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Source. Equation (1) (p. 2) and Section 5 (pp. 12--14: the table of M(x)M(x), p. 12; equation (23) and Figure 2, p. 13; Questions 5.1 and 5.2, p. 14) of Nathan McNew, The convex hull of the prime number graph, in: Irregularities in the Distribution of Prime Numbers, Springer, Cham (2018), 125--141, doi:10.1007/978-3-319-92777-0_7, cited at the page numbers 1--15 of the author's preprint named on the source card.

Statement

Midpoint convex primes (p. 2). Following Pomerance, a midpoint convex prime is a prime pnp_n with

2pn<pn−i+pn+ifor all positive i<n.(1)2p_n<p_{n-i}+p_{n+i}\quad\text{for all positive }i<n. \qquad (1)

Every convex prime (a prime whose point (n,pn)(n,p_n) is a vertex of the convex hull of the prime number graph) is a midpoint convex prime (p. 2).

Equation (23) (p. 13). Put

Mn=min⁡1≤i<n(pn+i+pn−i)−2pn.M_n=\min_{1\le i<n}(p_{n+i}+p_{n-i})-2p_n .

By (1), the midpoint convex primes are exactly the pnp_n with Mn>0M_n>0. This is a definition and a restatement of (1); the paper proves no theorem about MnM_n.

Data (pp. 12--13). The paper tabulates the count M(x)M(x) of midpoint convex primes up to xx for x=101,…,1011x=10^1,\ldots,10^{11}, with M(1011)=1195764M(10^{11})=1195764 and log⁡M(x)/log⁡x\log M(x)/\log x rising from 0.451540.45154 at x=102x=10^2 to 0.552510.55251 at x=1011x=10^{11}. Figure 2 shows the distribution of MnM_n for n<1.6×108n<1.6\times10^8, and separately its nonnegative part. On this data the paper says that "it appears likely that MnM_n can be arbitrarily large" (p. 13). It notes that MnM_n can be arbitrarily negative, since pn+1+pn−1−2pnp_{n+1}+p_{n-1}-2p_n is a difference of consecutive prime gaps, and that the values of MnM_n tend to avoid multiples of 6.

Questions (p. 14). Question 5.1 asks whether M(x)=o(π(x))M(x)=o(\pi(x)), and likewise whether the count G(x)G(x) of good primes is o(π(x))o(\pi(x)). Question 5.2 asks whether C(x)=o(M(x))C(x)=o(M(x)) or L(x)=o(G(x))L(x)=o(G(x)), where CC and LL count the convex and log-convex primes.

Read depth. Claims checked: equations (1) and (23), the table of M(x)M(x), the caption of Figure 2, the sentence quoted above and Questions 5.1 and 5.2 were read on the page images of the preprint. The computations were not reproduced here.

Dependencies

None; the data are the author's computation.

Bears on

  • Problem 454: MnM_n is the problem's f(n)−2pnf(n)-2p_n with the minimum taken over 1≤i<n1\le i<n, so the problem asks whether lim sup⁡nMn=∞\limsup_n M_n=\infty. The paper proves nothing about this. It presents histograms of MnM_n for n<1.6×108n<1.6\times10^8 and on them judges it likely that MnM_n can be arbitrarily large: numerical evidence for a yes answer, not a proof.