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a(n) = n + floor(ns/r) + floor(nt/r), where r = sqrt(2) - 1, s = sqrt(2), t = sqrt(2) + 1.
2

%I #5 Jan 05 2020 08:13:08

%S 9,19,30,40,51,60,70,81,91,102,112,121,132,142,153,163,174,183,193,

%T 204,214,225,235,244,255,265,276,286,297,306,316,327,337,348,357,367,

%U 378,388,399,409,418,429,439,450,460,471,480,490,501,511,522,532,541,552

%N a(n) = n + floor(ns/r) + floor(nt/r), where r = sqrt(2) - 1, s = sqrt(2), t = sqrt(2) + 1.

%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 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 a(n)=n+[ns/r]+[nt/r],

%C b(n)=n+[nr/s]+[nt/s],

%C c(n)=n+[nr/t]+[ns/t], where []=floor.

%C Taking r = sqrt(2) - 1, s = sqrt(2), t = sqrt(2) + 1 yields

%C a=A330171, b=A016789, c=A330172.

%F a(n) = n + floor(ns/r) + floor(nt/r), where r = sqrt(2) - 1, s = sqrt(2), t = sqrt(2) + 1.

%t r = Sqrt[2] - 1; s = Sqrt[2]; t = Sqrt[2] + 1;

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

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

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

%t Table[a[n], {n, 1, 120}] (* A330171 *)

%t Table[b[n], {n, 1, 120}] (* A016789 *)

%t Table[c[n], {n, 1, 120}] (* A330172 *)

%Y Cf. A016789, A330172.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jan 04 2020