Cut the Cards

[Albert Chan]

(07-30-2020 09:49 PM)Jim Horn Wrote:  So, it approaches x*(ln(x)+0.57721), where the latter constant is the Euler-Mascheroni constant.

Replacing ln(x) by ln(x+0.5) gives much better estimate for ψ(x+1).
ψ(x+1) = slope of ln(gamma(t)) at t = x+1. When x is large:

\(\psi(x+1) ≈ {\ln\Gamma(x+2)\;-\;\ln\Gamma(x)\over 2}=\ln\sqrt{(x+1)(x)} ≈ \ln(x+0.5)\)

>>> x = 52
>>> H = log(x+0.5) + 0.57722
>>> print format(x*H, 'g')
235.978

see https://en.wikipedia.org/wiki/Digamma_fu...roximation