The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A189386 a(n) = n+[ns/r]+[nt/r]; r=1, s=sqrt(2), t=1/sqrt(3), []=floor. 6

%I

%S 2,5,8,11,14,17,20,23,26,29,32,34,38,41,44,47,50,53,55,59,62,65,68,70,

%T 74,77,80,83,86,89,91,95,98,101,104,106,110,112,116,119,121,125,127,

%U 131,133,137,140,142,146,148,152,155,157,161,163,167,169,173,176,178,182,184,188,190,193,197,199,203,205,208,211,214,218,220,224,226,229,233,235,239,241,244,247

%N a(n) = n+[ns/r]+[nt/r]; r=1, s=sqrt(2), t=1/sqrt(3), []=floor.

%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 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=1, s=sqrt(2), t=1/sqrt(3) gives

%C a=A189386, b=A189387, c=A189388.

%H G. C. Greubel, <a href="/A189386/b189386.txt">Table of n, a(n) for n = 1..5000</a>

%t r=1; s=2^(1/2); t=3^(-1/2);

%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}] (*A189386*)

%t Table[b[n], {n, 1, 120}] (*A189387*)

%t Table[c[n], {n, 1, 120}] (*A189388*)

%Y Cf. A189387, A189388, A189361, A189383, A189395.

%K nonn

%O 1,1

%A _Clark Kimberling_, Apr 21 2011

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified September 16 23:53 EDT 2021. Contains 347477 sequences. (Running on oeis4.)