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A184812 n+floor(ns/r)+floor(nt/r), where r=sqrt(2), s=sqrt(3), t=sqrt(5). 37
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 (list; graph; refs; listen; history; text; internal format)
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
LINKS
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
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jan 22 2011
STATUS
approved

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Last modified July 15 21:59 EDT 2024. Contains 374334 sequences. (Running on oeis4.)