OFFSET
1,1
COMMENTS
Let x be the real solution of 1/x + 1/(1+x+x^2) = 1. Then (floor(n x)) and (floor(n*(x^2 + x + 1)))) are a pair of Beatty sequences; i.e., every positive integer is in exactly one of the sequences. See the Guide to related sequences at A329825.
LINKS
Eric Weisstein's World of Mathematics, Beatty Sequence.
FORMULA
a(n) = floor(n (1+x+x^2)), where x = 1.324717... is the constant in A060006.
MATHEMATICA
Solve[1/x + 1/(1 + x + x^2) == 1, x]
u = 1/3 (27/2 - (3 Sqrt[69])/2)^(1/3) + (1/2 (9 + Sqrt[69]))^(1/3)/3^(2/3);
u1 = N[u, 150]
RealDigits[u1, 10][[1]] (* A060006 *)
Table[Floor[n*u], {n, 1, 50}] (* A329974 *)
Table[Floor[n*(1 + u + u^2)], {n, 1, 50}] (* A329975 *)
Plot[1/x + 1/(1 + x + x^2) - 1, {x, -2, 2}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jan 02 2020
STATUS
approved