OFFSET
1,1
COMMENTS
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,
R = (..., -10, -9, -7, -6, -4, -3, -1, 0, 2, 3, 5, 6, 8, ...).
S = (..., -15, -11, -8, -5, -2, 1, 4, 7, 10, 14, 17, 20, ...)
By Fraenkel's theorem (Theorem XI in the cited paper); R and S partition the integers.
R is the set of integers n such that (cos n)*(cos(n + 1)) < 0;
S is the set of integers n such that (cos n)*(cos(n + 1)) > 0.
A246046 = (2,3,6,6,8,...), positive terms of R;
A062389 = (1,4,7,10,14,17,...), positive terms of S;
A258048 = (1,3,4,6,7,9,10,...), - (nonpositive terms of R).
A257984 = (2,5,8,11,15,...), - (negative terms of S);
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..10000
A. S. Fraenkel, The bracket function and complementary sets of integers, Canadian J. of Math. 21 (1969) 6-27.
FORMULA
a(n) = ceiling((n - 1/2)*Pi).
MATHEMATICA
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 15 2015
STATUS
approved