Can this be solved for $f(n)$? While working upon a partial sum formula for the harmonics, I came across a necessity for the function defined below
$f\left(\frac{n(n-1)}{2} +1\right) = n!$
Can it be solved for $f(n)$?

Let's define $f$ in terms of two functions $h, g$.
$$g(n) = \frac{n(n - 1)}{2} + 1$$
$$h(n) = n!$$
$$f\left(\frac{n(n - 1)}{2} + 1\right) = h\left(g^{-1}\left(\frac{n(n - 1)}{2} + 1\right)\right)$$
So we want to find the composition $h \circ g^{-1}$.
$$
g^{-1}(n) = \frac{1 \pm \sqrt{8n - 7}}{2}
$$
Since you're working with natural numbers $n$, it doesn't make sense to include the negative solutions. However, a problem arises when trying to use this for $n = 0$. We can rewrite the function $g^{-1}$ to accommodate this:
$$
g^{-1}(n) = \begin{cases}\frac{1 + \sqrt{8n - 7}}{2} & n > 0 \\
\ \ \quad 1 & n = 0
\end{cases}
$$
Now, let's form the composition:
$$
f(n) = h(g^{-1}(n)) = \begin{cases}\Gamma\left(\frac{1 + \sqrt{8n - 7}}{2}\right)& n > 0 \\
\ \ \quad 1 & n = 0
\end{cases}
$$
We need to use $\Gamma(n)$ because $g^{-1}$ is not guaranteed to be an integer. In fact, $g^{-1}(3) = \frac{1 + \sqrt{17}}{2}$. (There is no natural number $n$ for which $g(n) = 3$, so this makes sense).
