OFFSET
0,3
COMMENTS
Terms >= 2 give numbers n satisfying: floor(sqrt(2)*n) is even. - Benoit Cloitre, May 27 2004
Essentially equivalent to A120752, see Fried link. - Charles R Greathouse IV, Jan 20 2023
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..5000
Sela Fried, Equivalent conditions for the nth element of the Beatty sequence B_sqrt(2) being even, arXiv preprint arXiv:2301.00644 [math.HO], 2022.
FORMULA
It seems that a(n) = 2*n + o(n); conjecture : a(n) = 2*n + O(1).
MATHEMATICA
With[{r = Sqrt[2]}, Select[Range[0, 150], Floor[#*r*Floor[#/r]] == Floor[(#/r)*Floor[#*r]] &]] (* G. C. Greubel, Feb 06 2019 *)
PROG
(PARI) r=sqrt(2); for(n=0, 150, if(floor(n*r*floor(n/r))==floor(n/r*floor(n*r)), print1(n, ", "))) \\ G. C. Greubel, Feb 06 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Feb 15 2004
STATUS
approved