find formula to count of string I have to find pattern of count of series.
Lenght of series is $2n$. It is neccessary to use exaclty double times every number in range $[1...n]$ and all of neighboring numbers are different.
Look at example.
$a_0 = 0$
$a_1 = 0$
$a_2 = 2$    (because of $1212$ and $2121$)
And my idea is:
$$ a_{n+1}= {2n+1\choose 2}a_n $$
Is it properly ?

For $a_3$, you'll add two $3$'s somewhere to each one of those so they're not next to each other.  There are $5$ possible places: the two ends, and three in between two numbers, so there are ${5 \choose 2}$ ways of putting the two $3$'s in so they're not next to each other.
For case $a_n$:  There are $2n+1$ possible places to place $n+1$ in each of the ${a_n}$ series of length $2n$, so there are ${2n+1 \choose 2}$ ways to place the two $n$'s in each series.
Or,
$$a_{n+1} = {2n+1 \choose 2} a_n.$$
We can see the closed form by inspection:
$$a_{n} = {2n-1 \choose 2} a_{n-1} \\ = {2n-1 \choose 2}{2n-3 \choose 2}{2n-5 \choose 2} \cdots {7 \choose 2}{5 \choose 2}a_2 \\ = \frac{(2n-1)(2n-2)}{2}\frac{(2n-3)(2n-4)}{2}\frac{(2n-5)(2n-6)}{2} \cdots \frac{7 \cdot 6}{2}\frac{5 \cdot 4}{2}\cdot 2,$$
so
$$a_n =\frac{(2n-1)!}{3 \cdot 2^{n-2}}, n \geq 3.$$
