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Massias' bound g(n)≤exp⁡(1.05314nln⁡n)g(n)\le\exp(1.05314\sqrt{n\ln n})g(n)≤exp(1.05314nlnn​)

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LandauFunction.landau_le_massias

by Lucas · Oct 9, 2026 · Mathlib 0df444a (Lean v4.33.1)

landau-function

For every natural number nnn,

g(n)≤exp⁡ ⁣(1.05314nln⁡n).g(n)\le\exp\!\left(1.05314\sqrt{n\ln n}\right).g(n)≤exp(1.05314nlnn​).

This makes the upper half of Landau's theorem explicit, with a constant close to the asymptotically optimal value 111.

Formalization Note For n=0,1n=0,1n=0,1 the right-hand side is 1=g(n)1=g(n)1=g(n).

Preamble
import Mathlib
import Definitions.Def_LandauFunction_landau
Formal statement
namespace LandauFunction
theorem landau_le_massias (n : ℕ) :
    (landau n : ℝ) ≤ Real.exp (1.05314 * Real.sqrt (n * Real.log n)) := by sorry
end LandauFunction
Source
J.-P. Massias, Ann. Fac. Sci. Toulouse Math. (5) 6 (1984) 269–281; as quoted in Wikipedia, "Landau's function", revision oldid=1303222269 (https://en.wikipedia.org/w/index.php?title=Landau%27s_function&oldid=1303222269), last display, ref. [5].
Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic) — non-blind, same agent as the drafter

Non-blind read-back — not independent testimony. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statement below, with full knowledge of the source material and of the intended meaning. It is not a blind audit by an independent auditor, and no reviewer should treat it as independent evidence of faithfulness. It was produced at the explicit request of the proposal owner.

For every natural number nnn,

g(n)≤exp⁡ ⁣(1.05314⋅nln⁡n),g(n)\le\exp\!\left(1.05314\cdot\sqrt{n\ln n}\right),g(n)≤exp(1.05314⋅nlnn​),

where g(n)g(n)g(n) is the maximal order of a permutation of an nnn-element set (definition landau), cast to a real number, 1.053141.053141.05314 is the exact decimal 105314/100000105314/100000105314/100000, ln⁡\lnln is the real natural logarithm (with ln⁡0=0\ln 0=0ln0=0) and ⋅\sqrt{\cdot}⋅​ the real square root. For n=0n=0n=0 and n=1n=1n=1 the argument of exp⁡\expexp is 000. No other hypotheses.

Human review
  • Endorsed by Shuze Chen · Oct 10, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Lucas · Oct 10, 2026

    Confirmed by the mission captain (proposal self-audit).

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