Fast solution for Advent of Code 2023 Day 9: Mirage Maintenance.
https://adventofcode.com/2023/day/9

It's clear that the answer for a given row is some linear combination of the
values in the row, and the coefficients of the row only depend on the length of
the input row. This can be derived by simply writing down the equations of the
rows of differences without evaluating them.

For example, for a row of length 4:

  Given row:           a, b, c, d
  First differences:   (b-a), (c-b), (d-c)
  Second differences:  (c-b)-(b-a),  (d-c)-(c-b)
                       = a - 2b + c  = b - 2c + d
  Third differences:   (b - 2c + d) - (a - 2b + c) = -a + 3b - 3c + d

Then calculating missing values as instructed:

  Third row:   0 + (-a + 3b -3c + d) =
                -a + 3b - 3c + d
  Second row:  (-a + 3b - 3c + d) + (b - 2c + d) =
                -a + 4b - 5c + 2d
  First row:   (-a + 4b - 5c + 2d) + (d - c) =
                -a + 4b - 6c + 3d
  Answer:      (-a + 4b - 6c + 3d) + d =
                -a + 4b - 6c + 4d

It turns out that the coefficients are simply binomial coefficients nCr(n, k)
for 1 ≤ k < n where n is the length of the input, with alternating signs that
start at -1 if n is even, and +1 if n is odd.

The answer for a sequence [a, b, c, d, e] (length 5) is 1*a - 5*b + 10*c - 10*d + 5*e.
The answer for a sequence [a, b, c, d, e, f] of (length 6) is -1a + 6b - 15c + 20d - 15e + 6f.

For example, for [2, 3, 4, 5, 6], 1*2 - 5*3 + 10*4 - 10*5 + 5*6 = 7.

To evaluate the answer we need the binomial coefficients nCr(n, k) for a given
value of n. We can calculate them incrementally, given the definition:

  nCr(n, k) = n! / k! / (n - k)!

We can calculate the ratio:

  nCr(n, k) / nCr(n, k - 1)
      = n! / k! / (n - k)! / (n! / (k - 1)! / (n - (k - 1))!)
      = (k - 1)! × (n - (k - 1))! / k! / (n - k)!
      = (n - k + 1) / k

So we can calculate nCr(n, k) from nCr(n, k - 1) by multiplying by (n - k + 1)
and then dividing by k.