OFFSET
1,1
COMMENTS
This is one of three sequences that partition the positive integers. In general, suppose that r, s, t are positive real numbers for which the sets
{i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1} are disjoint.
Let a(n) be the rank of n/r when all the numbers in the three sets are jointly ranked. Define b(n) and c(n) as the ranks of n/s and n/t. It is easy to prove that
a(n)=n+[ns/r]+[nt/r],
b(n)=n+[nr/s]+[nt/s],
c(n)=n+[nr/t]+[ns/t], where []=floor.
Taking r=sqrt(2), s=sqrt(3), t=sqrt(5) yields
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..10000
FORMULA
a(n)=n+floor(ns/r)+floor(nt/r), r=sqrt(2), s=sqrt(3), t=sqrt(5).
MATHEMATICA
PROG
(PARI) sr=sqrt(3/2); tr=sqrt(5/2); for(n=1, 100, print1(n+floor(n*sr)+floor(n*tr)", ")) \\ Charles R Greathouse IV, Jul 15 2011
(Maxima) r:sqrt(2)$ s:sqrt(3)$ t:sqrt(5)$
makelist(n+floor(n*s/r)+floor(n*t/r), n, 1, 50); /* Martin Ettl, Oct 18 2012 */
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jan 22 2011
STATUS
approved