A190504 a(n) = n+[n*s/r]+[n*t/r]+[n*u/r]; r=golden ratio, s=r+1, t=r+2, u=r+3.

6, 14, 21, 29, 38, 45, 52, 59, 68, 76, 83, 91, 100, 106, 114, 121, 130, 138, 145, 153, 159, 168, 176, 183, 191, 200, 207, 214, 221, 230, 238, 245, 253, 262, 268, 276, 283, 291, 300, 307, 315, 321, 330, 338, 345, 353, 362, 368, 376, 383, 392, 400, 407, 415, 421, 430, 438, 445, 453, 462, 469, 476, 483, 492, 500, 507, 515, 524, 530

Offset

1,1

Comments

This is one of four sequences that partition the positive integers. In general, suppose that r, s, t, u are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1}, {h/u: h>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the four sets are jointly ranked. Define b(n), c(n), d(n) as the ranks of n/s, n/t, n/u, respectively.

It is easy to prove that

a(n)=n+[n*s/r]+[n*t/r]+[n*u/r],

b(n)=n+[n*r/s]+[n*t/s]+[n*u/s],

c(n)=n+[n*r/t]+[n*s/t]+[n*u/t],

d(n)=n+[n*r/u]+[n*s/u]+[n*t/u], where []=floor.

Taking r=golden ratio, s=r+1, t=r+2, u=r+3 gives a=A190504, b=A190505, c=A190506, d=A190507.

Links

Table of n, a(n) for n=1..69.

Mathematica

r=GoldenRatio; s=r+1; t=r+2; u=r+3;
a[n_] := n + Floor[n*s/r] + Floor[n*t/r]+Floor[n*u/r];
b[n_] := n + Floor[n*r/s] + Floor[n*t/s]+Floor[n*u/s];
c[n_] := n + Floor[n*r/t] + Floor[n*s/t]+Floor[n*u/t];
d[n_] := n + Floor[n*r/u] + Floor[n*s/u]+Floor[n*t/u];
Table[a[n], {n, 1, 120}]  (*A190504*)
Table[b[n], {n, 1, 120}]  (*A190505*)
Table[c[n], {n, 1, 120}]  (*A190506*)
Table[d[n], {n, 1, 120}]  (*A190507*)

Crossrefs

Cf. A190505, A190506, A190507 (the other three sequences in the partition of N).

Sequence in context: A063299 A184924 A110223 * A175582 A182081 A125086

Adjacent sequences: A190501 A190502 A190503 * A190505 A190506 A190507

Keyword

nonn,changed

Author

Clark Kimberling, May 11 2011

Status

approved