slug: oeis-a343502-divisor-count-primes-dequatre
bibkey: dequatre2021a343502
doi: null
url: https://oeis.org/A343502
triage: theorem
motivation_gids:
  - D5/S3/Arith/Divisors/OeisA343502DivisorCountPrimes

Dequatre’s OEIS A343502 divisor-count prime conjecture

Problem

C. H. R. Dequatre’s OEIS A343502 is titled “Numbers k such that tau(tau(k)) and tau(k+1) are both prime, where tau is the number of divisors function.” Its comment states: “Conjecture: all members except 2 and 3 are composite.” The entry also asks whether there are finitely many odd terms.

The formal claim is that every natural number $k$ with $1\le k$, with both divisor-count conditions prime, and with $k\ne 2$ and $k\ne 3$, satisfies $1<k$ and is not prime.

Statement echo

The membership predicate in D5/S3/Arith/Divisors/OeisA343502DivisorCountPrimes.lean expands tau literally as k.divisors.card, preserving the source convention. The settling declaration is result : claim; its Scribe formula expands the complete claim body.

Motivation

The conjecture isolates the only prime members visible in the published data and would explain the observed composite tail. A proof also reduces the remaining odd-term question to a more specific divisor-count problem without claiming that question is solved.

Gap

OEIS A343502 revision 27 presents the assertion as a conjecture and gives no proof. The preregistered literature screen in issue #18685 found no proof in the searched OEIS, formal-conjecture, and repository sources. Literature completeness is ASSUMED-UNVERIFIED.

Route

Assume a member $k\ne 2,3$ is prime. Then $k$ is odd, so $k+1$ is even. The prime value of the divisor count of $k+1$ forces all prime factors of $k+1$ to equal $2$, hence $k+1=2^e$ and its divisor count is $e+1$. The primality of $k=2^e-1$ makes $e$ prime; the primality of $e+1$ then forces $e=2$, giving $k=3$, a contradiction. The proof also derives $1<k$ from the membership condition.

Falsifier

A prime member other than $2$ or $3$, or a failure of the prime-power reduction for an even $k+1$, would falsify the settlement. The independent finite check is evidence for the stated range only and does not replace the universal proof.

Evidence

The kernel-checked settling declaration is D5.S3.Arith.Divisors.OeisA343502DivisorCountPrimes.result. Its proof uses the frozen divisor-factor bound D5.S3.Arith.DivisorCountRadicalCoincidence.primeFactors_card_le_cardFactors_card_divisors and the frozen divisor-count owner through the consumed correspondence lemma. The public result’s axiom closure is contained in {propext, Classical.choice, Quot.sound}.

The numerical check is preserved in the experiments repository at docs/reports/oeis-a343502-divisor-count-primes. Running python3 check.py 2000000 exits 0 and reports 37181 members through 2,000,000, with prime members exactly 2 and 3; the script SHA-256 is 57e9715d1209fafdf2d9090b4f1c3fa1f100069d45288e92e018d266f653f3c3.

The escape audit for the settling result and the published owner theorem is unfinished and tracked in #19974; no declared_validated registration is claimed.

Triage

What the settlement shows

Proved by result. Every A343502 member other than 2 and 3 is composite, with the exact universal quantifiers and divisor-count convention in the source claim.

Proved mechanism. For a hypothetical prime member, parity makes $k+1$ even; the prime divisor count collapses its prime-factor set to {2}, giving a power of two. The two adjacent prime conditions force the exponent to be 2, which is exactly the excluded member $k=3$.

Scope. The argument settles the conjecture for all natural $k$ under the stated membership predicate. It does not determine whether the sequence has finitely many odd terms.

Effect on the source. The OEIS conjecture is resolved as proved. The separate odd-term question remains open.

ASSUMED-UNVERIFIED

The literature search establishes only the bounded scope recorded in issue #18685; exhaustive worldwide completeness and priority remain ASSUMED-UNVERIFIED. The separate finiteness question for odd terms is open.