Solution to the Plouffe Puzzle

Yesterday I posted the following puzzle from Simon Plouffe: why are there so many zeroes in the base-10 expansion of sqrt(51)/14 = 0.51010203061020 ... ?

Here's the answer. Let f(x) = (sqrt(1-4x²) + 2x - 1)/(2x - 4x²). Then it is not hard to see that f can be expanded as a power series
f(x) = 1 + x + 2x² + 3x³ + 6x⁴ + 10x⁵ + 20x⁶ + ...
where the coefficients are binom(n, floor(n/2)), the central binomial coefficients.
It now follows that
f(1/t)/t = 1/t + 1/t² + 2/t³ + 3/t⁴ + ...
= sqrt(t²/4 - 1)/(t-2) - 1/2.
When we substitute t = 100, we get
sqrt(51)/14 - 1/2 = 1/100 + 1/100² + 2/100³ + 3/100⁴ + ...
which explains the expansion we see.

We can get even more of the central binomial coefficients by taking t to be larger powers of 10. For example, for t = 1000, we get
sqrt(249999)/998 = 0.501001002003006010020035070126252462 ... .