lean_workbook_plus_63475

⚠️ 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.

Define a new series $q_n=p_n-\frac{1}{2}$ . Substituting gives us

so the new series satisfies $q_{n+1}=\frac{q_n}{3}$ . The characteristic polynomial is simply $x-\frac{1}{3}$ which has $\frac{1}{3}$ as its only root. Now we can substitute $q_n=\frac{c}{3^n}$ for some constant $c$ , which gives us $p_n=\frac{c}{3^n}+\frac{1}{2}$ . Since $p_0=1$ , we find that $c=\frac{1}{2}$ , and therefore $p_n=\frac{1}{2\cdot 3^n}+\frac{1}{2}=\frac{1+3^n}{2\cdot 3^n}$ .

Why this node was retired

The posted statement is

theorem lean_workbook_plus_63475  (p : ℕ → ℚ)
  (h₀ : p 0 = 1)
  (h₁ : ∀ n, p (n + 1) = 1 / 3 + p n / 3 + 1 / 6) :
  ∀ n, p n = (1 + 3^n) / (2 * 3^n)   :=  by sorry

Use the unshifted recurrence p(n+1)=1/3+p(n)/3; the extra1/6 belongs to the substituted q recurrence and must not be added again.

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

Proposed corrected statement

Use the unshifted recurrence p(n+1)=1/3+p(n)/3; the extra1/6 belongs to the substituted q recurrence and must not be added again.

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.