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%I #29 May 08 2018 15:11:55
%S 1,4,7,10,14,17,20,23,26,29,32,36,39,42,45,48,51,54,58,61,64,67,70,73,
%T 76,80,83,86,89,92,95,98,102,105,108,111,114,117,120,124,127,130,133,
%U 136,139,142,146,149,152,155,158,161,164,168,171,174,177,180,183,186
%N a(n) = floor( (2n-1)*Pi/2 ).
%C In general, the complement of a nonhomogenous Beatty sequence [n*r + h] is given by [n*s + h - h*s], where s = r/(r - 1). As an example, the complement of this sequence is A246046. This sequence gives the positive integers k satisfying tan(k) > tan(k + 1), and A246046 gives those satisfying tan(k) < tan(k + 1). - _Clark Kimberling_, Aug 24 2014
%C Excluding a(1), a(n) = positive floored solutions to tan(x) = x. - _Derek Orr_, May 30 2015
%D M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 223.
%H Harry J. Smith, <a href="/A062389/b062389.txt">Table of n, a(n) for n=1..1000</a>
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%p seq(floor((2*n-1)*Pi/2), n=1..1000); # _Robert Israel_, Jun 01 2015
%t r = Pi; s = Pi/(Pi - 1); h = -Pi/2; z = 120;
%t u = Table[Floor[n*r + h], {n, 1, z}] (* A062389 *)
%t v = Table[Floor[n*s + h - h*s], {n, 1, z}] (* A246046 *)
%t (* _Clark Kimberling_, Aug 24 2014 *)
%o (PARI) j=[]; for(n=1,150,j=concat(j,floor(1/2*(2*n-1)*Pi))); j
%o (PARI) { default(realprecision, 50); for (n=1, 1000, write("b062389.txt", n, " ", (2*n - 1)*Pi\2); ) } \\ _Harry J. Smith_, Aug 06 2009
%Y Cf. A246046.
%K nonn,easy
%O 1,2
%A _Jason Earls_, Jul 08 2001
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