lean_workbook_plus_50109

⚠️ Retired — specification defect

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

Find $\tau(a_{10})$ where $a_1=6$ and $a_{n+1}=2a_n^2$ for $n\geq 1$.

Why this node was retired

The posted statement is

theorem lean_workbook_plus_50109 (a : ℕ → ℕ) (h : a 1 = 6) (ha : ∀ n, a (n + 1) = 2 * (a n)^2) : a 10 = 1025   :=  by sorry

Recurrence includes n=0, and the target is a(10) rather than the divisor count τ(a(10)) requested by the prose.

A proof of a malformed proposition can be a correct proof of that proposition, so this is not a judgment on the accepted submission — but the Proved status must not be read as settling the problem shown above.

Proposed corrected statement

Use the recurrence only for n≥1 and a(1)=6. Ask for the divisor count Nat.divisors(a(10)).card, rather than equating a(10) itself to the proposed answer; compute that count before fixing a numeral.

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.