Summation

The sum of the diagonals is actually the sum of the 4 corners of each subcube of the spiral matrix.

With:

\[ \begin{gather} \color{red}{21}\ 22\ 23\ 24\ \color{red}{25}\\ 20\ \ \ \color{red}{7}\ \ \ 8\ \ \ \color{red}{9}\ 10\\ 19\ \ \ 6\ \ \ \color{red}{1}\ \ \ 2\ 11\\ 18\ \ \ \color{red}{5}\ \ \ 4\ \ \ \color{red}{3}\ 12\\ \color{red}{17}\ 16\ 15\ 14\ \color{red}{13}\\ \end{gather} \]

The sum is \( (1) + (3 + 5 + 7 + 9) + (13 + 17 + 21 + 25) \)

For an \( n \times n \) grid, with n being odd, the top right corner is \( n^ 2 \), the top left corner is \( n^2 -n + 1 \), the bottom left corner is \( n^2 -2n + 2 \) and the bottom right corner is \( n^2 -3n + 3 \). The sum of these corners is \( 4n^2 -6n + 6 \).

We need to sum these corners for all odd \( n \) between 3 and \( l \). Odd numbers can be written in the form \( 2k + 1 \). By replacing \( n \) with \( 2k + 1 \) we have:

\[ 4n^2 -6n + 6 = 4(2k+1)^2 -6(2k+1) + 6 = 16k^2 + 4k + 4 \]

The sum become:

\[ \begin{align} S &= \sum _{k=1}^{m} 16k^{2} + 4k + 4\\ S &= 16\sum _{k=1}^ {m} k^{2} + 4\sum _{k=1}^{m} k + 4\sum _{k=1}^{m} 1\\ S &= \frac{16m(m+1) (2m+1)}{6} + \frac{4m(m + 1)}{2} + 4m\\ S &= \frac{16m^3}{3} + 10m^2 + \frac{26m}{3}\\ \end{align} \]

where \( m = \frac{l - 1}{2} \)

This sum plus \( 1 \) (the center of the spiral matrix) is the solution we are searching for:

def number_spiral_diagonals(n=1001):
    m = (n - 1) // 2
    return int(1 + (16 / 3) * m ** 3 + 10 * m ** 2 + (26 / 3) * m)