Lindstrom 2000 b h g sequences b h

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corollary_p659: Lindström's corollary that for g = m^(h-1) the interval [1,n] contains a B_h[g] sequence of size (gn)^(1/h)(1+o(1)) as n tends to infinity.

theorem_p658: Lindström's theorem that if A is a B_h sequence, m >= 2 is an integer and g = m^(h-1), then the union B of the sets {ma + i : a in A} for i = 0, ..., m-1 is a B_h[g] sequence with |B| = m|A|.

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Bernt Lindström, B_h[g]-sequences from B_h-sequences. Proceedings of the American Mathematical Society 128 (2000), no. 3, 657-659 (electronically published 9 September 1999). doi:10.1090/S0002-9939-99-05122-9. The copy read for this card is the publisher's PDF, which prints "©1999 American Mathematical Society" on its first page, every other right reserved.

The main theorem shows that if A is a B_h sequence and g = m^{h-1} for an integer m >= 2, then B = union_{i=0}^{m-1} {ma + i : a in A} is a B_h[g] sequence with |B| = m|A|; the corollary deduces that the interval [1,n] contains a B_h[g] sequence of size (gn)^{1/h}(1 + o(1)) as n -> infinity. The paper first records the necessary constraints m <= g (when |A| >= 2) and m^{h-1}/h <= g (when |A| >= h) on the construction, correcting a flawed attempt of Jia and generalizing Theorem 3 of Kolountzakis; the second constraint comes from counting solutions of x_1 + ... + x_h = c with 0 <= x_i <= m-1. The proof of the theorem is elementary: among g + 1 = m^{h-1} + 1 distinct representations of one integer, two share their first h - 1 residues mod m, hence all h residues, and the B_h property of A then makes them equal. For Problem 158 the case h = m = 2 turns an infinite Sidon set A into the infinite B_2[2] set 2A union (2A+1), which has at most 2A(N/2) elements up to N; by Erdős's theorem on infinite Sidon sets its counting function also has lower limit 0 against sqrt(N), so the construction yields no counterexample. For h = 2 the corollary's sets reach square-root order, but separately in each interval [1,n]. For Problem 863 the corollary with h = 2 and g = m = r gives B_2[r] sets in [1,N] of size (rN)^{1/2}(1+o(1)) for every integer r >= 2. The paper mentions neither problem.

Source: https://doi.org/10.1090/S0002-9939-99-05122-9.

Bears on. #158: the theorem with h = m = 2 gives infinite B_2[2] sets B = 2A union (2A+1) from infinite Sidon sets A; since B(N) <= 2A(N/2), Erdős's theorem for Sidon sets gives B(N)/sqrt(N) lower limit 0 as well (a deduction made here, not in the paper), so no counterexample arises; the corollary gives only finite B_2[2] sets, which do not address the liminf the problem asks about. #863: the corollary with h = 2, g = m = r gives a B_2[r] set in [1,N] of size (rN)^{1/2}(1+o(1)) for every integer r >= 2, so c_r >= sqrt(r) whenever the constant c_r exists; it does not separate c_r from c_r'.

Results. The paper numbers neither result; the pages are named by the page of the printed statement.

• Theorem (p. 658, also in the abstract on p. 657): for a B_h sequence A, an integer m >= 2 and g = m^{h-1}, the union B of mA + i over i < m is a B_h[g] sequence with |B| = m|A|; with the necessary conditions (1.3) m <= g when |A| >= 2 and (1.6) m^{h-1}/h <= g when |A| >= h (p. 658).
• Corollary (p. 659, also in the abstract on p. 657): for g = m^{h-1}, the interval [1,n] contains a B_h[g] sequence of size (gn)^{1/h}(1 + o(1)) as n -> infinity.

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