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

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

Author

Clark Kimberling, Jan 19 2011

Keywords

• nonn
• easy

Comments

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

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

Links

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

Crossrefs

Cf. A184675, A184676, A014616.

Programs

Magma

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

Maple

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

Mathematica

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 *)

Formula

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