Recent content by AtomSeven

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...

The identity is wrong, it should be \sum_{i=0}^{n}2^{n-i} {n+i \choose i}=2^{2 n}

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, As the title says I'm looking for a software package that computes the roots of complex functions. Thanks.

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?