Help_

[subtle bane]

@steady fern

@steady fern a partition is basically a sum of integers that sum up of a total n.

@steady fern

[steady fern]

I've never seen this before but it looks interesting.

looks like we need a construction methodology for generating symmetric partitions

techinically, this would be recursion not induction

[subtle bane]

the thing is we are asked to not do a generating function;

@steady fern We can utilize this so called Ferrers diagrams to consturct symmetric partitions

[steady fern]

ok

I'd start out examining simple symmetric partitions and looking for patterns

I think I got it

[subtle bane]

ok, I am still figuring out what k can be to make it a symmetric, cause like for n=9, I have k=5 a symmetric, can it be greater or less thatn 5 I do not know

[steady fern]

for n=9, k<=5

e.g. 9 = 6 + 1 + 1 + 1 //only 4 elements, fail

lets look at this example where n = 11 and k = 4

what does the pink represent?

[subtle bane]

Is the larget k

and green are the n-k

[steady fern]

n-2k+1

[subtle bane]

this is the formula of largest k ?

cause it gives 4 for , n=9 , when palguing in n-2k+1

oh, green is n-2k+1

For induction : We can calculate for n=11, and then extend to n+1 , but k should be fixed

[steady fern]

recursion not induction

[subtle bane]

ok

[steady fern]

lets look at n = 17, k = 5

[subtle bane]

I think the first one is 16 not 17 tho

[steady fern]

sorry

[subtle bane]

both are symmetric

[steady fern]

maybe this will help:

[subtle bane]

we see the diagonal stays fixed but we can move the last 2 blue into green

[subtle bane]

So we can deduce from this particular exmaple a recursion?

Is this the goal?

[steady fern]

ignoring pink, green and blue are the instances that solve P(8,4) and P(8,3) respectively

P(8,2) and P(8,1) have no solutions

[subtle bane]

But are not we suppsoed to fix k , and vary n

Also, are u saying that a symmetric Ferrers diagram can be built from a previous diagram if we remove the 1st row and column and so on?

Can we split k into cases when k is even and when k is odd?

[steady fern]

stop thinking induction, think recursion

[steady fern]

yes

[steady fern]

we dont have to

[subtle bane]

by doing this we vary k, we need where k is fixed i think ,and only n varies

[steady fern]

think about it. I fear I've already been a little too heavy handed with my hints

but feel free to ask questions

[subtle bane]

@steady fern : I understand, but I am saying we are solving for different condition it seems like. I have been provided an exmaple for fix k, let k =3, then all possibel partitions where n varies, but k stays same , are : 3 1 1; 3 2 1; 3 3 2 ;

and for k =2: we have : 2 1; 2 2

This is the formual we need to deduce

[steady fern]

aren't we trying to find P(n,k)?

[subtle bane]

we are looking for formual to count symmetric P(n,k), such that k is fixed but n is varying

[steady fern]

I'm defining P(n,k) as the number of symmetric partitions of n into k bins

if P(n,k) is all partitions, rename my function S(n,k)

if you want every symmetric partition where k is fixed but n can vary, then sum(S(n,k) for all n)

or bound it, it doesn't need to be an infinite sum since most of them will be 0

[subtle bane]

got it

Thank you !