A190329 - OEIS

%I #15 Oct 20 2024 02:08:55

%S 2,5,9,11,15,18,20,24,27,31,33,36,40,42,46,49,53,55,58,62,64,68,71,73,

%T 77,80,84,86,90,93,95,99,102,106,108,111,115,117,121,124,126,130,133,

%U 137,139,143,146,148,152,155,159,161,164,168,170,174,177,181,183,186,190,192,196,199,201,205,208,212,214,217,221,223,227,230

%N a(n) = n + [n*s/r] + [n*t/r]; r=1, 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 a(n) = n + [n*s/r] + [n*t/r],

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

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

%C Taking r=1, s=sqrt(2), t=1/s gives

%C a=A190326, b=A190327, c=A190328.

%H G. C. Greubel, <a href="/A190329/b190329.txt">Table of n, a(n) for n = 1..10000</a>

%F A190329:  a(n) = n + [n*sqrt(2)] + [n/sqrt(2)].

%F A190330:  b(n) = n + [n/sqrt(2)] + [n/2].

%F A187338:  c(n) = 3*n + [n*sqrt(2)].

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

%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}]  (*A190329*)

%t Table[b[n], {n, 1, 120}]  (*A190330*)

%t Table[c[n], {n, 1, 120}]  (*A187338*)

%o (PARI) for(n=1,50, print1(n + floor(n*sqrt(2)) + floor(n/sqrt(2)), ", ")) \\ _G. C. Greubel_, Jan 29 2018

%o (Magma) [n + Floor(n*Sqrt(2)) + Floor(n/Sqrt(2)): n in [1..50]]; // _G. C. Greubel_, Jan 29 2018

%Y Cf. A190330, A187338.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, May 08 2011