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A190364
n + [n*s/r] + [n*t/r] + [n*u/r]; r=sqrt(2), s=1/r, t=sqrt(3), u=1/t.
4
2, 5, 8, 11, 15, 18, 20, 24, 27, 31, 33, 36, 39, 43, 46, 49, 51, 56, 58, 62, 64, 67, 71, 74, 77, 80, 84, 87, 89, 93, 95, 100, 102, 105, 108, 112, 115, 118, 120, 124, 127, 131, 133, 136, 140, 143, 146, 149, 153, 156, 158, 162, 164, 169, 171, 174, 177, 181, 184, 187, 189, 193, 196, 200, 202, 205, 209, 212, 215, 218, 220, 225, 227, 231
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=sqrt(2), s=1/r, t=sqrt(3), u=1/t gives
LINKS
FORMULA
A190364: a(n) = n + [n/2] + [n*sqrt(3/2)] + [n*sqrt(1/6)].
A190365: b(n) = 3*n + [n*sqrt(6)] + [n*sqrt(2/3)].
A190366: c(n) = n + [n*sqrt(2/3)] + [n*sqrt(1/6)] + [n/3].
A190367: d(n) = 4*n + [n*sqrt(6)] + [n*sqrt(3/2)].
MAPLE
r:=sqrt(2): s:=1/r: t:=sqrt(3): u:=1/t: seq(n + floor(n*s/r) + floor(n*t/r) + floor(n*u/r), n=1..10^3); # Muniru A Asiru, Feb 01 2018
MATHEMATICA
Table[n + Floor[n/2] + Floor[n*Sqrt[3/2]] + Floor[n*Sqrt[1/6]], {n, 1, 30}] (* G. C. Greubel, Jan 31 2018 *)
PROG
(PARI) for(n=1, 30, print1(n + floor(n/2) + floor(n*sqrt(3/2)) + floor(n*sqrt(1/6)), ", ")) \\ G. C. Greubel, Jan 31 2018
(Magma) [n + Floor(n/2) + Floor(n*Sqrt(3/2)) + Floor(n*Sqrt(1/6)): n in [1..30]]; // G. C. Greubel, Jan 31 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, May 09 2011
STATUS
approved