A330173 a(n) = n + floor(ns/r) + floor(nt/r), where r = sqrt(2), s = sqrt(2) + 1, t = sqrt(2) + 2.

4, 9, 15, 19, 25, 30, 34, 40, 45, 51, 55, 60, 66, 70, 76, 81, 87, 91, 96, 102, 106, 112, 117, 121, 127, 132, 138, 142, 148, 153, 157, 163, 168, 174, 178, 183, 189, 193, 199, 204, 208, 214, 219, 225, 229, 235, 240, 244, 250, 255, 261, 265, 270, 276, 280, 286

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(2) + 1, t = sqrt(2) + 2 yields

a=A330173, b=A016789, c=A091087.

Links

Table of n, a(n) for n=1..56.

Formula

a(n) = n + floor(ns/r) + floor(nt/r), where r = sqrt(2), s = sqrt(2) + 1, t = sqrt(2) + 2.

Mathematica

r = Sqrt[2]; s = Sqrt[2] + 1; t = Sqrt[2] + 2;
a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
c[n_] := n + Floor[n*r/t] + Floor[n*s/t]
Table[a[n], {n, 1, 120}]  (* A330173 *)
Table[b[n], {n, 1, 120}]  (* A016789 *)
Table[c[n], {n, 1, 120}]  (* A091087 *)
Table[n+Floor[(n(Sqrt[2]+1))/Sqrt[2]]+Floor[(n(Sqrt[2]+2))/Sqrt[2]], {n, 60}] (* Harvey P. Dale, Apr 16 2023 *)

Crossrefs

Cf. A016789, A091087.

Sequence in context: A313213 A313214 A313215 * A313216 A189388 A184916

Adjacent sequences: A330170 A330171 A330172 * A330174 A330175 A330176

Keyword

nonn,easy

Author

Clark Kimberling, Jan 04 2020

Status

approved