A183867 a(n) = n + floor(2*sqrt(n)); complement of A184676.

3, 4, 6, 8, 9, 10, 12, 13, 15, 16, 17, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 31, 32, 33, 35, 36, 37, 38, 39, 40, 42, 43, 44, 45, 46, 48, 49, 50, 51, 52, 53, 54, 56, 57, 58, 59, 60, 61, 63, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 77, 78, 80, 81, 82, 83

Offset: 1

Author

Clark Kimberling, Jan 07 2011

Keywords

• nonn
• easy

Comments

Also equals n + floor(sqrt(n) + sqrt(n+1/2)). Proof: floor(2*sqrt(n)) is the largest k such that k^2/4 <= n, while floor(sqrt(n) + sqrt(n+1/2)) is the largest k such that (k^2 - 1)/4 + 1/(16*k^2) <= n. All perfect squares are 0 or 1 (mod 4). In either case, it is easily verified that one of the inequalities is satisfied if and only if the other inequality is satisfied. - Nathaniel Johnston, Jun 26 2011

Links

• Nathaniel Johnston, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A179272.

Programs

Magma

[n+Floor(2*Sqrt(n)): n in [1..100]]; // Vincenzo Librandi, Dec 09 2015

Maple

seq(n+floor(2*sqrt(n)), n=1..67); # Nathaniel Johnston, Jun 26 2011

Mathematica

a=4; b=0;
Table[n+Floor[(a*n+b)^(1/2)],{n,100}]
Table[n-1+Ceiling[(n*n-b)/a],{n,70}]

PARI

a(n) =  n+sqrtint(4*n); \\ Michel Marcus, Dec 08 2015, Jul 28 2025