The Bernoulli numbers seem to grow very fast after about B20 or B30. What I'm trying to figure out is whether or not \(\alpha = \left(\frac{z!}{B_z}\right)^{\frac{1}{z}}\) has any \(\alpha\) less than 1 for sufficiently high z. Of course the exponent shouldn't really matter so it's just the numerator vs. the denominator. Maybe I should program out the solution, but I should really study this a little bit more.
No. For even z which are not divisible by 4 that ratio approaches \(2^{\tiny 1 - \frac{1}{z}} \pi\) \(\begin{array}{r|ll} z & 2^{\tiny 1-z^^{-1}} \pi & \alpha \\ \hline \\ 2 & 4.44288 & 3.4641 \\ 6 & 5.59768 & 5.58166 \\ 10 & 5.86242 & 5.86184 \\ 14 & 5.97968 & 5.97965 \\ 18 & 6.04583 & 6.04583 \\ 22 & 6.08831 & 6.08831 \\ 26 & 6.11789 & 6.11789 \\ 30 & 6.13968 & 6.13968 \\ 34 & 6.15639 & 6.15639 \\ 38 & 6.16961 & 6.16961 \\ 42 & 6.18034 & 6.18034 \\ 46 & 6.18922 & 6.18922 \\ 50 & 6.19668 & 6.19668 \\ 54 & 6.20305 & 6.20305 \\ 58 & 6.20854 & 6.20854 \end{array}\) http://functions.wolfram.com/IntegerFunctions/BernoulliB/06/02/ http://functions.wolfram.com/GammaBetaErf/Factorial/06/02/ With a minus sign inserted in the right place in the definition of \(\alpha\), the same behavior is seen when z is divisible by 4. Odd z, with the exception of z=1, need not apply. \(\alpha\) when it exists, is an algebraic number, while my approximation is transcendental.
