%I #6 Jul 29 2015 13:54:19
%S 1,3,4,6,7,9,10,12,13,14,16,17,19,20,22,23,25,26,28,29,31,32,34,35,36,
%T 38,39,41,42,44,45,47,48,50,51,53,54,56,57,58,60,61,63,64,66,67,69,70,
%U 72,73,75,76,78,79,80,82,83,85,86,88,89,91,92,94,95,97
%N Nonhomogeneous Beatty sequence: ceiling((n + 1/2)*Pi/(Pi- 1))
%C Let r = Pi, s = r/(r-1), and t = 1/2. Let R be the ordered set {floor[(n + t)*r] : n is an integer} and let S be the ordered set {floor[(n - t)*s : n is an integer}; thus,
%C R = (..., -10, -9, -7, -6, -4, -3, -1, 0, 2, 3, 5, 6, 8, ...).
%C S = (..., -15, -11, -8, -5, -2, 1, 4, 7, 10, 14, 17, 20, ...)
%C By Fraenkel's theorem (Theorem XI in the cited paper); R and S partition the integers.
%C R is the set of integers n such that (cos n)*(cos(n + 1)) < 0;
%C S is the set of integers n such that (cos n)*(cos(n + 1)) > 0.
%C A246046 = (2,3,6,6,8,...), positive terms of R;
%C A062389 = (1,4,7,10,14,17,...), positive terms of S;
%C A258048 = (1,3,4,6,7,9,10,...), - (nonpositive terms of R).
%C A257984 = (2,5,8,11,15,...), - (negative terms of S);
%C A062389 and A246046 partition the positive integers, and A258048 and A257984 partition the nonnegative integers.
%H Clark Kimberling, <a href="/A258048/b258048.txt">Table of n, a(n) for n = 0..10000</a>
%H A. S. Fraenkel, <a href="http://dx.doi.org/10.4153/CJM-1969-002-7">The bracket function and complementary sets of integers</a>, Canadian J. of Math. 21 (1969) 6-27.
%F a(n) = ceiling((n + 1/2)*Pi/(Pi - 1)).
%t Table[Ceiling[(n - 1/2) Pi], {n, 1, 120}] (* A257984 *)
%t Table[Ceiling[(n + 1/2) Pi/(Pi - 1)], {n, 0, 120}] (* A258048 *)
%Y Cf. A257984 (complement), A246046, A062380, A258833.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Jun 15 2015
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