lean_workbook_plus_44328

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Disproved status carries no information about that problem. Do not import this node or use it as a dependency.

We want to construct a sequence such that $a_i > 2a_{i-1}$ and then take one more than the number of terms to find the least $m$ we need. We could start at $1$ and do $2a_i + 1$ but to maximize the terms we work backwards. Note that operating $a_i = \lceil \frac{a_{i+1}}{2} \rceil - 1$ will ensure $a_{i+1}$ is the least integer greater than $2a_i$. We start at $2023$ and work backwards to $2$ over 10 of these divisions, so we have $m = 11 + 1 = \boxed{12}$.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_44328 :
  IsLeast {m : ℕ | ∃ a : ℕ → ℕ,
    StrictMono a ∧
    ∀ n, 2 * a n < a (n + 1) ∧ a (m - 1) = 2023 ∧ a m = 2} 12   :=  by sorry

The endpoint order is reversed and the original extremal quantifiers are absent.

The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.

Proposed corrected statement

Model a finite increasing chain bounded by2023 and the original guarantee defining the least m; do not force the later term to2 after a term2023.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.