03/11/2013, 03:16 PM
(03/11/2013, 11:38 AM)Balarka Sen Wrote: I invented a function, somewhat similar to that of the Riemann zeta, but replacing the reciprocal of powers by superexponentiation. Namely,
\( \mathfrak{H}(z) = \sum_{n=1}^{\infty} \frac{1}{{}^z n} \) where by definition \( {}^z 1 = 1 \)
(...)
I've also posted this in a forum : http://www.mymathforum.com/viewtopic.php?f=15&t=38901
Hi Balarka,
I've looked at your computation of H(3), where your method seems to work incorrect. Look at the first five summands only, we get
Code:
vectorv(5,k,1/k^k^k)*1.0
%977 = [1.00000000000, 0.0625000000000, 1.31137265240 E-13, 7.45834073120 E-155, 5.23282787918 E-2185]~Gottfried
Gottfried Helms, Kassel

