A330172 a(n) = n + floor(nr/t) + floor(ns/t), where r = sqrt(2) - 1, s = sqrt(2), t = sqrt(2) + 1.

1, 3, 4, 6, 7, 10, 12, 13, 15, 16, 18, 21, 22, 24, 25, 27, 28, 31, 33, 34, 36, 37, 39, 42, 43, 45, 46, 48, 49, 52, 54, 55, 57, 58, 61, 63, 64, 66, 67, 69, 72, 73, 75, 76, 78, 79, 82, 84, 85, 87, 88, 90, 93, 94, 96, 97, 99, 100, 103, 105, 106, 108, 109, 111

Offset

1,2

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) - 1, s = sqrt(2), t = sqrt(2) + 1 yields

a=A330171, b=A016789, c=A330172.

Links

Table of n, a(n) for n=1..64.

Formula

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

Mathematica

r = Sqrt[2] - 1; s = Sqrt[2]; t = Sqrt[2] + 1;
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}]  (* A330171 *)
Table[b[n], {n, 1, 120}]  (* A016789 *)
Table[c[n], {n, 1, 120}]  (* A330172 *)
Table[n+Floor[(n(Sqrt[2]-1))/(Sqrt[2]+1)]+Floor[(n Sqrt[2])/(Sqrt[2]+1)], {n, 70}] (* Harvey P. Dale, Feb 10 2023 *)

Crossrefs

Cf. A016789, A330171.

Sequence in context: A101299 A143837 A354005 * A206903 A157611 A147609

Adjacent sequences: A330169 A330170 A330171 * A330173 A330174 A330175

Keyword

nonn,easy

Author

Clark Kimberling, Jan 04 2020

Status

approved