OpenFrontiers
← Back to the index
OP-6836OpenNumber theory

Third-level log-concavity of the Riemann Xi kernel

TheoremDB P11158 · analytic number theory

Inferred from the community field; finer classification has not been reviewed.

Source catalogue assessmentAs of 2026-09-08Source & evidence ↓

Problem statement

Let

Φ(r)=∑m≥1πm2(2πm2e4r−3)e5r−πm2e4r,r≥0, \Phi(r)=\sum_{m\ge1}\pi m^2\bigl(2\pi m^2e^{4r}-3\bigr)e^{5r-\pi m^2e^{4r}},\qquad r\ge0,

be the classical Riemann Ξ\Xi-kernel, with its usual even extension, and put S0(t)=Φ(t)S_0(t)=\Phi(\sqrt t) for t>0t>0. Recursively define

Sj+1(t)=(Sj′(t))2−Sj(t)Sj′′(t). S_{j+1}(t)=(S_j'(t))^2-S_j(t)S_j''(t).

It is known that S0>0S_0>0 and (log⁡S0)′′<0(\log S_0)''<0, and Planat--Solé (2026) proved (log⁡S1)′′<0(\log S_1)''<0; hence S1,S2>0S_1,S_2>0. Determine whether

S3(t)>0for every t>0. S_3(t)>0\qquad\text{for every }t>0.

Equivalently, prove that S2S_2 is strictly log-concave on (0,∞)(0,\infty), i.e. (log⁡S2(t))′′<0(\log S_2(t))''<0 for all t>0t>0, or disprove this by giving a rigorous point or interval on which S3≤0S_3\le0.

Lean 4

Not yet attached

No Lean statement or proof is attached yet. Have a formalization? Share its source to add it to this problem.

Background

Additional formulation recorded by the source

With S0(t)=Φ(t)S_0(t)=\Phi(\sqrt t) and Sj+1=(Sj′)2−SjSj′′S_{j+1}=(S_j')^2-S_jS_j'', prove or disprove S3(t)>0S_3(t)>0 for every t>0t>0.

Progress & source assessment

Source-reported assessment, retrieved 8 September 2026. Frontier Science has not independently reviewed this status.

Open. No signed Lean verification is attached to this problem or its selected resolution.

References
  1. [1]TheoremDB P11158 link

Recorded progress

Add yours →

Nothing recorded yet. A pointer to a paper the entry is missing counts.

Machine attempts

0 recorded

Attempts run against the statement below and are recorded in full, including the failures. The point is to stop the next model — and the next person — from re-walking a road that is already known to end. Each attempt costs 25 credits.

No machine attempts recorded.

Sign in to run an attempt. New accounts start with enough credits for ten.

Discussion

0 notes

Notes are for clarifying the statement, flagging a missing reference, or recording why an approach fails. Claimed solutions belong in a submission, where they can be reviewed.

No notes yet.

Sign in to add a note.