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Reciprocals and square roots of the naturals are not Benford

Proved
BenfordLaw.not_isBenfordSeq_inv_sqrt

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

benford-lawequidistributionleading-digits

Neither the reciprocals 1/1,1/2,1/3,…1/1, 1/2, 1/3,\dots1/1,1/2,1/3,… nor the square roots 1,2,3,…\sqrt1,\sqrt2,\sqrt3,\dots1​,2​,3​,… of the successive natural numbers satisfy Benford's law in base 101010: for each of the two sequences there is a digit d∈{1,…,9}d \in\{1,\dots,9\}d∈{1,…,9} for which the relative frequency of leading digit ddd among the first NNN terms does not converge to log⁡10(1+1/d)\log_{10}(1+1/d)log10​(1+1/d).

Preamble
import Mathlib
import Definitions.Def_BenfordLaw_Defs
Formal statement
namespace BenfordLaw

theorem not_isBenfordSeq_inv_sqrt :
    ¬ IsBenfordSeq 10 (fun n => 1 / ((n : ℝ) + 1)) ∧
      ¬ IsBenfordSeq 10 (fun n => Real.sqrt ((n : ℝ) + 1)) := by sorry

end BenfordLaw
Source
Wikipedia, "Benford's law" (https://en.wikipedia.org/wiki/Benford%27s_law), snapshot uploaded by the proposer, section "Distributions known to disobey Benford's law" ("The square roots and reciprocals of successive natural numbers do not obey this law", ref. Raimi 1976).
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, with full knowledge of the source article and of the intended meaning. It is not a blind audit by an independent reader, and no reviewer should treat it as one. Please compare it against the Lean code yourself.

Both of the following hold: (i) it is not the case that IsBenfordSeq(10,a)\mathrm{IsBenfordSeq}(10, a)IsBenfordSeq(10,a) for an=1/(n+1)a_n = 1/(n+1)an​=1/(n+1); (ii) it is not the case that IsBenfordSeq(10,a)\mathrm{IsBenfordSeq}(10, a)IsBenfordSeq(10,a) for an=n+1a_n = \sqrt{n+1}an​=n+1​ (n=0,1,…n = 0,1,\dotsn=0,1,…, real arithmetic). Unfolded, each says: there exists a natural 1≤d<101\le d<101≤d<10 such that the proportion of n∈{0,…,N−1}n\in\{0,\dots,N-1\}n∈{0,…,N−1} with ⌊10{log⁡10an}⌋=d\lfloor 10^{\{\log_{10}a_n\}}\rfloor = d⌊10{log10​an​}⌋=d does not tend to log⁡10(1+1/d)\log_{10}(1+1/d)log10​(1+1/d) (it may converge to something else or not converge at all).

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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