OFFSET
0,4
COMMENTS
a(n) = floor(n*sqrt(2)) - floor(n/sqrt(2)). Indeed, the equation {(nearest integer to n/r) = floor(nr) - floor(n/r) for all n>=0} has exactly two solutions: sqrt(2) and -sqrt(2). - Clark Kimberling, Dec 18 2003
Let s(n) = zeta(3) - Sum_{k=1..n} 1/k^3. Conjecture: for n >=1, s(a(n)) < 1/n^2 < s(a(n)-1), and the difference sequence of A049473 consists solely of 0's and 1, in positions given by the nonhomogeneous Beatty sequences A001954 and A001953, respectively. - Clark Kimberling, Oct 05 2014
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..10000
MATHEMATICA
Round[Range[0, 70]/Sqrt[2]] (* Harvey P. Dale, Feb 17 2015 *)
PROG
(PARI) a(n)=round(n/sqrt(2)) \\ Charles R Greathouse IV, Sep 02 2015
(Magma) [0] cat [Round(n/Sqrt(2)): n in [1..100]]; // G. C. Greubel, Jan 27 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved