OFFSET
1,1
COMMENTS
Let u = 2^(1/4) + 2^(-1/4) and v = u/(u-1); let b(n) = floor[n*v], so that (b(n)) is the Beatty sequence for v. Then (a(n)) and (b(n)) partition the positive integers. Also, 0 <= a(n) - A378142(n) <= 1 for every n.
LINKS
FORMULA
a(n) = floor(n*(2^(1/4) + 2^(-1/4))).
MATHEMATICA
Table[Floor[n (2^(1/4) + 2^(-1/4))], {n, 1, 120}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jan 11 2025
STATUS
approved
