|
| |
|
|
A189524
|
|
a(n) = n + [n*s/r] + [n*t/r]; r=2, s=1+sqrt(2), t=1+sqrt(3).
|
|
3
|
|
|
|
3, 6, 10, 13, 17, 21, 24, 27, 31, 35, 39, 42, 45, 49, 53, 56, 60, 63, 66, 71, 74, 78, 81, 84, 89, 92, 95, 99, 103, 106, 110, 113, 117, 121, 124, 128, 131, 134, 139, 142, 146, 149, 152, 157, 160, 163, 167, 170, 174, 178, 181, 185, 188, 192, 196, 199, 202, 207, 210, 213, 217, 220, 225, 228, 231, 235, 238, 242, 246, 249, 252, 256, 260, 264, 267, 270, 274, 278, 281, 285, 288, 292, 296
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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 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
f(n) = n + [n*s/r] + [n*t/r],
g(n) = n + [n*r/s] + [n*t/s],
h(n) = n + [n*r/t] + [n*s/t], where []=floor.
|
|
|
LINKS
|
|
|
|
MATHEMATICA
|
r=2; s=1+2^(1/2); t=1+3^(1/2);
f[n_] := n + Floor[n*s/r] + Floor[n*t/r];
g[n_] := n + Floor[n*r/s] + Floor[n*t/s];
h[n_] := n + Floor[n*r/t] + Floor[n*s/t];
Table[f[n], {n, 1, 120}] (* A189524 *)
Table[g[n], {n, 1, 120}] (* A189525 *)
Table[h[n], {n, 1, 120}] (* A189526 *)
|
|
|
PROG
|
(PARI) for(n=1, 100, print1(n + floor(n*(sqrt(2)+1)/2) + floor(n*(1+sqrt(3))/2), ", ")) \\ G. C. Greubel, Apr 20 2018
(Magma) [n + Floor(n*(Sqrt(2)+1)/2) + Floor(n*(1+Sqrt(3))/2): n in [1..100]]; // G. C. Greubel, Apr 20 2018
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
