Richter 1976 uber die monotonie von differenzenfolgen

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Bernd Richter, Über die Monotonie von Differenzenfolgen. Acta Arithmetica 30 (1976), 225-227. doi:10.4064/aa-30-3-225-227. The image-only scan shows no copyright or license line on its rendered first or last page; the journal's record offers the PDF under the download link "Pobierz zgodnie z CC-BY", rendered "Free download under CC-BY license" on the English site, and names no version or URL for it (https://www.impan.pl/get/doi/10.4064/aa-30-3-225-227, read 2026-10-02): the Creative Commons Attribution license, with no version stated.

Richter's single Theorem states that if q_1, q_2, ... are primes with q_(n+1) - q_n >= q_n - q_(n-1) > 0 for all n >= 2, then liminf q_n / n^2 >= 1/S, where S = sum over r of (p_(r+1) - 1)^2 / (p_2 ... p_(r+1)) = 2.84010..., the sum running over the odd primes. The proof builds an extremal minorant sequence (Q_n): for each d, P(d) is the least prime not dividing d, and any arithmetic progression of primes with common difference d has at most P(d) - 1 terms (at most P(d) if the first term is P(d) itself), so the gap value d can repeat at most P(d) - 1 times; defining d_(n+1) = d_n + 2 once that quota is used and Q_(n+1) = Q_n + d_n gives q_n >= Q_n for all n. Counting how many d <= x have p_2...p_r | d but p_(r+1) not dividing d yields n ~ Sx and Q_(n+1) ~ S x^2, hence Q_n ~ n^2/S, which is estimate (2) and the theorem. The prime number theorem is used for P(d) = O(log d) and for the two asymptotics; the paper notes both extreme progression lengths for d = 2 and d = 6 occur, and that it is unknown whether the maximal length is attained for infinitely many d. This is the primary source for problem 455: the constant S = 2.84010... and the bound liminf q_n/n^2 >= 1/S are exactly Richter's, and the argument is the prime-AP-length obstruction P(d) applied to an extremal minorant.

Source: https://doi.org/10.4064/aa-30-3-225-227.

Bears on. #455

Results to transcribe.

• Theorem: If q_1, q_2, ... are primes with non-decreasing positive consecutive differences, then liminf q_n / n^2 >= 1/S with S = sum (p_(r+1)-1)^2/(p_2...p_(r+1)) = 2.84010...
• Estimate (3): P(d), the least prime not dividing d, satisfies P(d) = O(log d) by the prime number theorem.
• Estimates (4) and (5): For the minorant sequence, n ~ Sx and Q_(n+1) = S_1 x^2 + O(x log x) ~ S x^2, whence Q_n ~ n^2/S.