activity
20102014
most citedAlgorithms for determining integer complexity

8 citations · 17 across the 7 of their papers we have counts for

collaborators

7 papers

math.NT2014★ 8 cited

Algorithms for determining integer complexity

J. Arias de Reyna, J. van de Lune

We present three algorithms to compute the complexity of all natural numbers . The first of them is a brute force algorithm, computing all these complexitie…

math.NT2014★ 2 cited

"How many 's are needed?" revisited

J. Arias de Reyna, J. van de Lune

We present a rigorous and relatively fast method for the computation of the "complexity" of a natural number (sequence A005245), and answer some "old and new" questions related to…

math.NT2013

On the exact location of the non-trivial zeros of Riemann's zeta function

Juan Arias de Reyna, Jan van de Lune

In this paper we introduce the real valued real analytic function kappa(t) implicitly defined by exp(2 pi i kappa(t)) = -exp(-2 i theta(t)) * (zeta'(1/2-it)/zeta'(1/2+it)) and kapp…

math.NT2012★ 1 cited

On the sign of the real part of the Riemann zeta-function

Juan Arias de Reyna, Richard P. Brent, Jan van de Lune

We consider the distribution of on fixed lines , and in particular the density \[d(σ) = \lim_{T \rightarrow +\infty} \frac{1}{2T} |\{t \in [-T,+T]: |\argζ…

math.NT2011★ 3 cited

A note on the real part of the Riemann zeta-function

Juan Arias de Reyna, Richard P. Brent, Jan van de Lune

We consider the real part of the Riemann zeta-function in the half-plane . We show how to compute accurately the constant which…

math.NT2011★ 1 cited

Some bounds and limits in the theory of Riemann's zeta function

J. Arias de Reyna, J. van de Lune

For any real a>0 we determine the supremum of the real σ such that ζ(σ+it) = a for some real t. For 0 < a < 1, a = 1, and a > 1 the results turn out to be quite different.} We also…

math.CA2010★ 2 cited

A glimpse inside the mathematical kitchen

Juan Arias-de-Reyna, Jan van de Lune

We prove the inequality sum_{k=1}^infty (-1)^{k+1} r^k cos(k*phi) (k+2)^{-1} < sum_{k=1}^infty(-1)^{k+1} r^k (k+2)^{-1} for 0 < r <= 1 and 0 < phi < pi. For the case r = 1 we give…