OFFSET
0,2
COMMENTS
Complement of A258834.
Let r = sqrt(2) and s = r/(r-1) = 2 + sqrt(2). Let R be the ordered set {floor[(n + 1/4)*r] : n is an integer} and let S be the ordered set {floor[(n - 1/4)*s : n is an integer}; thus,
R = (..., -8, -7, -5, -4, -2, -1, 1, 2, 3, 5, 6, ...)
S = (..., -13, -10, -6, -3, 0, 4, 7, 11, 14, ...).
By Fraenkel's theorem (Theorem XI in the cited paper); R and S partition the integers.
A184580 = (1,2,3,5,6,...), positive terms of R;
A184581 = (4,7,11,14,...), positive terms of S;
A258833 = (1,2,4,5,6,...), - (negative terms of R);
A258834 = (0,3,6,10,...), - (nonpositive terms of S).
LINKS
Clark Kimberling, Table of n, a(n) for n = 0..10000
A. S. Fraenkel, The bracket function and complementary sets of integers, Canadian J. of Math. 21 (1969) 6-27.
Clark Kimberling, Beatty sequences and trigonometric functions, Integers 16 (2016), #A15.
FORMULA
a(n) = ceiling((n + 1/4)*sqrt(2)) = floor((n + 1/4)*sqrt(2) + 1).
MATHEMATICA
PROG
(Magma) [Ceiling((n + 1/4)*Sqrt(2)): n in [0..80]]; // Vincenzo Librandi, Jun 13 2015
(PARI) for(n=0, 50, print1(ceil((n + 1/4)*sqrt(2)), ", ")) \\ G. C. Greubel, Feb 08 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 12 2015
STATUS
approved