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

1,1

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

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}]

Prog

(PARI) a(n) = n+floor(2*sqrt(n)); \\ Michel Marcus, Dec 08 2015
(Magma) [n+Floor(2*Sqrt(n)): n in [1..100]]; // Vincenzo Librandi, Dec 09 2015

Crossrefs

Cf. A179272.

Sequence in context: A263098 A317391 A134745 * A182829 A112800 A294456

Adjacent sequences: A183864 A183865 A183866 * A183868 A183869 A183870

Keyword

nonn,easy

Author

Clark Kimberling, Jan 07 2011

Status

approved