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Determining a sequence from a symmetric index relation

Proved
WorkbookCorrected.plus_66265

by wamlart · Sep 13, 2026 · Mathlib 0df444a (Lean v4.33.1)

corrected-formalizationlean-workbooksource-checked

The sequence {an}n≥0\{a_n\}_{n\geq 0}{an​}n≥0​ of real numbers satisfies the relation: am+n+am−n−m+n−1=12(a2m+a2n)a_{m+n} + a_{m-n} - m + n -1 = \frac12 (a_{2m} + a_{2n})am+n​+am−n​−m+n−1=21​(a2m​+a2n​) for all non-negative integers mmm and nnn , m≥nm \ge nm≥n . If a1=3a_1 = 3a1​=3 find a2004a_{2004}a2004​ .

The required value is a₂₀₀₄=4018021.

Formalization Note: The source condition m≥n is an antecedent restricting when the recurrence applies. The original formalization asserted m≥n for every pair of natural numbers, making its assumptions inconsistent. This correction restores the implication and proves the requested value from the original recurrence and initial value.

Source: InternLM Lean-Workbook, record lean_workbook_plus_66265 (Apache-2.0).

Preamble
import Mathlib
Formal statement
theorem WorkbookCorrected.plus_66265 (a : ℕ → ℝ) (h0 : a 1 = 3)
    (h : ∀ m n : ℕ, n ≤ m → a (m+n)+a (m-n)-(m:ℝ)+(n:ℝ)-1 = 1/2*(a (2*m)+a (2*n))) : a 2004 = 4018021 := by sorry
Source
https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_plus_66265; Apache-2.0

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