 0. Anyone who co-authored a paper with Paul Erdos has an Erdos number of 1. Anyone who co-authored a paper with someone with an Erdos number n has an Erdos number n+1. It's basically someone's shortest distance to Paul Erdos in terms of co-authoring a paper. Meaning this dude co-authored a paper with someone who has worked with Paul Erdos himself. I didn't check lately, but my Erdos number is likely 5.

What I want to know is how was the "fingerprint formula" picked as "the" formula? Why is it that versus something else?

Wow, Erdos number 2. Nice!

We need more videos with Jared, this guy is a greater presenter!

I find it oddly heartwarming that one of the most kickass mathematicians in the world has to take a few seconds to calculate 100-24.7.

He won a Numberphile video, Brady!

What a great man! Obviously very mathematically gifted, but also humble, well spoken and he had great handwriting! I look forward to hearing from him more!

TWO!

I really like the phrase "vanishingly small"

Two great videos. Bravo, and thank you!

I love how he is a huge fan of Numberphile (or just did his homework before the interview). But he knows all the videos and their progression and everything. lol

I like this guy.

555 you asked about is Erdos number, and I posted comment about it on main video just before watching that one. It recalls me the 15-theorem, for the idea of primitive sets....maybe a minimal number of elements and a rule providing them all in the primitive set would be great. In dynamical systems, computing n-orbit, we like the property than m doesn't divide n in the same set.

If a quarter of the numbers are abundant because the sum of their factors is greater than the original, I wonder how much you need to subtract from the original number so that the statistic would be 50-50 between "pseudo"-deficient/abundant numbers.

@oresteszoupanos

2 жыл бұрын

You'll make a fine mathematician, asking questions like that! :-D

@wiseSYW

2 жыл бұрын

maybe we could divide the deficient numbers between those that sums above half the number and below

@danielyuan9862

2 жыл бұрын

I don't think adding or subtracting would be helpful. It's more like multiplying and dividing here.

@RaiinWing

2 жыл бұрын

​@@wiseSYW the threshold probably needs to be a.bit higher given that all even numbers satisfy that, but I reckon that's pretty close

Huh. I'm looking forward to the actual video.

awesome!!!

the secret of the perfect number is the number (6) and the number (4) they are opposites start with (6) which is part of the family 3.6.9 then (4) which goes through all other perfect numbers

To get a low Erdős number is getting more and more difficult. Erdős is dead and many people with Erdős number 1 have died... My Erdős number is 3 😵

Hi i have an unrelated math question, ive been trying to figure out what this series converges to. Well, idk if it can be called a series but it does converge, to around 1.47. Basically, you get all the odd numbers between two powers of 2, like (3), or (5, 7), or (9, 11, 13, 15). Then, you make these the numerators of fractions over the power of 2 right under it. (3/2) (5/4, 7/4) (9/8, 11/8, 13/8, 15/8). Finally, you take the geometric mean of all these numbers. (3/2)^(1/1), or (5/4 * 7/4)^(1/2), or (9/8 * 11/8 * 13/8 * 15/8)^(1/4) As you take larger and larger geometric means the value converges, and i have found an explicit form for the function using double factorials and a limit to infinity. lim as n-->infinity ( (2^(n+2) - 1)!!) / (2^(n+1) - 1)!!) )^(2^-n) ( _____________________________ ) ( (2^(n+1))^(2^n) ) (the 3 parentheses pairs are supposed to be 1 big parentheses pair) This fraction is much simpler than it looks, the first part on the top left would just be 1 under a power of 2, lets say n=2, then the top left gives you 2^(2+2) - 1, or 15. This number is then double factorialized to give you (15*13*11*9*7*5*3) Then the number it gets divided by, which is the part on the top right, is just -1 of the power of 2 right before, and if n=2 as it was in the last example, itll be 2^3 - 1, or 7. 7!! is (7*5*3) This leaves only 2 numbers in the top half of the fraction, 15!! divided by 7!!. When you divide the 15!! by the 7!!, the (7*5*3) gets removed from the equation, and youre left with just the (15*13*11*9) which is the product of the numerators for the fraction. The bottom half is just for the powers of 2 multiplied by themselves, so in this same case where n=2, 2^3 is 8, which then gets multiplied by itself 2^2 or 4 times. Both of these line up with the original idea, 15*13*11*9/8*8*8*8. Finally the exponent outside of the big parentheses is just to make it a geometric mean now, being the square root of the number of terms. So in the case where n=2, you get the whole fraction to the power of 1/4, or the 4th root of the whole fraction, which again lines up with the original idea of a geometric mean for the product. This formula works for any positive integer "n" as far ive tested, reals/complex dont matter to me here. The only thing i want to know is an explicit answer as n goes to infinity that removes the limit and is exactly what the convergence is, an explicit solution to the limit that is only 1 number. I know this number has to be about 1.47 as n goes to infinity, and its irrational, i just dont know what it could be. I was thinking it could related to pi since its double factorial, it could just be an exponent or a root of a number, it could just be a solution to some polynomial, it could be proportional to e or be a log, honestly anything. I just dont know enough math to turn a limit or a series into a finite product. Thanks even for reading it i just thought this was interesting, and wondered if the solution led to an important constant/irrational number.

@bethmmq3840

2 жыл бұрын

Hi there! I got intrigued by your comment and I've found the answer to your question... but in a kind of shameful way. I've computed the first 9 terms of your series, seemingly converging to about 6 correct decimals, and after a few unfruitful manipulations to try and obtain something familiar, I've simply googled it. :-/ I'm not proud of it, but there you go: your constant is 4/e. It's far too late at night for me to start digging into why that should be the case, but at least now you know what to aim for. "The proof is left as an exercise to the reader", as they say! :-)

@aitorcazalis2307

2 жыл бұрын

@@bethmmq3840 THANK YOU SO MUCH

@aitorcazalis2307

2 жыл бұрын

@@bethmmq3840 Hey do you have any links to where you found it? I cannot find it.

@bethmmq3840

Жыл бұрын

@@aitorcazalis2307 Oops! Sorry for the very late response. It seems that I log into my YT account only about twice a year and I'm only now seeing your question. :-) I'll answer anyway. As far as I remember, as I mentioned in my first message, I just googled the numerical approximation I had computed, including the first 6 correct decimals or so, and some of the search results mentioned 4/e, which I then probably checked at higher accuracy (computing more terms) for confirmation. Doing that search again today, with 1.471517, one of the results is a discussion on Mathematics Stack Exchange titled "Factorial limit convergence on e", which, at first sight, might very well be related to your own question. I'll read it more in depth later, but I hope it could already be interesting to you. (That's assuming my response is still relevant 8 months later and that you haven't read this discussion yet, so a pretty big assumption.)

What's the fingerprint number for the perfect numbers?

@conanpickford4617

2 жыл бұрын

0.104... . the perfect numbers is sequence A000396 in the OEIS and you can just use a calculator to find an approximate sum

@GuzmanTierno

2 жыл бұрын

@@conanpickford4617 Yes, thank you, you're right ... 1 / (6 * ln(6) ) +1 / (28 * ln(28) ) +1 / (496 * ln(496) ) +1 / (8128 * ln(8128) ) +1 / (33550336 * ln(33550336) ) +1 / (8589869056 * ln(8589869056)+1 / (137438691328 * ln(137438691328) ) = 0.104... I don't think however that this number has any particular meaning ...

@GuzmanTierno

2 жыл бұрын

And ... is the set of perfect numbers a maximal primitive set??

@jordanlinus6178

2 жыл бұрын

@@GuzmanTierno No, you can add 434 to it. All even perfect numbers are of the form 2^(p-1)(2^p - 1) where 2^p - 1 is prime (and all number of that form are perfect). Which means on one hand that no even perfect number has more than two different prime factors (thus there are no multiples of 434), and all even perfect numbers greater than 6 are divisible by 4 (thus there is no divisor of 434). Now odd numbers are not divisible by 2 (proof left to the reader), which means there can be no odd multiple of 434. All that is left to show is that there are no odd perfect numbers dividing 434, but for this I will use the fact that 434 < 10^1500, and there are no odd perfect numbers < 10^1500 (and probably there are none at all).

@GuzmanTierno

2 жыл бұрын

@@jordanlinus6178 thank you, very clear

If we throw out both trivial divisors (i.e. the number itself and 1), then we still don't know whether any perfect numbers exist, do we?

3:49 hey, umh, are you the person shown on the left right?

About proving that if you have two perfect numbers, one does not divide the other. I can't even begin to tackle this, but I smell a proof by contradiction. Assume that PerfA = k * PerfB, explore what k has to be, and something awful happens. Did I guess right?

@Zach8877

2 жыл бұрын

Yeah I think you’re on the right track. Then you notice that {k*f for all factors f of b} are all factors of PerfB*k. The sum of that set is equal to PerfB*k. But since 1 is also a factor of PerfB*k, the sum of the factors of PerfB*k is greater than PerfB*k. So the sum of the factors of PerfA is greater than PerfA making it impossible for a to be a perfect number.

👍👍

How sweet, that took him a while to get 75.3. Proves that math is more about ideas, not numerology. Too bad I got it right away without any effort..;) Props to him though, I look forward to reading his paper

@scottdebrestian9875

2 жыл бұрын

I think there's added pressure when you're on camera -- you don't want to be responsible for the next Parker square by accident, after all!

Didn't know his Erdős number is already 2. Still think he should get an honorary 1.

@renerpho

2 жыл бұрын

Lichtman has been working closely with Carl Pomerance for a number of years. Pomerance, of course, has Erdős number 1. He tells the story of how he came to collaborate with Erdős in this Numberphile video: kzread.info/dash/bejne/k3elkpaHaczaaLg.html

What's an example of a maximal primitive set that is not the set of k prime-factors numbers ?

@jordanlinus6178

2 жыл бұрын

Start with all the perfect numbers, and go through all the natural numbers in order, adding all the numbers which keep the set primitive (i.e. that are not a divisor or a multiple of a number already taken). Or you could do the same but start with {2, 9}, or any set containing at least two numbers with a different number of prime factors (but it's a lot cooler to start with perfect numbers).

@GuzmanTierno

2 жыл бұрын

@@jordanlinus6178 yes, ok, I was looking for a set with a more descriptive definition ... of course any primitive set can be enlarged (in multiple ways) to reach a maximal set ...

@Conflux4761

2 жыл бұрын

The primes, but with 4 instead of 2. I think that should work, since any other even number is a multiple of 4 or one of the other primes?

@danielyuan9862

2 жыл бұрын

@@Conflux4761 I agree that it should work

Does anyone know what it means here to say "the probability of a number being perfect/abundant/deficient is x%"? Are we taking the weak limit of uniform measures on the first n natural numbers as n tends to infinity?

@danielyuan9862

2 жыл бұрын

I don't know what a "weak" limit is compared to a regular limit, but that's what I had in mind.

@pigworts2

2 жыл бұрын

I think this is probably in the sense of natural density. I'm not sure that the uniform measures converge at all, weakly or otherwise (since any "uniform" measure over the whole set of integers cant satisfy the countable additivity property, and so can't be a probability measure).

@gammaknife167

2 жыл бұрын

@@pigworts2 don't take my word 100% on this but I think you can take a weak limit, it just won't converge to a probability measure. It might even coincide with natural density if you take a weak limit of uniform measures.

@pigworts2

2 жыл бұрын

@@gammaknife167 ahh okay, that makes sense. Yeah, it does seem like it should match natural density.

this guy is a don

Ha-ha. There is no such thing as "picking a random number" (a random integer from ℕ to be precise) 😀

Numbers of the form n^n - n