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A190344
a(n) = n + [n*s/r] + [n*t/r]; r=1, s=sqrt(5/2), t=sqrt(2/5).
3
2, 6, 8, 12, 15, 18, 22, 25, 28, 31, 34, 37, 41, 44, 47, 51, 53, 57, 61, 63, 67, 69, 73, 76, 79, 83, 86, 89, 92, 95, 99, 102, 105, 108, 112, 114, 118, 122, 124, 128, 130, 134, 137, 140, 144, 147, 150, 153, 156, 160, 163, 166, 169, 173, 175, 179, 183, 185, 189, 191, 195, 199, 201, 205, 208, 211, 214, 218, 221, 224, 227, 230, 234, 237
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.
Taking r=1, s=sqrt(5/2), t=sqrt(2/5) gives f=A190344, g=A190345, h=A190346.
LINKS
FORMULA
A190344: f(n) = n + [n*sqrt(5/2)] + [n*sqrt(2/5)].
A190345: g(n) = n + [n*sqrt(2/5)] + [2*n/5].
A190346: h(n) = 3*n + [n*sqrt(5/2)] + [n/2].
MATHEMATICA
r=1; s=(5/2)^(1/2); t=1/s;
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}] (* A190344 *)
Table[g[n], {n, 1, 120}] (* A190345 *)
Table[h[n], {n, 1, 120}] (* A190346 *)
PROG
(PARI) for(n=1, 100, print1(n + floor(n*sqrt(5/2)) + floor(n*sqrt(2/5)), ", ")) \\ G. C. Greubel, Apr 05 2018
(Magma) [n + Floor(n*Sqrt(5/2)) + Floor(n*Sqrt(2/5)): n in [1..100]]; // G. C. Greubel, Apr 05 2018
CROSSREFS
Sequence in context: A138626 A178406 A189515 * A287000 A282358 A064796
KEYWORD
nonn
AUTHOR
Clark Kimberling, May 09 2011
STATUS
approved