For those interested here is a refference:
M. Lukkarinen, The Mellin transform of the square of Riemann’s zeta-function and Atkinson’s formula, Doctoral Dissertation, Annales Acad. Sci. Fennicae, No. 140, Helsinki, 2005
Well, this
\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}
is an identity it is true for all n but, if I understand correctly, you may ask for the values of n that make
\sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2^{n}}
true. In this case we have the equation 2n=2^{2^{n}}, and the solutions...
Hy everyone,
I think that some time ago I've seen D(x) expressed in terms of the roots of the \zeta(s) function. Does anyone knows of references about this?
Hi,
I've done this by a different approach considering that d(n)=D(n)-D(n-1) and D(0)=0 it follows that
\begin{align}
\zeta^{2}(s)&=\sum_{n=1}^{\infty} \frac{\sigma_{0}(n)}{n^{s}}=\sum_{n=1}^{\infty} \frac{D(n)-D(n-1)}{n^{s}} \nonumber\\
&=\sum_{n=1}^{\infty}...
Hi,
The divisor summatory function, D(x), can be obtained from \zeta^{2}(s) by D(x)=\frac{1}{2 \pi i} \int_{c-i \infty}^{c+i \infty}\zeta^{2}(w)\frac{x^{w}}{w}dw and I was trying to express \zeta^{2}(s) in terms of D(x) but I didnt succeed, could someone help?