A184812 n+floor(ns/r)+floor(nt/r), where r=sqrt(2), s=sqrt(3), t=sqrt(5).

3, 7, 10, 14, 18, 22, 26, 29, 34, 37, 41, 44, 48, 53, 56, 60, 63, 68, 72, 75, 79, 82, 87, 90, 94, 98, 102, 106, 109, 113, 117, 121, 125, 128, 132, 136, 140, 144, 147, 151, 155, 159, 162, 166, 171, 174, 178, 181, 186, 190, 193, 197, 200, 205, 208, 212, 216

Offset

1,1

Comments

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

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

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

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

Taking r=sqrt(2), s=sqrt(3), t=sqrt(5) yields

a=A184812, b=A184813, c=A184815.

Links

G. C. Greubel, Table of n, a(n) for n = 1..10000

Formula

a(n)=n+floor(ns/r)+floor(nt/r), r=sqrt(2), s=sqrt(3), t=sqrt(5).

Mathematica

r=2^(1/2); s=3^(1/2); t=5^(1/2);
a[n_]:=n+Floor[n*s/r]+Floor[n*t/r];
b[n_]:=n+Floor[n*r/s]+Floor[n*t/s];
c[n_]:=n+Floor[n*r/t]+Floor[n*s/t]
Table[a[n], {n, 1, 120}]  (* A184812 *)
Table[b[n], {n, 1, 120}]  (* A184813 *)
Table[c[n], {n, 1, 120}]  (* A184814 *)

Prog

(PARI) sr=sqrt(3/2); tr=sqrt(5/2); for(n=1, 100, print1(n+floor(n*sr)+floor(n*tr)", ")) \\ Charles R Greathouse IV, Jul 15 2011
(Maxima) r:sqrt(2)$ s:sqrt(3)$  t:sqrt(5)$
makelist(n+floor(n*s/r)+floor(n*t/r), n, 1, 50); /* Martin Ettl, Oct 18 2012 */

Crossrefs

Cf. A184813, A184814. Associated partition of the primes: A184815, A184816, A184817.

Sequence in context: A189460 A172323 A190080 * A140487 A310189 A293788

Adjacent sequences: A184809 A184810 A184811 * A184813 A184814 A184815

Keyword

nonn

Author

Clark Kimberling, Jan 22 2011

Status

approved