Pomerance's Questions on Good Primes

Pomerance’s Good-Prime Questions

Definitions and mathematical statement

Shared notation.

Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

Let $p_n$ be the increasing sequence of primes. A good prime has $p_n^2>p_{n-i}p_{n+i}$ for every $1\leq i<n$. Density in the first question is the natural density of the qualifying indices $n$. The second assertion concerns strict inequalities $p_np_{n+1}>p_{n-i}p_{n+1+i}$ simultaneously over $1\leq i<n$, for infinitely many $n$. [S1] [S2]

Statement recorded in the supplied source.

Call prime $p_n$ good if $p_n^2 > p_{n-i}p_{n+i}$ for all $i$, $1 \le i \le n-1$. Is it true that the set of $n$ for which $p_n$ is good has density 0? Are there infinitely many $n$ with $p_n p_{n+1} > p_{n-i} p_{n+1+i}$ for all $i$, $1 \le i \le n-1$? [S1]

Residual task reported by the archive.

The embedded review $2026-08-17$ records: Prove G(x)=o(pi(x)) for all good primes and prove or refute infinitely many indices satisfying the adjacent-pair product inequalities. [S1]

Short English statement

Are good primes sparse among prime indices, and do infinitely many adjacent prime pairs dominate every symmetrically placed outer pair in the stated product comparison?

Research attempt

Scope and separation of the two questions.

The first target is zero natural density of all good-prime indices, not of a smaller convex-hull subset. The second is infinitude of indices satisfying every adjacent-pair product inequality. McNew's primary paper [E1], inspected on 1 October 2026, explicitly distinguishes these related sets. This attempt derives exact gap restrictions and reconstructs the support-line reason that there are infinitely many good primes, then explains why that does not settle either target.

A necessary neighboring-gap inequality.

Let $p=p_n$, $a=p_n-p_{n-1}$, and $b=p_{n+1}-p_n$ for $n\ge2$. The case $i=1$ of goodness is \[ p^2>(p-a)(p+b) \quad\Longleftrightarrow\quad b(p-a)<pa \quad\Longleftrightarrow\quad b<\frac{pa}{p-a}. \] Equivalently $p(b-a)<ab$. Thus a rise from $a$ to a noticeably larger following gap obstructs goodness when the prime is large relative to the gap product. For instance if $b\ge a+2$ and $p\ge ab/2$, the strict inequality fails. This is a necessary local test only; passing it leaves all more distant symmetric pairs to check.

For the second question put additionally $c=p_{n+2}-p_{n+1}$. Its $i=1$ inequality becomes \[ p(p+b)>(p-a)(p+b+c) \quad\Longleftrightarrow\quad c<\frac{a(p+b)}{p-a}. \] It compares the two outer gaps while retaining the middle gap. There is no implication from the neighboring-gap test for one good prime to all adjacent-pair inequalities, so the two targets cannot be conflated.

The logarithmic geometric reformulation.

Set $f(n)=\log p_n$. Then $n$ is good exactly when \[ f(n)>\tfrac12\left(f(n-i)+f(n+i)\right)\qquad(1\le i<n). \] A sufficient condition is that a line of slope $s$ through $(n,f(n))$ lie strictly above every other point $(m,f(m))$. Indeed adding its two strict inequalities at $n-i$ and $n+i$ cancels the slope terms. Such a supporting line imposes conditions at all indices, whereas goodness only compares symmetric pairs, so the support-line subset may be smaller.

Why the support-line subset is infinite.

Use the standard prime-number estimate $p_n\asymp n\log n$, which implies $f(n)/n\to0$ while $f(n)\to\infty$. For every $s>0$, the function $f(n)-sn$ tends to minus infinity and attains a maximum at a finite index. Avoiding the countable set of slopes $(f(m)-f(n))/(m-n)$ ensures that the maximizing index is unique. It then gives a strict supporting line and hence a good prime.

The maximizing indices become unbounded as $s\downarrow0$ through such slopes. Otherwise they would remain in a finite set with maximal value $C$ of $f$. Choose $N$ with $f(N)>C+1$ and then $s>0$ small enough that $sN<1$; the value at $N$ would exceed every value at that finite set, a contradiction. This reproduces the underlying infinitude mechanism, using the stated prime-growth input.

The result is consistent with density zero. An infinite set can have zero density, and even a zero-density theorem for this support-line subset would not bound the larger set of all good indices. To settle the first catalogue target one must control good points lying away from the global supporting hull as well.

What would suffice for the adjacent-pair problem.

If a single supporting line touches the logarithmic graph exactly at consecutive indices $n,n+1$ and lies strictly above it at all other indices, then for every $1\le i<n$ adding the inequalities at $n-i$ and $n+1+i$ gives \[ f(n)+f(n+1)>f(n-i)+f(n+1+i). \] This would certify the requested adjacent-pair property at $n$. But infinitely many supporting vertices do not imply infinitely many edges whose endpoint indices differ by one. The gaps between consecutive supporting vertices can grow. The preceding maximization proof gives no control of such adjacency and therefore does not answer the second question.

Executed exact checks.

For every $1\le n\le2000$, the script checks all allowed $i<n$ using integer prime products, rather than rounded logarithms. It finds $240$ good indices and $249$ indices satisfying the adjacent-pair condition. The case $n=1$ is included in both counts because its list of comparisons is empty. Enough primes are generated to include the largest required index $2n$. Files are checks_second.py and results_second.json in _Local/Erdos_problems/UnsolvedMath/attempt_audit_2026_10_01/number_theory/. These finite proportions neither prove convergence of a density nor certify an infinite supply of adjacent-pair indices.

Remaining gap and outcome.

The attempt obtains useful necessary gap tests and a complete support-line infinitude argument for a related subclass. It has no upper bound of order $o(N)$ for all good indices up to $N$, and no infinitude proof for the adjacent-pair condition. Outcome: UNRESOLVED for both catalogue questions. The distinction between the larger target set and a controlled subset is essential.

Sources

• S1 ULAM.AI, UnsolvedMath, supplied problems.json, record 1867 (GUY-A14a), statement and background. The embedded literature-review date is 2026-08-17; that is the archive’s review date, not a fresh certification. Dataset landing page: https://huggingface.co/datasets/ulamai/UnsolvedMath.

• S2 Nathan McNew, The Convex Hull of the Prime Number Graph, in Irregularities in the Distribution of Prime Numbers, Springer $2018$, 125-141. https://www.nathanmcnew.com/Convex.pdf. Bibliographic information from the supplied record.

• S3 Carl Pomerance, The prime number graph, Mathematics of Computation 33 $1979$, 399-408. https://doi.org/10.1090/S0025-5718-1979-0514821-1. Bibliographic information from the supplied record.

• E1 Nathan McNew, The Convex Hull of the Prime Number Graph, author manuscript of the 2018 chapter. https://www.nathanmcnew.com/Convex.pdf. Definitions, introductory distinctions, and stated scope inspected 1 October 2026. The paper's result for log-convex primes is not substituted for the question about all good primes.