OFFSET
1,1
COMMENTS
Let x be the least positive solution of 1/(2 + sin x) + 1/(2 + cos x) = 1. Then (floor(n*(2 + sin x))) and (floor(n*(2 + cos x))) 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*(2 + sin x)), where x = 2.058943... is the constant in A329960.
MATHEMATICA
Solve[1/(2 + Sin[x]) + 1/(2 + Cos[x]) == 1, x]
u = ArcCos[-(1/2) + 1/Sqrt[2] - 1/2 Sqrt[-1 + 2 Sqrt[2]]]
u1 = N[u, 150]
RealDigits[u1, 10][[1]] (* A329960 *)
Table[Floor[n*(2 + Sin[u])], {n, 1, 50}] (* A329961 *)
Table[Floor[n*(2 + Cos[u])], {n, 1, 50}] (* A329962 *)
Plot[1/(2 + Sin[x]) + 1/(2 + Cos[x]) - 1, {x, -1, 3}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jan 02 2020
STATUS
approved