lean_workbook_plus_12173

⚠️ 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.

Let $p_n$ be the probability when the numbers $1$ to $\frac{n(n+1)}{2}$ are used with $n$ columns; the problem here asks for $p_5$ . Obviously $p_1 = 1$ . For a given $n$ , select the $n$ numbers in the $n$ th column first. If the largest number, $\frac{n(n+1)}{2}$ , is not selected for this column, we can't possibly satisfy the inequality. If it is, then probability we can fill in the remaining columns in a good way is $p_{n-1}$ , since the remaining numbers are all distinct. The probability of putting $\frac{n(n+1)}{2}$ in the last column is $\frac{2}{n+1}$ , so we get $p_n = \frac{2 p_{n-1}}{n+1}$ . This gives that $p_n = \frac{2^{n}}{(n+1)!}$ and $p_5 = \frac{2}{45}$ .

Why this node was retired

The posted statement is

theorem lean_workbook_plus_12173  (p : ℕ → ℚ)
  (h₀ : p 1 = 1)
  (h₁ : ∀ n, p (n + 1) = (2 * p n) / (n + 1)) :
  p 5 = 2 / 45   :=  by sorry

The recurrence denominator was shifted incorrectly.

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

Proposed corrected statement

Take p₁=1,pₙ₊₁=2pₙ/(n+2) for n≥1; prove pₙ=2ⁿ/(n+1)! for n≥1 and p₅=2/45.

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.