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RESEARCH / Focused target

The A008474 prime-exponent sum iteration reaches five

Literature status: not recheckedRepository record: proved

RELEASED FOUNDATIONS / PROPOSED CONNECTION

Research connections

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Problem

The question

OEIS A008474, NAME (%N, verbatim):

If n = Product (p_j^k_j) then a(n) = Sum (p_j + k_j).

COMMENT (%C, verbatim):

Conjecture: for m > 4, by iterating the map m -> A008474(m) one always reaches 5 [tested up to m = 320000]. - Ivan N. Ianakiev, Nov 10 2014

FORMULA (%F, verbatim):

Additive with a(p^e) = p + e.

AUTHOR (%A, verbatim):

Olivier Gérard

The first terms (%S, verbatim excerpt) are 0,3,4,4,6,7,8,5,5,9,12,8,14,....

For every natural n, define F(n) = Sum (p + factorization(n)(p)) over p in primeFactors(n). The literal proved statement is forall m : Nat, 4 < m -> exists t : Nat, F^[t](m) = 5. Thus "reaches 5" means that some finite iterate equals 5. The values follow the cycle 5 -> 6 -> 7 -> 8 -> 5; four is a fixed point, so m > 4 is sharp; and 2 -> 3 -> 4 never reaches five. The generating function and the identities with A001222 and A008472 are NOT claimed.

Motivation

Our foothold

The conjecture asks for a global termination statement about an arithmetic map defined from the complete prime factorization of its input. The proof turns the orbit question into strict descent while preserving the lower bound five.

Gap

Missing bridges

The dated surfaces recorded in preregistration issue #7647 and the probe were all 64 revisions of OEIS A008474. The conjecture is unchanged in revision #64 of 2025-12-23, which added Krueger's formula. Daniel Tsai, Integers 21 (2021), Article #A32, was read in full: its v-function omits exponent one and the paper does not study this iteration. Searches returned arXiv 0, MathOverflow 0, Crossref 0, and GitHub 0. OpenAlex returned HTTP 429 - ASSUMED-UNVERIFIED. No priority claim is made.

Route

Proposed approach

(a) Establish coprime additivity of F and the value F(p^e) = p + e for prime p and positive e. (b) Prove F(n) <= n + 1 for every n >= 2. (c) Prove F(n) >= 5 for n >= 5, and one-step descent F(n) < n for composite n > 4 other than six. (d) Prove F(2n) <= F(n) + 3 and the prime two-step descent F(F(p)) < p for primes p >= 11. (e) Verify the four-cycle and close the theorem by strong induction.

Falsifier

What would falsify this route

A natural m greater than four whose every finite F-iterate differs from five would refute the theorem.

Evidence

Evidence to collect

  • Lean module: D5/S3/Factorization/IanakievPrimeExponentSumIterationReachesFive.lean.
  • Main theorem: ianakiev_a008474, with std3 axiom closure [propext, Classical.choice, Quot.sound].
  • Kernel profile after normalization inlining and import minimization: 9.85 s wall time, 266 ms cumulative type checking, and 1706754048 bytes maximum RSS.
  • Orchestrator check: every 5 <= m <= 10^5 reaches five, with maximum first hitting time 13 at m = 26833.
  • Probe check: every 5 <= m <= 10^6 reaches five, with maximum first hitting time 15 at m = 461938.
  • Search-seat check: the same range through 10^6, maximum 15 steps at m = 461938.
  • The proof asserts no generating-function formula, no A001222 or A008472 identity, and no theorem about Tsai's v-palindromes.

Triage

Scope assessment

theorem

ASSUMED-UNVERIFIED

Unverified assumptions

The OpenAlex search was rate-limited, the bounded orbit searches are finite, and no priority claim is made. The literature search is not an exhaustive proof of novelty.

Dossier & anchor history

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