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

1,2

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).

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

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 *)

Prog

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

Crossrefs

Cf. A184675, A184676, A014616.

Sequence in context: A088236 A194244 A014616 * A227353 A183136 A144873

Adjacent sequences: A184671 A184672 A184673 * A184675 A184676 A184677

Keyword

nonn,easy

Author

Clark Kimberling, Jan 19 2011

Status

approved