TITLE: How to compute coefficients in Trinomial triangle at specific position?
QUESTION [5 upvotes]: I need to compute coefficients of $i$-th power of $x$ from simplifying of $$(x^2 + x + 1)^n$$ 
From that site i know about trinomial triangle. 
But how can compute coeficients of $i$-th element at $n$-th row without computing all upper elements?

REPLY [4 votes]: Denote by
$$\sum_{k=0}^{m(s-1)}{\binom{m}{k}}_{s}x^{k}=(1 + x + ... + x^{s-1})^{m}\,$$ 
the expansion of multinomial of degree $s-1$
From that relation follow that 
$$ \sum_{k=0}^{m(s-1)}{\binom{m}{k}}_{s}x^{k}=\sum_{k=0}^{\infty}{\binom{m}{k}}_{s}x^{k}=\left(\frac{1-x^{s}}{1-x}\right)^{m}=(1-x^{s})^{m}(1-x)^{-m}\,$$ 
from binomial formula we have
$$(1-x^{s})^{m}=\sum_{i=0}^{m}(-1)^{i}\binom{m}{i}x^{si}$$ 
and from Taylor formula 
$$(1-x)^{-m}=\sum_{j=0}^{\infty}\binom{m+j-1}{j}x^{j}$$ 
after substitutions we get 
$$ \sum_{k=0}^{\infty}{\binom{m}{k}}_{s}x^{k}=\sum_{i=0}^{m}(-1)^{i}\binom{m}{i}x^{si}\sum_{j=0}^{\infty}\binom{m+j-1}{j}x^{j}=$$ 
$$=\sum_{j=0}^{\infty}\sum_{i=0}^{m}(-1)^{i}\binom{m}{i}\binom{m+j-1}{j}x^{si+j}=$$ 
$$ =\sum_{k=0}^{\infty}\sum_{i=0}^{m}(-1)^{i}\binom{m}{i}\binom{m+k-si-1}{k-si}x^{k}\,$$ 
for $si+j=k$ follow that $j=k-si$ then equate the coefficients next to x follow formula 
$${\binom{m}{k}}_{s}=\sum_{i=0}^{m}(-1)^{i}\binom{m}{i}\binom{m+k-si-1}{m-1}\,$$ 
put $s=3$ you get what you ask