lean_workbook_plus_20662

⚠️ 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.

Prove that $x_{n+1}=ax_n+x_{n-1}$ with $x_0=1$ and $x_1=a$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_20662 (a : ℝ) (n : ℕ) : ∃ x : ℕ → ℝ, x 0 = 1 ∧ x 1 = a ∧ ∀ n, x (n + 1) = a * x n + x (n - 1)   :=  by sorry

The second-order recurrence includes n=0 and reuses x₀ as x_(-1), contradicting initial data.

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

Proposed corrected statement

Construct x₀=1,x₁=a and x_(n+2)=a·x_(n+1)+x_n for every n≥0.

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.