In 1985 Odlyzko and te Riele disproved the Mertens conjecture with an amplitude sum over 2,000 zeta zeros. The same sum can now be taken over two million zeros, with the derivative ζ′(ρ) computed at every zero rather than replaced by an average-order law. This post is about what that sum does: where it crosses 1, where it crosses 2, how it grows, and one independent check that the wave picture behind it is real.
The current version of the paper is on Zenodo: doi:10.5281/zenodo.22714412. Code and data are on GitHub.
The Starting Point: Odlyzko–te Riele, 1985
The Mertens function M(x) is the running sum of the Möbius function. The Mertens conjecture said that |M(x)| stays below √x forever. Odlyzko and te Riele showed it does not.
Their method rests on the explicit formula. Under the Riemann hypothesis and simplicity of the zeros, M(x)/√x is a sum of oscillations, one per non-trivial zero ρn = ½ + iγn, and the wave belonging to ρn has amplitude 2/|ρn ζ′(ρn)|. The absolute amplitudes add up to a ceiling on what the waves can reach if they ever align. If that ceiling is above 1, the conjectured bound can be breached; the remaining work, which is where Odlyzko and te Riele spent their effort, is to find a value of x where enough of the waves actually do align. They used 2,000 zeros and a lattice-reduction search.
The Sum
The truncated amplitude sum is
S(K) = Σn ≤ K 2 / |ρn ζ′(ρn)|.
Computing it requires ζ′(ρn) at each zero. I compute it directly for all 2,001,052 zeros in Odlyzko’s tables, from γ1 = 14.13 up to γ2,001,052 ≈ 1.13 × 106. That is the difference between this computation and one that substitutes a growth law for |ζ′(ρ)|: the early behaviour of S(K) depends on the individual zeros, and it is the early behaviour that says how many zeros the Mertens problem needs.
What the Sum Does
K = 58
S(K) crosses 1
K ≈ 1,040
S(K) crosses 2
(log T)5/4
growth, consistent with the conjectural law
Fifty-eight zeros are enough for the amplitude ceiling to pass 1. That is far fewer than the 2,000 Odlyzko and te Riele used, and the reason is not that the sum is different. It is the same sum. Their bottleneck was the alignment search, not the amplitude budget. By about 1,040 zeros the budget passes 2.
From there the growth is slow. Summed over the zeros up to height T, S is consistent with the conjectural (log T)5/4 law across the full range of the data: each factor of ten in the number of zeros adds a nearly constant increment to a fractional power of the logarithm. There is no sign of saturation over six orders of magnitude in K, and no sign of the sum running ahead of the conjectured rate either.
Validation: The Zeros Are in the Arithmetic
How do I know the wave decomposition is the right picture and not curve fitting? Run it backwards. Sample M(x)/√x at log-spaced points and take a discrete Fourier transform, without using the explicit formula and without evaluating ζ(s) anywhere. The frequencies present in the raw arithmetic data are the first zeta zeros: the peak at γ1 = 14.13 and its successors appear where the explicit formula says they should.
Two independent routes agree. The explicit formula says the frequencies should be there; the transform of the arithmetic data finds them. The Mertens function really is a superposition of oscillations at zeta-zero frequencies, and the amplitude sum is a sum over the actual components of the signal.
What This Clarifies
The Mertens amplitude budget is not marginal. It exceeds 1 almost immediately and keeps growing at the conjectured rate. The difficulty in the Mertens problem was never the size of the amplitude sum; it was, and is, finding where the phases align, which is why the 1985 disproof needed lattice reduction and why the smallest counterexample is still unknown.
The computed ζ′(ρn) matter for exactly this reason. An average-order substitute for |ζ′(ρ)| gets the growth rate right and the crossing points wrong, and the crossing points are what tell you how much of the spectrum a breach needs.
The Numbers
2,001,052
zeta zeros, ζ′(ρ) computed at each
58
zeros for S(K) to pass 1
≈ 1,040
zeros for S(K) to pass 2
5/4
exponent of log T in the growth law
14.13
first zero, recovered from M(x)/√x by DFT
Read the Paper
A Unified Spectral Framework for Arithmetic Oscillations
The Mertens amplitude sum over two million zeta zeros with computed ζ′(ρ)