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 A189380 a(n) = n + floor(n*s/r) + floor(n*t/r); r=1, s=-1+sqrt(2), t=1+sqrt(2). 3

%I

%S 3,6,11,14,19,22,25,30,33,38,41,44,49,52,57,60,65,68,71,76,79,84,87,

%T 90,95,98,103,106,111,114,117,122,125,130,133,136,141,144,149,152,155,

%U 160,163,168,171,176,179,182,187,190,195,198,201,206,209,214,217,222,225,228,233,236,241,244,247,252,255,260,263,266,271,274,279,282,287,290,293,298,301,306,309,312,317

%N a(n) = n + floor(n*s/r) + floor(n*t/r); r=1, s=-1+sqrt(2), t=1+sqrt(2).

%C 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

%C f(n) = n + [n*s/r] + [n*t/r],

%C g(n) = n + [n*r/s] + [n*t/s],

%C h(n) = n + [n*r/t] + [n*s/t], where []=floor.

%C Taking r=1, s=-1+sqrt(2), t=1+sqrt(2) gives f=A189380, g=A189381, h=A189382.

%H G. C. Greubel, <a href="/A189380/b189380.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = 5*floor(n * (sqrt(2) - 1)) + 3*floor(n * (2 - sqrt(2))) + 3. - _Miko Labalan_, Dec 04 2016

%t r=1; s=-1+2^(1/2); t=1+2^(1/2);

%t f[n_] := n + Floor[n*s/r] + Floor[n*t/r];

%t g[n_] := n + Floor[n*r/s] + Floor[n*t/s];

%t h[n_] := n + Floor[n*r/t] + Floor[n*s/t]

%t Table[f[n], {n, 1, 120}] (* A189380 *)

%t Table[g[n], {n, 1, 120}] (* A189381 *)

%t Table[h[n], {n, 1, 120}] (* A189382 *)

%o (PARI) for(n=1, 100, print1(n + floor(n*(sqrt(2) -1)) + floor(n*(sqrt(2)+1)), ", ")) \\ _G. C. Greubel_, Apr 20 2018

%o (MAGMA) [n + Floor(n*(Sqrt(2) -1)) + Floor(n*(Sqrt(2) + 1)): n in [1..100]]; // _G. C. Greubel_, Apr 20 2018

%Y Cf. A189381, A189382.

%K nonn

%O 1,1

%A _Clark Kimberling_, Apr 20 2011

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Last modified September 26 13:25 EDT 2021. Contains 347668 sequences. (Running on oeis4.)