Let's determine the general term of the sequence whose formation rule is:
a) $u_{n+3}=u_{n+2}+u_{n+1}-u_{n}$,
b) $v_{n+3}=v_{n+2}-v_{n+1}+v_{n}$
c) $w_{n+3}=w_{n+2}+w_{n}$

$$
\begin{aligned}
& u_{0}=0, u_{1}=1, u_{2}=2: \\
& v_{0}=v_{1}=v_{2}=1 \\
& w_{0}=w_{1}=1, w_{2}=2
\end{aligned}
$$

(See the article by the proposer in the 45th volume of our journal, pages 193-203.)
We will use the method described in János Surányi's article "Polynomials and Infinite Polynomials" (K. M. L. 45th volume, 109117, pages 193-203). Let the rule for generating the sequence be

$$
u_{n+3}=\alpha u_{n+2}+\beta u_{n+1}+\gamma u_{n}
$$

and let the first three terms of the sequence, $u_{0}, u_{1}, u_{2}$, be given. Similarly to the method described in the article, consider the following infinite polynomial:

$$
U=u_{0}+u_{1} x-u_{2} x^{2}+\ldots+u_{n} x^{n}+\ldots
$$

Multiply the recursive relation (2) by $x^{n+3}$ and sum for $n=0,1,2, \ldots$:

$$
\begin{aligned}
u_{3} x^{3}+u_{4} x^{4}+\ldots= & \alpha x\left(u_{2} x^{2}+u_{3} x^{3}+\ldots\right)+\beta x^{2}\left(u_{1} x+u_{2} x^{2}+\ldots\right)+ \\
& +\gamma x^{3}\left(u_{0}+u_{1} x+u_{2} x^{2}+\ldots\right)
\end{aligned}
$$

Thus, using (3), we get

$$
U-\left(u_{0}+u_{1} x+u_{2} x^{2}\right)=\alpha x\left[U-\left(u_{0}+u_{1} x\right)\right]+\beta x^{2}\left(U-u_{0}\right)+\gamma x^{3} U
$$

which simplifies to

$$
U\left(1-\alpha x-\beta x^{2}-\gamma x^{3}\right)=\left(u_{0}+u_{1} x+u_{2} x^{2}\right)-\alpha x\left(u_{0}+u_{1} x\right)-\beta x^{2} u_{0}
$$

or

$$
U=\frac{u_{0}+\left(u_{1}-\alpha u_{0}\right) x+\left(u_{2}-\alpha u_{1}-\beta u_{0}\right) x^{2}}{1-\alpha x-\beta x^{2}-\gamma x^{3}}
$$

We apply these two relations to the sequences in the problem. In case a), $\alpha=1, \beta=1$ and $\gamma=1$, so

$$
\begin{aligned}
U & =\frac{0+(1-0) x+(2-1-0) x^{2}}{1-x-x^{2}+x^{3}}=\frac{x+x^{2}}{1-x-x^{2}+x^{3}}=\frac{x(1+x)}{(1-x)^{2}(1+x)}= \\
& =\frac{x}{(1-x)^{2}}
\end{aligned}
$$

In the article, we saw that

$$
\frac{1}{(1-x)^{2}}=1+2 x+3 x^{2}+\ldots+(n+1) x^{n}+\ldots
$$

Thus, $U$ is the $x$-multiple of this:

$$
U=0+x+2 x^{2}+3 x^{3}+\ldots+n x^{n}+\ldots
$$

So, according to (3), $u_{n}=n$.

In case b), $\alpha=1, \beta=-1, \gamma=1, v_{0}=v_{1}=v_{2}=1$, so

$$
U=\frac{1+(1-1) x+(1-1+1) x^{2}}{1-x+x^{2}-x^{3}}=\frac{1+x^{2}}{(1-x)\left(1+x^{2}\right)}=\frac{1}{1-x}
$$

We also saw in the article that

$$
\frac{1}{1-x}=1+x+x^{2}+\ldots+x^{n}+\ldots
$$

Thus, in this case, $v_{n}=1$ for all $n$.

Finally, in case c), $\alpha=1, \beta=0, \gamma=1$, so

$$
U=\frac{1+(1-1) x+(2-1) x^{2}}{1-x-x^{3} \mid}=\frac{1+x^{2}}{1-\left(x+x^{3}\right)}
$$

Using (5), we first convert $1 /\left(1-x-x^{3}\right)$ into an infinite polynomial by substituting $\left(x+x^{3}\right)$ for $x$ in (5):

$$
\frac{1}{1-\left(x+x^{3}\right)}=1+\left(x+x^{3}\right)+\left(x+x^{3}\right)^{2}+\ldots+\left(x+x^{3}\right)^{n}+\ldots
$$

Let's see how the $x^{n}$ term appears. In the expansion of $\left(x+x^{3}\right)^{k}=x^{k}\left(1+x^{2}\right)^{k}$, we find only even exponents that match $k$, and the $x^{k+2 t}$ term appears with the coefficient $\binom{k}{i}$. Thus, in $\left(x+x^{3}\right)^{n}$, the coefficient of $x^{n}$ is $1=\binom{n}{0}$, and in $\left(x+x^{3}\right)^{n-2}$, it is $\binom{n-2}{1}$, so the coefficient of $x^{n}$ is

$$
\binom{n}{0}+\binom{n-2}{1}+\ldots+\binom{n-2 i}{1}+\ldots
$$

If we multiply the infinite polynomial (6) by $\left(1+x^{2}\right)$, then the coefficient of $x^{n}$ in the resulting polynomial is the sum of the coefficients of $x^{n}$ and $x^{n-2}$ in (6), i.e., according to (7) (rearranged immediately):

$$
\begin{aligned}
\binom{n}{0}+\left[\binom{n-2}{0}\right. & \left.+\binom{n-2}{1}\right]+\left[\binom{n-4}{1}+\binom{n-4}{2}\right]+\ldots+\left[\binom{n-2 i}{i-1}+\binom{n-2 i}{i}\right]+\ldots= \\
& =\binom{n+1}{0}+\binom{n-1}{1}+\binom{n-3}{2}+\ldots+\binom{n+1-2 i}{i}+\ldots
\end{aligned}
$$

Since we have thus obtained the coefficients of the corresponding $U$ polynomial, we have

$$
w_{n}=\binom{n+1}{0}+\binom{n-1}{1}+\ldots+\binom{n+1-2 i}{i}+\ldots
$$

Remarks. 1. The sum obtained in (8) is clearly finite, since if $n+1-2 i<i$, then the binomial coefficient is 0, and if a binomial coefficient is 0, then all subsequent values are also 0. The last non-zero binomial coefficient is

$$
\binom{n-2\left[\frac{n+1}{3}\right]+1}{\left[\begin{array}{c}
n+1 \\
3
\end{array}\right]}
$$

2. The part a) and b) of the problem can be solved more easily due to the simplicity of the initial terms and the recursion, but our goal here was to apply the method described in the article. Of course, other solutions (e.g., by complete induction) were also accepted.