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%I #19 Jul 06 2020 20:22:29
%S 1,3,7,9,13,15,19,21,25,29,31,35,37,41,43,47,51,53,57,59,63,65,69,73,
%T 75,79,81,85,87,91,95,97,101,103,107,109,113,117,119,123,125,129,131,
%U 135,139,141,145,147,151,153,157,161,163,167,169,173,175,179,183,185,189
%N Beatty sequence 1+2*floor(n*Pi/2), which contains infinitely many primes.
%C The primes in this entirely odd sequence begin 3, 7, 13, 19, 29, 31, 37, 41, 43, 47, 53, 59, 73, 79, 97, 101. By the theorems in Banks, there are an infinite number of primes in this sequence.
%H William D. Banks and Igor E. Shparlinski, <a href="http://arXiv.org/abs/0708.1015">Prime numbers with Beatty sequences</a>, arXiv:0708.1015 [math.NT], 2007.
%F a(n) = 1 + 2*floor(n*Pi/2).
%F a(n) = 1 + 2*A140758(n). - _L. Edson Jeffery_, Mar 16 2013
%e a(0) = 1 because 1 + 2*floor(0*Pi) = 1 + 2*0 = 1 + 0 = 1.
%e a(1) = 3 because 1 + 2*floor(1*Pi/2) = 1 + 2*floor(1.5707963) = 1 + 2*1 = 3.
%e a(2) = 7 because 1 + 2*floor(2*Pi/2) = 1 + 2*floor(3.1415926) = 1 + 2*3 = 7.
%e a(3) = 9 because 1 + 2*floor(3*Pi/2) = 1 + 2*floor(4.7123889) = 1 + 2*4 = 9.
%e a(4) = 13 because 1 + 2*floor(4*Pi/2) = 1 + 2*floor(6.2831853) = 1 + 2*6 = 13.
%e a(5) = 15 because 1 + 2*floor(5*Pi/2) = 1 + 2*floor(7.8539816) = 1 + 2*7 = 15.
%e a(7) = 21 because 1 + 2*floor(7*Pi/2) = 1 + 2*floor(10.995574) = 1 + 2*10 = 21.
%t Table[1 + 2*Floor[n*Pi/2], {n, 0, 60}] (* _Stefan Steinerberger_, Sep 02 2007 *)
%Y Cf. A019669, A130568, A140758.
%K easy,nonn
%O 0,2
%A _Jonathan Vos Post_, Aug 14 2007
%E More terms from _Stefan Steinerberger_, Sep 02 2007