Suppose
$$\begin{cases}f(x)=g(x+1)\\f(y)=2018y+2016y+\cdots+2y\\g(x)=h(2x)-x\end{cases}$$
If $h(2018)=a^3$ and $a\in\mathbb Z$, what is the value of $a$?
The answer is $1009$. I found it by solving for $h(2x)$ in terms of $f(x)$,
$$h(2018)=f(1008)+1009$$
then applying Faulhaber's formula to $f(y)$,
$$f(y)=y\sum_{i=0}^{1008}(2018-2i)=1\,019\,090y$$
$a^3=1\,027\,243\,729$ happens to be a perfect cube, and $a=1009$.
But given the relationship between the solution $1009$ and $2018$, I suspect there may be a more elegant way to find $a$. Is there one?
Disclaimer: I'm assuming this comes from a $2018$ contest purely because the number appears in the question.
$$ \begin{split} h(2018) &= h(2\cdot 1009) \\ &= g(1009)+1009 = \\ &= g(1008 + 1)+1009 = \\ &= f(1008)+1009 = \\ &= \left( \sum_{i = 1}^{1009}2\cdot i \cdot 1008\right)+1009 = \\ &= 2\cdot 1008\cdot\left(\sum_{i = 1}^{1009} i\right)+1009 = \\ &= 2\cdot 1008\cdot\left(\frac{1009\cdot 1010}{2}\right)+1009 = \\ &= 1008\cdot 1009\cdot 1010+1009 = \\ &= 1009\cdot (1008\cdot 1010+1) = \\ &= 1009\cdot \left((1009-1)\cdot (1009+1)+1\right) = \\ &= 1009\cdot (1009^2 - 1 +1) = \\ &= 1009^3. \end{split}$$
Hence $a = 1009$.