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The Negative Regime of Singleton Prime Insertion

Abstract

Every increasing singleton prime insertion except (2,3) gives a negative full first compensated Laplace difference at every positive damping.

For actual primes q<p, use the original finite Euler product on the singleton {q} and on the two-prime set {q,p}. Both normalized ratios use the common clock log(p), and each numerator subtracts its exact slope at zero. The complete integrals extend over the whole positive half-axis.

Theorem 1.1 (All increasing prime pairs except (2,3) have a strictly negative difference).

Lean statement: D5/S3/Arith/Robin/SingletonPrimeInsertionNegativeRegime.result

Proof. Machine-checked in Lean as D5/S3/Arith/Robin/SingletonPrimeInsertionNegativeRegime.result (✓ std3). ∎

Source. Repository-derived.

Commentary.

For every prime pair q<p other than (2,3), and every sigma>0, both literal compensated integrands are absolutely integrable. The inserted integral minus the singleton integral is strictly negative.

Let c=log(q)/log(p). The exact difference numerator is G(v)=((q-exp(-cv))(1-exp(-v))-(q-1)v)/((p-1)(q-1)). The curvature bracket is strictly negative for v>0 whenever 2*c<=q-1, including equality. Zero value and slope then give a strictly negative numerator on the entire positive axis. A global quadratic bound pays absolute integrability; positivity of the measure gives the strict integral sign.

The already proved (2,3) transition is isolated within this increasing singleton family. A singleton {q} is not the full cutoff of all primes <=q. This result does not classify consecutive primorial insertions, the other factorial-density terms, or the complete Robin pairing, and it does not prove RH.

References