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A182773
Beatty sequence for 1+2^(2/3).
5
2, 5, 7, 10, 12, 15, 18, 20, 23, 25, 28, 31, 33, 36, 38, 41, 43, 46, 49, 51, 54, 56, 59, 62, 64, 67, 69, 72, 75, 77, 80, 82, 85, 87, 90, 93, 95, 98, 100, 103, 106, 108, 111, 113, 116, 119, 121, 124, 126, 129, 131, 134, 137, 139, 142
OFFSET
1,1
COMMENTS
Let u=2^(1/3). Jointly rank {j*u} and {k/u} as in the first comment at A182760; a(n) is the position of n*u.
FORMULA
a(n) = floor(n*(1+2^(2/3))).
MATHEMATICA
Floor[Range[100]*(1 + 2^(2/3))] (* Paolo Xausa, Jul 09 2024 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Nov 30 2010
STATUS
approved