How to compute quintinomial coefficients? I'm looking for a way to compute elements of a quintinomial triangle.
Is there a general case?
To be more specific I'm looking for a way to compute the coefficients of the polynomial $(x^4 + x^3 + x^2 +x + 1)^n $ , $\forall x\in\Bbb R , \forall n \in \Bbb N$ for consecutive values of n. 
I'm working on project Euler problem 588.
 A: Here is a derivation of  an  expression which could be  implemented. We use  the coefficient   of operator $[x^p]$ to denote the coefficient  of $x^p$  in  a series.

We obtain
  \begin{align*}
[x^p]&(x^4+x^3+x^2+x+1)^n\\
&=[x^p]\left(\frac{1-x^5}{1-x}\right)^n\tag{1}\\
&=[x^p](1-x^5)^n\sum_{k=0}^\infty\binom{-n}{k}(-x)^k\tag{2}\\
&=[x^p]\sum_{j=0}^n\binom{n}{j}\left(-x^5\right)^j\sum_{k=0}^\infty\binom{n+k-1}{k}x^k\tag{3}\\
&=\sum_{j=0}^{\min\{n,\lfloor{p/5}\rfloor\}}\binom{n}{j}(-1)^j[x^{p-5j}]\sum_{k=0}^\infty\binom{n+k-1}{k}x^k\tag{4}\\
&=\sum_{j=0}^{\min\{n,\lfloor{p/5}\rfloor\}}(-1)^j\binom{n}{j}\binom{n+p-5j-1}{p-5j}\tag{5}
\end{align*}

Comment:


*

*In  (1) we apply the formula of the finite geometric series.

*In (2) we apply the binomial series expansion of $(1-x)^{-n}$.

*In (3) we use the identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (4) we use the linearity of the coefficient of operator  and use the rule $$[x^p]x^qA(x)=[x^{p-q}]A(x)$$ We also restrict the upper limit of the sum, since the exponent of $x^{p-5j}$ has to be non-negative.

*In (5) we select the coefficient of $x^{p-5j}$.
A: Sorry, stupid question, and the site is blocking me from adding a comment....
What is P ?  For example, would p=8 produce the coefficient for X^8 when n = 2?
