lean_workbook_plus_24127

⚠️ 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.

Given function $f(x+1)+f(x)=2^x$ where as $f(0)=2013$ , find the last two digit of $f(2013)$ and $f(2014)$

Why this node was retired

The posted statement is

theorem lean_workbook_plus_24127 (f : ℕ → ℕ) (h₁ : f 0 = 2013) (h₂ : ∀ x, f (x + 1) + f x = 2 ^ x) : f 2013 ≡ 18 [ZMOD 100] ∧ f 2014 ≡ 66 [ZMOD 100]   :=  by sorry

The natural-valued codomain makes f₁+2013=1 impossible. The source recurrence needs negative intermediate values.

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

Define f:ℕ→ℤ by f₀=2013 and f_(n+1)+f_n=2^n. Determine f_2013 and f_2014 modulo100 from this consistent recurrence; verify rather than assume the inserted residues18 and66.

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.