A184676 -id:A184676 - cp's OEIS frontend

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

Clark Kimberling, Jan 07 2011

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

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

Cf. A179272.

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

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

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

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

A184674 a(n) = n+floor((n/2-1/(2*n))^2); complement of A184675.

Original entry on oeis.org

1, 2, 4, 7, 10, 14, 18, 23, 28, 34, 40, 47, 54, 62, 70, 79, 88, 98, 108, 119, 130, 142, 154, 167, 180, 194, 208, 223, 238, 254, 270, 287, 304, 322, 340, 359, 378, 398, 418, 439, 460, 482, 504, 527, 550, 574, 598, 623, 648, 674, 700, 727, 754, 782, 810, 839, 868, 898, 928, 959, 990, 1022, 1054, 1087, 1120, 1154, 1188, 1223, 1258, 1294, 1330, 1367, 1404

Offset: 1

Clark Kimberling, Jan 19 2011

Conjecture: a(n) = A014616(n-1). - R. J. Mathar, Jan 29 2011

The above conjecture is true. - Stefano Spezia, Apr 04 2023

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A184675, A184676, A014616.

[n+Floor((n/2-1/(2*n))^2): n in [1..80]]; // Vincenzo Librandi, Jul 10 2011

A184674:=n->n+floor((n/2-1/(2*n))^2): seq(A184674(n), n=1..100); # Wesley Ivan Hurt, Feb 22 2017

a[n_]:=n+Floor[(n/2-1/(2n))^2];
b[n_]:=n+Floor[n^(1/2)+(n+1)^(1/2)];
Table[a[n],{n,1,120}]   (* A184674 *)
Table[b[n],{n,1,120}]   (* A184675 *)
FindLinearRecurrence[Table[a[n],{n,1,120}]]
Join[{1},LinearRecurrence[{2,0,-2,1},{2,4,7,10},72]] (* Ray Chandler, Aug 02 2015 *)

a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n>=6.
G.f.: x*(x^4 - x^3 - 1)/((x + 1)*(x - 1)^3). - Álvar Ibeas, Jul 20 2021
a(n) = (2*n^2 + 8*n - 9 + (-1)^n)/8 for n > 1. - Stefano Spezia, Apr 04 2023

cp's OEIS Frontend

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

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