|
%I #10 Sep 08 2022 08:45:57
%S 1,3,6,7,9,12,13,15,18,20,21,24,26,27,30,32,35,36,38,41,42,44,47,48,
%T 50,53,55,56,59,61,62,65,67,70,71,73,76,77,79,82,83,85,88,90,91,94,96,
%U 97,100,102,105,106,108,111,112,114,117,119,120,123,125,126,129,131,132,135,137,140,141,143,146,147,149,152,154,155
%N a(n) = n + [n*s/r] + [n*t/r]; r=2, s=sqrt(2), t=1/s.
%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=2, s=sqrt(2), t=1/s gives
%C f=A190335, g=A190336, h=A190337.
%H G. C. Greubel, <a href="/A190335/b190335.txt">Table of n, a(n) for n = 1..10000</a>
%F A190335: f(n) = n + [n*sqrt(2)] + [n/sqrt(8)].
%F A190336: g(n) = n + [n/sqrt(2)] + [n/2].
%F A190337: h(n) = 3*n + [n*sqrt(8)].
%t r=2; s=2^(1/2); t=1/s;
%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}] (*A190335*)
%t Table[g[n], {n, 1, 120}] (*A190336*)
%t Table[h[n], {n, 1, 120}] (*A190337*)
%o (PARI) for(n=1,100, print1(n + floor(n*sqrt(2)) + floor(n/sqrt(8)), ", ")) \\ _G. C. Greubel_, Apr 04 2018
%o (Magma) R:=RealField(); [n + Floor(n*Sqrt(2)) + Floor(n/Sqrt(8)): n in [1..100]]; // _G. C. Greubel_, Apr 04 2018
%Y Cf. A190336, A190337.
%K nonn
%O 1,2
%A _Clark Kimberling_, May 08 2011
|
