Twin-Prime Sigma-Gcd Divisibility

Abstract

Every twin-prime center with prime gcd(k, sigma(k)) is divisible by 18.

Here sigma(k) is the sum of the positive divisors of k, represented by ArithmeticFunction.sigma 1 k. Subtraction is natural-number subtraction; the hypotheses ensure k is greater than one.

Theorem 1.1 (Twin-prime centers are even).

$$\forall k\in \mathbb{N},1<k⟹\mathrm{Prime}(k-1)⟹\mathrm{Prime}(k+1)⟹2\mid k$$

Proof. Machine-checked in Lean as D5/S3/Factorization/TwinPrimeSigmaGcdDivisibility.even_center (✓ std3). ∎

Source. Repository-derived.

Acknowledgement. OEIS Foundation Inc. (2026). OEIS A394757. URL: https://oeis.org/A394757.

Commentary.

If k were odd, both neighboring primes would be even, so both would equal 2. Their difference is 2, a contradiction.

Theorem 1.2 (Divisibility by three).

$$\forall k\in \mathbb{N},1<k⟹\mathrm{Prime}(k-1)⟹\mathrm{Prime}(k+1)⟹\mathrm{Prime}(\gcd (k,\mathrm{sigma}(k)))⟹2\mid k⟹3\mid k$$

Proof. Machine-checked in Lean as D5/S3/Factorization/TwinPrimeSigmaGcdDivisibility.three_center (✓ std3). ∎

Source. Repository-derived.

Acknowledgement. OEIS Foundation Inc. (2026). OEIS A394757. URL: https://oeis.org/A394757.

Commentary.

The proof treats k below 6 explicitly. At k=4, sigma(4)=7 and the gcd is 1, contradicting its primality. Above this range both neighboring primes exceed 3. Neither can be divisible by 3, so k must be.

Theorem 1.3 (A composite divisor of the sigma-gcd).

$$\forall k\in \mathbb{N},2\mid k⟹3\mid k⟹\neg (9\mid k)⟹(4\mid \gcd (k,\mathrm{sigma}(k))\vee 6\mid \gcd (k,\mathrm{sigma}(k)))$$

Proof. Machine-checked in Lean as D5/S3/Factorization/TwinPrimeSigmaGcdDivisibility.four_or_six_dvd_gcd (✓ std3). ∎

Source. Repository-derived.

Acknowledgement. OEIS Foundation Inc. (2026). OEIS A394757. URL: https://oeis.org/A394757.

Commentary.

Write k=3m. Since 9 does not divide k, 3 does not divide m, and multiplicativity gives sigma(k)=4 sigma(m). If 4 divides k, it divides the gcd. Otherwise write k=2r with r odd. Multiplicativity now gives sigma(k)=3 sigma(r). Thus both 2 and 3 divide the gcd, so 6 divides it. This argument applies to every k satisfying the three divisibility hypotheses, independently of the neighboring integers.

Theorem 1.4 (Every A394757 term is divisible by 18).

$$\forall k\in \mathbb{N},1<k⟹\mathrm{Prime}(k-1)⟹\mathrm{Prime}(k+1)⟹\mathrm{Prime}(\gcd (k,\mathrm{sigma}(k)))⟹18\mid k$$

Proof. Machine-checked in Lean as D5/S3/Factorization/TwinPrimeSigmaGcdDivisibility.sigma_gcd_divisibility (✓ std3). ∎

Resolves. Problems/oeis-a394757-twin-prime-sigma-gcd (proved) by D5/S3/Factorization/TwinPrimeSigmaGcdDivisibility.sigma_gcd_divisibility.

Source. Repository-derived.

Acknowledgement. OEIS Foundation Inc. (2026). OEIS A394757. URL: https://oeis.org/A394757.

Commentary.

The first two results give 2 and 3 dividing k. If 9 did not divide k, the preceding result would give a composite divisor of a prime gcd. Hence 9 divides k, and coprimality of 2 and 9 gives 18 dividing k. The result holds for all natural k; no finite search bound is used.

References

• Truth anchor: D5/S3/Factorization/TwinPrimeSigmaGcdDivisibility.even_center
• Truth anchor: D5/S3/Factorization/TwinPrimeSigmaGcdDivisibility.four_or_six_dvd_gcd
• Truth anchor: D5/S3/Factorization/TwinPrimeSigmaGcdDivisibility.sigma_gcd_divisibility
• Truth anchor: D5/S3/Factorization/TwinPrimeSigmaGcdDivisibility.three_center