This wiki page page inspired me to do discover with being a natural number;
I will just start using so I can discover the next ones;
Using the same method that the linked page had done with and , I will discover the polynomial that is equal to ;
First, I will use binomial theorem:
The will be always 1, so it will not change the result if not written
(I took the "" out of the sum notation)
As the page did, I will substitute n by 1,2,3....n
Then I will sum everything, notice that everything on the left side, except and cancels themselves, for the right side, every sum has now another sum inside it!
I will work the left side now, remember what is, but now, we will have to take the , because it will cancel with :
Now, I will take the term that appears on the sum of the right when i=a out of the notation:
Notice that this term is exactally what we are searching, so I will isolate it:
Remember that :
Now, Remember that, when , we have a polynomial expression , then if also "is"(As in: there is a polynomial expression that has the same values) a polynomial expression, and, by a strong induction, if a is natural, then the expression also "is".
I also made a program, in Python, that returned the expression of , but to much more values other than 0, 1, 2, 3(Which the original page did), some of the results are:
There are patterns there, but I will make another note to that, as this is getting huge