退出

Do repeated convergences in the prime-gap sequence contain predictive information about future primes?

数论 Math StackExchange -1 票 1 回答 60 浏览 提问者: cristian migar 2026-08-17 19:01
number-theory prime-numbers prime-gaps

问题内容

Let $p_n$ be the $n$-th prime and let

$$ g_n = p_n-p_{n-1}. $$

Define

$$ S_n=\sum_{i=2}^{n} g_i=p_n-2 $$

and

$$ V_n=S_n+g_n=p_n+g_n-2. $$

I call a convergence the occurrence of the same value $V$ at two or more distinct positions.

For example, for $V=103$:

$$ 95+8=103,\qquad 99+4=103,\qquad 101+2=103. $$

Thus the associated gap sequence is

$$ (8,4,2). $$

A convergence can also produce a composite value. For example,

$$ 395+8=403,\qquad 399+4=403, $$

while

$$ 403=13\cdot31. $$

I therefore do not interpret convergence itself as a proof of primality. Instead, I performed a chronological out-of-sample computational experiment in which candidate announcements were fixed before their future primality labels were evaluated.

In the future block there were 1,536,434 announcements, of which 171,556 were prime (11.165855481%). A matched control contained 152,384 primes (9.918030973%).

The difference was 1.247824508 percentage points, with a one-sided paired permutation $p$-value of 0.00009999.

The mathematical question I would like to ask is:

Is there a known number-theoretic explanation for why repeated values of

$$ V_n=p_n+g_n-2 $$

could produce this kind of out-of-sample enrichment in future primality?

In particular,

$$ V_n=V_{n+1}\iff g_n=2g_{n+1}, $$

which produces structures such as

$$ (8,4,2),\qquad (16,8,4,2),\ldots $$

I am interested in whether these configurations, or the associated omitted composite positions, have a known interpretation in terms of prime constellations, congruence restrictions, or sieve methods.

The complete manuscript, computational code and reproducibility material are available here:

https://doi.org/10.5281/zenodo.21970844

I would particularly welcome attempts to refute the observed effect or derive a mathematical explanation for it.

回答 (1)

Empy2 2 票 2026-08-17 19:33 原文

The possible remainders when you divide $(p_{n-1},p_n,g_n,V_n)$ by $3$ are

$$(1,1,0,2)\\(1,2,1,1)\\(2,1,2,1)\\(2,2,0,0)$$ So $V_n$ has a one in $4$ chance of being a multiple of $3$. That makes it more likely to be prime than similar numbers of the same size.