Why do Catalan numbers show up in this square root expression?

It has all to do with the fact that the Taylor series expansion of (1 - sqrt(1 - 4x))/2 equals x + x^2 + 2x^3 + 5x^4 + 14x^5 + ... (i.e. Catalan numbers) = \sum_{i=0}^{infinity} x^i * c_i

Say r = 10^N / 2 (i.e. 5 * 10^(N-1))

Then the expression r - sqrt(r^2-1) evaluates to 10^N / 2 - sqrt(10^2N/4 - 1).

Using some manipulation, this can be rewritten to 10^N * (1 - sqrt(1 - 4 * 10^(-2N))) / 2.

Now I'm not at all in a mood to write this out any further, but if we substitute 10^(-2N) with x, we get the above function, thus working with its Taylor expansion will eventually yield some expression like 10^-N * \sum_{i=0}^{infinity} c_i * 10^(-2N * i), which is the general form leading to the number A you have written above!

The mathematical argument could go as follows:

A = R - sqrt(R^2 -1) = R ( 1- sqrt(1-1/R^2 ))

Let's define R' = 2 R, so we get A = R'/2 (1-sqrt(1-4/R'^2 ))

Since 1/R'^2 << 1, we do a Taylor series expansion. The series is equivalent to sqrt(1-x) = 1+sum from m=1 to infinity of (-x)^m * (1/2 over m), where (a over b) = a! / b! (a-b)!

From the identities of (a over b) it follows that (1/2 over m) = 1/2/(1/2-m) (-1/2 over m) = 1/(1-2m) (-4)^(-m) (2m over m) = (m+1)/(1-2m) (-4)^(-m) C(m), where C(m) is the m-th Catalan number.

Then we get sqrt(1-4/R'^2 ) = 1 + sum (-4/R'^2 )^m (m+1)/(1-2m) (-4)^(-m) C(m) Simplifying, we get sqrt(1-4/R'^2 ) = 1 + sum R'^(-2m) (m+1)/(1-2m) C(m)

Finally, we have A = sum from m=1 to infinity of R'^(1-2m) (m+1)/(2m-1) C(m)/2

Now C(m-1) = (m+1)/(2*(2m-1)) C(m) Then finally we have A= sum from m=1 to infinity of R'^(1-2m) C(m-1) QED

The R'^(1-2m) terms space the Catalan numbers in A's decimal representation, since R' = (10)^(n+1) The rest you can just plug in. Cheers.

Sources: https://en.m.wikipedia.org/wiki/Catalan_number https://functions.wolfram.com/GammaBetaErf/Binomial/17/02/04/ https://functions.wolfram.com/GammaBetaErf/Binomial/17/01/01/

Edit: Formatting

My maths is not good, but I think what you have after the decimal point is just adding up the catalan numbers with each subsequent catalan number being multiplied by 10^(-34). Try working out the sum from 0 to infinity of Cn(10^(-34n)) where Cn is the nth catalan number, and see if your square root expression pops up.

I once tried this but with the powers of 2 instead of catalan numbers: the fraction 50/49 generates 1.0204081632... and 500/499 generates 1.002004008016032064128256... etc etc. Because the sum from 0 to infinity of (2^n )(100^-n )= Σ(50^-n ) = 1/(1-1/50) = 50/49.

Fibonacci numbers also work similarly; 1/89=0.011235... and 1/9899=0.000101020305081321... A similar expression can be derived for the fibonacci numbers.

Apologies that I do not possess sufficient skill in manipulating the expression for catalan numbers to fit into your square root expression, but I am sure that if you sum up the expression Cn(10^(-34n)) from zero to infinity, you will be able to prove this identity.

[removed]

[removed]

[removed]