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After asking a question here let us consider on the following case which is simpler, this might help me to make some progress: suppose we have the following polynomial: $$f(x)=(x^0+x^1+x^2+x^3)^3$$ expanding this gives: $$f(x)=1 + 3 x + 6 x^2 + 10 x^3 + 12 x^4 + 12 x^5 + 10 x^6 + 6 x^7 + 3 x^8 + x^9$$ Aim is to find the coefficients for a given power. Now let us use multinomial theorem. Let us now call: $$a=x^0$$ $$b=x^1$$ $$c=x^2$$ $$d=x^3$$ then the polynomial becomes: $$f(a,b,c,d)=(a+b+c+d)^3$$ which has the following expansion: $$f(a,b,c,d)=a^3 + 3 a^2 b + 3 a b^2 + b^3 + 3 a^2 c + 6 a b c + 3 b^2 c + 3 a c^2 + 3 b c^2 + c^3 + 3 a^2 d + 6 a b d + 3 b^2 d + 6 a c d + 6 b c d + 3 c^2 d + 3 a d^2 + 3 b d^2 + 3 c d^2 + d^3$$ surely upon substituting we recover the original polynomial or its expansion. Now as a matter of example say we want to determine the coefficient of $a^2d$, we can see that the coefficient is 3, we can derive this as: $$a^2d: \frac{3!}{2!1!}=3$$ which is the write coefficient, now we have $$3a^2d=3(x^0)^2(x^3)=3x^3$$ which is not the coefficient of the expanded original polynomial, the expanded original polynomial has coefficient equal to 10, now question is what am I missing here? why there is discrepancy in here?

Wiliam
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    The reason why the multinomial theorem does not work is because $x^3$ can be written as $a^2d$ or as $abc$ or as $b^3$. These terms have coefficients 3,6 and 1 and you can notice that 3+6+1=10 – michalis vazaios Jun 04 '19 at 16:14
  • @michailvazaios awesome, is there any way to generalise this? meaning if you have $f(x)=(1+x+x^2+\cdots+x^n)^n$ is this is a known problem? – Wiliam Jun 04 '19 at 16:17
  • The coefficients $x^k$ in $f(x)=\left(1+x+x^2+\cdots+x^{n-1}\right)^m$ are the number of ways of rolling a sum of $k$ on $m$ fair $n$-sided dice labeled $0,1.\dots,n-1.$ This can be solved by inclusion-exclusion. – Thomas Andrews Jun 04 '19 at 16:23

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$$f(x)=\frac{(1-x^4)^3}{(1-x)^3}$$

We have that $$\frac{1}{(1-x)^3}=\sum_{k=0}^{\infty}\binom{k+2}{2}x^k.$$

And $(1-x^4)^3=1-3x^4+3x^8-x^{12}.$

So $f(x)=\sum_{k=0}^{\infty} a_kx^k$ where:

$$a_k = \binom{k+2}{2}-3\binom{k-2}{2}+3\binom{k-6}{2}-\binom{k-10}{2}.$$

More general, if $$\begin{align}f_{n,m}(x)&=\left(1+x+x^2+\cdots+x^n\right)^m\\&=\frac{\left(1-x^{n+1}\right)^m}{(1-x)^m}\\&=\left(1-x^{n+1}\right)^m\sum_{k=0}^{\infty}\binom{k+m-1}{m-1}x^k \end{align}$$

From this we get that $$a_k=\sum_{j=0}^{m}\left(-1\right)^j\binom{m}{j}\binom{k+m-1-j(n+1)}{m-1}$$

When $n=m,$ this gives:

$$a_k=\sum_{j=0}^{n}\left(-1\right)^j\binom{n}{j}\binom{k-(n+1)(j-1)-2}{n-1}$$

There is no simpler form, I strongly suspect.

Thomas Andrews
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