cp's OEIS Frontend

A330173 a(n) = n + floor(ns/r) + floor(nt/r), where r = sqrt(2), s = sqrt(2) + 1, t = sqrt(2) + 2.

%I A330173 #7 Apr 16 2023 19:11:02
%S A330173 4,9,15,19,25,30,34,40,45,51,55,60,66,70,76,81,87,91,96,102,106,112,
%T A330173 117,121,127,132,138,142,148,153,157,163,168,174,178,183,189,193,199,
%U A330173 204,208,214,219,225,229,235,240,244,250,255,261,265,270,276,280,286
%N A330173 a(n) = n + floor(ns/r) + floor(nt/r), where r = sqrt(2), s = sqrt(2) + 1, t = sqrt(2) + 2.
%C A330173 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 A330173 a(n)=n+[ns/r]+[nt/r],
%C A330173 b(n)=n+[nr/s]+[nt/s],
%C A330173 c(n)=n+[nr/t]+[ns/t], where []=floor.
%C A330173 Taking r = sqrt(2), s = sqrt(2) + 1, t = sqrt(2) + 2 yields
%C A330173 a=A330173, b=A016789, c=A091087.
%F A330173 a(n) = n + floor(ns/r) + floor(nt/r), where r = sqrt(2), s = sqrt(2) + 1, t = sqrt(2) + 2.
%t A330173 r = Sqrt[2]; s = Sqrt[2] + 1; t = Sqrt[2] + 2;
%t A330173 a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
%t A330173 b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
%t A330173 c[n_] := n + Floor[n*r/t] + Floor[n*s/t]
%t A330173 Table[a[n], {n, 1, 120}]  (* A330173 *)
%t A330173 Table[b[n], {n, 1, 120}]  (* A016789 *)
%t A330173 Table[c[n], {n, 1, 120}]  (* A091087 *)
%t A330173 Table[n+Floor[(n(Sqrt[2]+1))/Sqrt[2]]+Floor[(n(Sqrt[2]+2))/Sqrt[2]],{n,60}] (* _Harvey P. Dale_, Apr 16 2023 *)
%Y A330173 Cf. A016789, A091087.
%K A330173 nonn,easy
%O A330173 1,1
%A A330173 _Clark Kimberling_, Jan 04 2020