Could someone please enlighten me on why the sequence of sopfr(n!) (the sum of prime factors of n!) is seemingly related to the number of diagonals in $n$-polygons? What is the relation between them? How do these puzzle pieces connect, intuitively?
As a refresher, the sopfr(n) function (or the integer log of n) calculates the sum of all distinct prime factors of a positive integer n, each raised to the power of its multiplicity. For eg., 12 in terms of its prime factors is 2^2 * 3^1. sopfr(n), then, would be 2^2 + 3^1. In other words, sopfr(n) = Σ(p^k) for each prime factor p^k of n, where p is a prime number and k is its multiplicity in the prime factorization of n.
Also, just for clarity, $n$-polygons refer to regular polygons with n sides starting from 3. The number of diagonals in an $n$-polygon is the count of all the possible line segments that can be drawn between non-adjacent vertices (that is excluding the sides) of the polygon. In a triangle, the count is 0, in a quadrilateral, it is 2, in a pentagon, it is 5, and so on.
The first 24 terms of the sopfr(n!) sequence are:-
0, 0, 2, 5, 9, 14, 19, 26, 32, 38, 45, 56, 63, 76, 85, 93, 101, 118, 126, 145, 154, 164, 177, 200, ...
While the first 24 terms of the sequence of the number of diagonals of $n$-polygons are:-
0, 2, 5, 9, 14, 20, 27, 35, 44, 54, 65, 77, 90, 104, 119, 135, 152, 170, 189, 209, 230, 252, 275, 299, ...
As is pretty clear, the first few terms are similar but then they begin to diverge.
Here is a graph showing their relation:-
All insights are extremely appreciated!
The first is sequence $A025281$ in $OEIS$ and in year $1977$ Alladi and Erdős gave the asymptotic $$a_n\sim \frac{\pi ^2\, n^2}{12\, \log (n)}$$ Have a look at this paper.
The second one is $$b_n=\frac {n(n+3)} 2$$
If you plot $a_n$ as a function of $b_n$ (I used $10^2 \leq n \leq 10^6$ by steps of $100$), a quick and dirty linear regression with no intercept gives with $R^2=0.999663$
$$a_n= k\, b_n$$
$$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \hline k & 6.87345 & 0.00399 & \{6.86561,6.88128\} \\ \end{array}$$
If, instead of the asymptotics, you use the exact values, with $R^2=0.999721$ $$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ \hline k & 7.56504 & 0.00126 & \{7.56256,7.56751\} \\ \end{array}$$
If, to get rid of very large numbers, you compute instead the ratio $\frac {b_n}{a_n}$ and curve fit
$$\frac {b_n}{a_n}=\alpha\,\log(n)-\beta$$
with $R^2=0.99999975$
$$\begin{array}{l|lll} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \hline \alpha & 0.608193 & 0.000036 & \{0.608123,0.608263\} \\ \beta & 0.712634 & 0.000460 & \{0.711732,0.713537\} \\ \end{array}$$ for an average absolute error of $0.002$ and a maximum absolute error of $0.171$.
Notice that $\frac {12} {2\pi^2}=0.607927$