Sometimes it's useful to have a number which has lots of factors. Like the number 12 which can be divided by 2 and 3 and 4 and 6 and 12 itself. We find these things useful sometimes. The ancient Greek philosopher Plato He thought the best choice for something like this was the number 5040. He thought this was the best choice because it had loads of divisors, lots of factors. So, 1 divides into it obviously. 2 divides, 3, 4, 5, 6, 7 divides, 8 divides, 9 divides, 10 divides So all the numbers up to 10 divide, 12 divides into it as well, And it's got 60 divisors all together. So Plato was thinking this is the best number you could have for, like say, a city. If the population of a city was 5040 you could divide that up into all kind of different groups. If you wanted to divide up the land Then you'd divide it up into units of 5040. 5040, lots of divisors, plus it has more divisors then all the numbers less than 5040. We now call this a Highly Composite Number. [Brady Haran] It's like an Antiprime [James Grime] It's like, yeah, it's like an Antiprime it's the most divisible number, lots of divisors In fact the first guy who really properly studied this was the famous indian mathematician Ramanujan did it about 100 years ago Let's take a look at properties of Highly Composite Numbers. Antiprime Numbers (laughs) I don't think it's gonna catch on Brady, I think you are 100 years too late for that! So the definition of a Highly Composite Number is one that has more divisors, factors if you wanna call it factors, more divisors than any number smaller than it let's just run through the numbers, let's find some so these are like the previous title holders yeah, exactly! How many divisors for 1, it's just one. So, 2, it's a prime, so like primes do, the
only things that divide primes are 1 and itself it has two divisors, primes have
two divisors so 3 would be the same, it has two divisors 4, now 4 is gonna be different it can divide by 1, 2 and 4 so it has
three divisors, and hey! This has more divisors than the others before it so now this is the current winner This is the highly composite number so far. Let's see what 5 does, well 5 is a prime so 5 loses all right, way down there, 2. So 4 is still the title holder? Yeah, 4 is still winning, but then 6 comes along oh dear, 6, it can be divided by 1, 2, 3 and 6. Four divisors, and now 6 is ahead of the game. We keep going, oh 7, ah no
7 is no good, is a prime it has two divisors 8, does 8 do any better? So we can
divide by 1, 2, 4 and 8 is has four divisors and it's not better than 6 then so four divisors, no it hasn't won
anything, and 9 has three divisors so it's 1, 3 and 9 10 has four divisors 11, prime, two divisors. 12 has six divisors, cause, I told you 12 is one of these numbers that have lots of
things that divide into it 1, 2, 3, 4, 6 and 12. So that has six divisors, 12 is really good, really up there and then so on. So then yeah, well let's have a look at the title holders there, ok highly composite numbers let's write out the sequence 1, you're correct, 1 is there 2 is there 4, 6, 12 and if we carry on, 24, 36, 48, 60 60 is a good number, that's why we have 60 minutes in an hour,
60 seconds... 120, 180, 240 360's there like degrees in a
circle lots of things divide into 360, it's a good number to do 720 and 840 and so on. right and they carry on like that. So these are, this is our sequence of highly composite numbers There's is a very important
theorem in maths called the Fundamental Theorem of Arithmetic if you want its fancy name. It means all
positive whole numbers can be written as a product of primes by multiplying
prime numbers together this is why primes are important, they're our building blocks for other numbers they're our atoms for other numbers So all
other numbers can be written like this if you had a number you'd have primes, let's just call them just call them prime 1 to a power, prime 2 to some power, prime 3 to
some power, and that would go on and then and you would end at some point you have
the last prime here prime K to some power let's do an example so if you had the
number 30 it's built up from primes it's 2 times 3 times 5. Primes that divide 30
has to be either 2 3 or 5. 7 doesn't divide yeah so if i had if i do another example
had 550 it be 2 times 5 squared times 11 up three doesn't divide into it, seven
doesn't divide into it 19 cannot divide into this number I know if you want to do you want to
have a factors you just use that idea repeatedly so the factors are the primes that
divide into it repeatedly so I could divide this by 25, 'cause I can divide by 5 and 5 again or
if I want to divide by 10, I can divide it by 2 and then divide by 5. So the factors are just all the possible
combinations or permutations of these prime atoms that you've been given to
work with They would all look like this all the
factors would look like this would be the same primes to be a P1 and P2 P3 to Pk you would have powers here, let's call them B1 and B2 and B3 Just these powers would be 0 or 1
or 2 or 3 or up to and including the final thing So anything less than what I've called a
pair Now those are your factors so how many factors are there just to show you how many factors that are then the factors
of a number of, let's call it n, are the divisors of n is a well how many
choices do I get for these powers and it's this it could be 0, 1, 2, 3, 4, anything look to A1 so it be at A1 plus 1 multiplied by how many choices for the second prime power 0, 1, 2, 3, 4, anything up to A2 that's 1
plus A2 choices You just do the same thing for each of
these prime powers so that would go up to the last one which is Ak plus 1 and
that's how many divisorss the I'll do an example of these shall I So if I do like 30 and look for the divisors
of 30 I can use this here formula look called the powers are just one How many choices would I have for each of
these powers there's two choices there two for that one two for that one So all the powers are one and this is going
to be 2 times 2 times 2 which is 8 and there are 8 divisors of 30 and if I
did it for 550 slightly different 'cause I've got this square in it so many choices
here for the first prime power it is 2 and so I would have 12 factors of 550 so we've seen how to work out divisors
and we should just check 5040 mentally great 5040 let's see how many devices
that has if you break it down into primes then it's going to look like this It's 2 to the power 4 multiplied by 3 squared multiplied by 5 multiplied by 7
and so let's look at the divisor formula so we want divisors of 5040 and we can
use the powers to help us work out So it's just gonna be 5 by 3 by 2 by 2 and
that's going to be 60 so there are 60 devices are 5040 which is greater than
the number smaller and 5040 that makes it highly composite a hundred years ago
Ramanujan's those studying these and notice three properties that highly
composite numbers have to have which I'll show you now and the not too
difficult to understand but the first property is the primes of the
factorization of our highly composite number have to be consecutive primes I mean look that's what happened here
you've got 5040 at the where they were consecutive primes they were 2, 3, 5, and 7 if you look at 550 just to compare it to something that doesn't
work and that wasn't consecutive primes and that was 2 by 5 squared by 11 now I know that this is not a highly
composite number because I could replace this 11 for one of the missing
primes I could replace it for 7 and it
would be a smaller number but with the same number of divisors same number of
factors so it's not going to be highly composite or the best choice would be if
I picked a number which had consecutive primes so if I picked i'm going to use the same
powers they like they are there so I'm going to use this 2 by 3 squared by
5 yeah that's better that has consecutive
primes it's a lot smaller number is 90 and 90 we can see has the same number of
devices than 550 so this is much better choice than 550
so it failed so yes if you've got a highly composite
number of primes are consecutive that's the that's kinda nice the second thing Ramanujan noticed is the
powers in their prime factorization they have to be weakly decreasing they have
to be going down like this so you can see it here in 540
look 4, 2, 1, and 1 so they're going weakly down so they're
not increasing but it didn't happen here with my 550 didn't happen
here with my 90 either I tell you why because I can make a
better choice if I swapped the powers around and if i
put them in a decreasing order it would look like this I can have a 2 squared multiplied by 3 multiplied by 5 so it's the
same powers but in a decreasing order that makes the number smaller as to why
is that a 60 so now you've got 60 there has the same
number of divisors and hey this is the most better choice than 19 in fact 60
there's a highly composite number at the third thing Ramanujan noticed was these highly
composite numbers all end with a with with 1 as the last power so they always end
up with a single prime there at the end and so it doesn't end with a square and
then there's actually a couple of exceptions for that there are two
exceptions these are highly composite numbers that
break that rule 4 is a highly composite number and that's 2 squared and the other number is
a 36 which is a 2 squared by 3 squared what Ramanujan showed is that highly
composite numbers have to end with a prime that has a power one or two and
two has these two exceptions there and everything else they all end with one
with their prime power at the end oh that's less obvious less obvious than
hear the facts i showed you i took some while to prove but it's another
necessary condition for a highly composite number [Brady Haran] In addition to our usual supporters we'd
like to thank audible.com for supporting this video if you have to cover a few
miles in planes or trains or automobiles one of the best ways to pass the time is
listening to audiobooks so while you watch me driving here from
Bristol to Nottingham let me tell you about audible.com they've got a huge
range of titles a great app and a good offer for new customers but before I
tell you about that first recommendation and I'm going to suggest airframe by
michael crichton it deals with the topic i always find fascinating that is air
crash investigations maybe not want to listen to on a plane
if you're a nervous flyer but definitely worth your time now audible are offering a free 30-day trial
of their service which includes your first book if you go to
audible.com/numberphile
using that URL will mean they know you came from here that
address again audible.com/numberphile for the free trial I use audible I think they're well worth a look and
I'd like to thank them again for supporting this numberphile video [Brady Haran] Not very anti-prime
[James Grime] Stop trying to make anti-prime a thing Stop trying to make fetch a thing
If I formulate the fundamental theorem of algebra to say that any number can be written as a product of unique prime number multiset (set with repetitions) P, am I correct to say that the set of factors (divisors) F is equal to the products of individual sets in the power set of P?
Just panicked thinking I've been pronouncing "Ramanujan" incorrectly all these years. Google reassured me, it is "ra-MAHN-uh-jahn," not "ra-mahn-OO-jahn."
Wouldn't normally have a problem with a little mispronunciation, except that I was inclined to believe him over myself.
So couldn't I construct my own highly composite number by having a ton of consecutive primes raised to insane powers?
Here is a pattern I noticed, does anyone know anything about it?
Is this chain infinitely long? Let us define d_1 = 4 and conjecture that for every i, there is a highly composite number d_{i+1} with d_i divisors.
Are there other infinite chains (i.e., choose different initial condition for d_1)? For instance, 24 -> 360 -> 3603600 -> ... Maybe all chains are infinite? Define an infinite graph whose vertices are highly composite numbers, with edges { u->v | d(v)=u }. Let's conjecture that the graph is a disjoint union of infinite paths (meaning every chain is infinite). What is the relationship between a highly composite number and its rank along its infinite path?
edit: Well, that was short-lived. According to A009287, the smallest number with 293318625600 divisors is not highly composite. That sure is weird. It rules out my conjectured infinite chain above, but the existence of some infinite chain is still unknown to me, though.
Can you explain why this is true?
There seems to be some pattern in the spacing between HCNs but my math skills aren't good enough to make anything of it:
Is there anything that can be discerned by or about this pattern?
God they are so beautiful! Just looking at highly composite numbers has always made me feel so good, it's weird. I don't know why... They're like make-me-happy-numbers!ο»Ώ
This video caused me to write a mini-python program to generate anti-primes up to whatever value you want!
Don't make fun of my coding, I've never had much formal training :P Here's an example of output
For what it's worth, I find the most interesting fact that 4 & 36 are the ONLY highly prime numbers with an odd number of factors (i.e. it's a perfect square). Seems arbitrary to me!
And yes I know this is literally the least optimal way to calculate this. This is totally brute force. A much easier way would be to manually compute a list of primes up to sqrt(num), and use this list to find the prime factorization, and then find the number of factors via the method James used.
But that's a harder program. Maybe I'll do it later.
EDIT1: Made the program a heck of a lot more efficient just using some super basic common sense. I.e. all highly composite numbers must be even, as they all have at least one factor of 2. AND once you get past 60, all highly composite numbers have at least a 2 & 5 (so they are a multiple of 10)
So I added in
So it goes roughly 10 times as fast now. I'm guessing I could do this for a few of the bigger breakpoints (i.e. when they all become multiples of 30, 150, etc.)
I'm curious about the sequence of the number of divisors of each successive HCN (highly composite number). In this table at wiki you can see the number of divisors, d(n), for the first several HCN's. The successive higher numbers of divisors d(n) has weird jumps.
Here's the first several jumps:
1,1,1,2,2,1,1,2,4,2,2,4,6,2,4,4,8,12,4,8,8,4,6,6,4,8,12,8,16,16,8,12,12,8,16,8,16
Just curious if anyone has some insights.