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Statement

Setting (pp. 1, 4). $a\prec b$ means $b\equiv1\pmod a$, and $H(p)$ is the length of the longest prime chain $p_1\prec\cdots\prec p_k=p$.

Passage in Section 1.3 (p. 4; the print gives it no label). Define the chain $2=q_1\prec q_2\prec\cdots$ by letting $q_{j+1}$ be the smallest prime congruent to $1$ modulo $q_j$. Linnik's theorem gives $q_{j+1}\le q_j^L$ for some constant $L$, hence $H(q_j)\ge\frac{\log\log q_j}{\log L}$. The paper adds that it is conjectured that $q_{j+1}\le q_j(\log q_j)^C$ for some fixed $C$, and that this would imply the much stronger bound $H(q_j)\gg\frac{\log q_j}{\log\log q_j}$.

The paper uses the chain only to exhibit large values of $H(p)$; it states no theorem about it and does not name the source of the conjecture.

Proof pointer

The Linnik bound is a citation (Linnik's theorem on the least prime in an arithmetic progression); the deduction $\log q_j\le L^{j-1}\log2$ is immediate by induction. The conjectured bound is not proved.

Read depth

Claims checked: the remark was read clause by clause on the print (p. 4). Nothing here is independently reviewed.

Dependencies

None in the corpus. External input: Linnik's theorem.

Source. Kevin Ford, Sergei V. Konyagin and Florian Luca, Prime chains and Pratt trees, Geom. Funct. Anal. 20 (2010), no. 5, 1231--1258, doi:10.1007/s00039-010-0089-0, arXiv:0904.0473; page numbers are those of the arXiv version 4 named on the source card.

Bears on

• Problem 695: the chain $q_1<q_2<\cdots$ is a sequence of the kind the problem considers. The paper does not mention the problem. Reading the remark against its second question (whether some such sequence has $p_k\le\exp(k(\log k)^{1+o(1)})$): the conjectured $q_{j+1}\le q_j(\log q_j)^C$ gives $\log q_{j+1}\le\log q_j+C\log\log q_j$, so by induction $\log q_k=O(k\log k)$ and the chain would answer that question yes. The conjecture is unproved, and the unconditional Linnik bound gives only $\log q_k\le L^{k-1}\log2$, far from the required growth. The remark does not bear on the first question.