A330177 a(n) = n + floor(ns/r) + floor(nt/r), where r = e - 2, s = e - 1, t = e.

6, 13, 21, 28, 34, 42, 49, 57, 64, 70, 78, 85, 93, 99, 106, 114, 121, 129, 135, 142, 150, 157, 165, 171, 178, 186, 193, 199, 207, 214, 222, 229, 235, 243, 250, 258, 265, 271, 279, 286, 294, 300, 307, 315, 322, 330, 336, 343, 351, 358, 366, 372, 379, 387, 394

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 = e - 2, s = e - 1, t = e yields

a=A330177, b=A016789, c=A330178.

Links

Table of n, a(n) for n=1..55.

Formula

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

Mathematica

r = E - 2; s = E - 1; t = E;
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}]  (* A330177 *)
Table[b[n], {n, 1, 120}]  (* A016789 *)
Table[c[n], {n, 1, 120}]  (* A330178 *)

Crossrefs

Cf. A016789, A330178.

Sequence in context: A172330 A017053 A046040 * A227359 A056115 A173358

Adjacent sequences: A330174 A330175 A330176 * A330178 A330179 A330180

Keyword

nonn,easy

Author

Clark Kimberling, Jan 05 2020

Status

approved