On the numerical evaluation of factorials X

The last post in this sequence was about some more general way to work with the prime factorization of the factorials and about the Narayana numbers:

{\displaystyle{ N(n,k) = \frac{1}{n}{ \binom nk}{\binom n{k-1}} }}

Which expands to

{\displaystyle{ {{n!^2}\over{\left(k-1\right)!\,k!\,n\,\left(n-k\right)!\,\left(n-k +1\right)!}}  }}

Using (n+1)! = n(n!) and vice versa (n-1)! = \frac{n!}{n} we can shove more things together and get

{\displaystyle{ \frac{k n!^2}{k!^2 n \left(n-k+1\right) \left(n-k\right)!^2}  }}

Get the non-factorials out of the way

{\displaystyle{  \frac{k}{n \left(n-k+1\right)}\; \frac{n!^2}{k!^2 \left(n-k\right)!^2}  }}

With P_n the prime factorization of n! and with their arithmetic resembling the relation of the arithmetic of numbers and the arithmetic of their logarithms (e.g.:x\cdot y = \exp\left(\log\left(x\right)+\log\left(y\right) \right) ) we get (dropping the fraction with the non-factorials)

{\displaystyle{ \begin{aligned} \phantom{\Leftrightarrow\;} &2P_n - \left( 2P_k + 2P_{n-k} \right) \\ \Leftrightarrow\; &2P_n  - 2P_k - 2P_{n-k}  \\ \Leftrightarrow\;  &2\left(P_n - P_k - P_{n-k} \right) \end{aligned} }}

We have to be carefull with the term n \left(n-k+1\right) for it can overflow, so we should divide the prime factorization by and by instead of multiplying n \left(n-k+1\right) out and divide by that all together.

The actual implementation is left as an exercise for the reader.
Or just wait until I do it myself in the next days 😉

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