lean_workbook_plus_1126

⚠️ 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.

So $3a(n)=a(3n)+a(3n+1)+a(3n+2)$ and so, since $a(n)\in\{1,3\}$ : $\boxed{a(3n+2)=a(3n+1)=a(3n)=a(n)}$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_1126 (a : ℕ → ℕ) (h : a 1 = 1) (h' : a 2 = 3) (h'' : ∀ n, a (3 * n) + a (3 * n + 1) + a (3 * n + 2) = 3 * a n) : ∀ n, a (3 * n + 2) = a (3 * n + 1) ∧ a (3 * n + 2) = a (3 * n) ∧ a (3 * n + 2) = a n   :=  by sorry

The crucial source range a(n)∈{1,3} was dropped; the posted initial values also conflict with the n=0 averaging conclusion.

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

Proposed corrected statement

For an index n with a(n),a(3n),a(3n+1),a(3n+2)∈{1,3}, prove3a(n)=their sum implies all four values equal. Recover the original index range before imposing global initial values a(1)=1,a(2)=3.

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.