, 2019

Solution:

Fix any prime $p$, and let $a_n=f\left(p^n\right)$ for $n \geq 0$. Notice that using the relation for $p^n$, we obtain
$$\sum_{i=0}^{n} a_i a_{n-i}=1$$
which means that if we let $g(x)=\sum_{n \geq 0} a_n x^n$, then $g(x)^2=1+x+x^2+\cdots=\frac{1}{1-x}$ as a generating function. Thus $g(x)=(1-x)^{-\frac{1}{2}}$, and this is well-known to have generating function with coefficients $a_n=\frac{\binom{2 n}{n}}{4^{n}}$. One way to see this is using the Taylor series and then reorganizing terms; it is also intimately related to the generating function for the Catalan numbers. In particular, $a_n$ is independent of our choice of $p$.

Now if we define $f_0(n)=\prod_{p \mid n} a_{v_p(n)}$, then we see that $f=f_0$ on the prime powers.

If we define the Dirichlet convolution of two functions $\chi_1, \chi_2: \mathbb{N} \rightarrow \mathbb{R}$ as $\chi_3$ such that
$$\chi_3(n)=\sum_{d \mid n} \chi_1(d) \chi_2\left(\frac{n}{d}\right)$$
then it is well-known that multiplicative functions ($\chi(m) \chi(n)=\chi(m n)$ if $\operatorname{gcd}(m, n)$, so e.g. $\phi(n)$, the Euler totient function) convolve to a multiplicative function.

In particular, $f_0$ is a multiplicative function by definition (it is equivalent to only define it at prime powers then multiply), so the convolution of $f_0$ with itself is multiplicative. By definition of $a_n$, the convolution of $f_0$ with itself equals 1 at all prime powers. Thus by multiplicativity, it equals the constant function 1 everywhere.

Two final things to note: $f_0(1)=a_0=1>0$, and $f$ satisfying the conditions in the problem statement is indeed unique (proceed by induction on $n$ that $f(n)$ is determined uniquely and that the resulting algorithm for computing $f$ gives a well-defined function). Therefore $f_0$, satisfying those same conditions, must equal $f$.

At last, we have
$$f\left(p^{2019}\right)=f_0\left(p^{2019}\right)=\frac{\binom{4038}{2019}}{4^{2019}}$$
so
$$f\left(2018^{2019}\right)=f\left(2^{2019}\right) f\left(1009^{2019}\right)=\frac{\binom{4038}{2019}^{2}}{4^{4038}}=\frac{\binom{4038}{2019}^{2}}{2^{8076}}$$