A189518 n+[ns/r]+[nt/r]; r=1, s=1/e, t=1/(e+1).

1, 2, 4, 6, 7, 9, 10, 12, 14, 15, 17, 19, 20, 22, 24, 25, 27, 28, 30, 32, 33, 35, 37, 38, 40, 41, 43, 45, 46, 49, 50, 51, 53, 55, 56, 58, 59, 61, 63, 64, 67, 68, 69, 71, 73, 74, 76, 77, 80, 81, 82, 84, 86, 87, 89, 91, 92, 94, 95, 98, 99, 100, 102, 104, 105, 107, 109, 111, 112, 113, 116, 117, 118, 120, 122, 123, 125, 126, 129, 130, 131, 134, 135, 136, 138

Offset

1,2

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 pairwise 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=1, s=1/e, t=1/(e+1) gives

a=A189518, b=A189519, c=A189520.

Links

Table of n, a(n) for n=1..85.

Formula

The three sequences a=A189518, b=A189519, c=A189520 are given by

a(n)=n+[n/e]+[n/(e+1)],

b(n)=n+[ne]+[ne/(e+1)],

c(n)=3n+[ne]+[n/e].

Is there a simple formula for the complement of a?

Mathematica

r=1; s=1/E; t=1/(E+1);
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}]  (*A189518*)
Table[b[n], {n, 1, 120}]  (*A189519*)
Table[c[n], {n, 1, 120}]  (*A189520*)

Crossrefs

Cf. A189519, A189520.

Sequence in context: A184658 A247910 A189407 * A213733 A218608 A288475

Adjacent sequences: A189515 A189516 A189517 * A189519 A189520 A189521

Keyword

nonn

Author

Clark Kimberling, Apr 23 2011

Status

approved