Question

Can anyone explain this part of a partition congruence?

Answer 1

i don't understand how is can be used to get the exact values of partitions ending in 4 or 9

Answer 2

@satellite73 @Michele_Laino

Answer 3

@Kainui

Answer 4

@ganeshie8 Any luck with the congruence?

Answer 5

.

Answer 6

haha no?

Answer 7

It can't be used to get the exact values of a partition ending in 4 or 9. What the congruence tells use is that the number of partitions will be congruent to, or divisible by, 5. So while it can't tell us that, say 9 has 30 partitions, it DOES tell us that 9 cannot have 29 or 31 partitions.

Answer 8

Can you substitute a value and demonstrate an example? I think that would really help me out

Answer 9

Sure. Let's consider 4. We don't know how many partitions 4 has, but Ramanujan's Congruence tells us that it must be divisible by 5, so the number of partitions must be in the set {0,5,10,15,....5x}. Therefore, it cannot be 4 or 6 or any other value not in this set. The actual number of partitions can be iterated for a number as small as 4:
4
3 + 1
2 + 2
2 + 1 + 1
1 + 1 + 1 + 1
for a total of 5 partitions. This is a number congruent to 0 mod 5, so Ramanujan's congruence holds.

The actual partition function, the function that gives the number of partitions for a given value, is this:
https://upload.wikimedia.org/math/f/6/6/f667eb0488ccae6fcabb705e39df2eba.png

Answer 10

But do you know the symbols in the formula represent? I do know what you mean, but like what does k represent?

Answer 11

And also, are you familiar with how that generating function works for the exact amount of partitions?

Answer 12

like if you could explain pretty much how the theorem was made in the congruence one and generally how the generating function one works (that one I have no idea how it gives you perfect numbers of partitions) that would really help

Answer 13

In the congruence p(5k+4) == 0 (mod 5), k is any whole number. 5k + 4 is just a symbolic way of saying any number that's 4 plus a multiple of 5.

In the equation I linked, n and k are used as the indices of the addition and multiplication functions and range from 0 to infinity and 1 to infinity, respectively.

How to actually calculate p(k) is complicated, but details can be found here:
https://en.wikipedia.org/wiki/Partition_(number_theory)#Partition_function

Answer 14

The number theory that gives these results is quite complicated and draws on a number of other theorems and I'm not sure I could explain them if I wanted to, so unfortunately I'm not going to be able to help you there.

Answer 15

http://www.math.uiuc.edu/~berndt/articles/partitions2.pdf

Answer 16

Well this seems like it should be the first thing to understand, \[\sum_{k=0}^\infty p(k)x^k=\frac{1}{(x)_\infty}\] Really the infinite product is a bunch of geometric series, and you can exactly calculate p(k) for small k by only looking at finitely many terms of the geometric series.

Answer 17

also does anyone know why the hardy-ramujan formula gives decent approximations?

Answer 18

my main problem is now the generating function, i need to refresh on geometric series