A179207 a(n) = n - 1 + ceiling((-3 + n^2)/2) if n > 1 with a(1)=1, complement of A182835.

1, 2, 5, 10, 15, 22, 29, 38, 47, 58, 69, 82, 95, 110, 125, 142, 159, 178, 197, 218, 239, 262, 285, 310, 335, 362, 389, 418, 447, 478, 509, 542, 575, 610, 645, 682, 719, 758, 797, 838, 879, 922, 965, 1010, 1055, 1102, 1149, 1198, 1247, 1298, 1349, 1402, 1455

Offset

1,2

Links

Guenther Schrack, Table of n, a(n) for n = 1..10010

Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Formula

a(n) = n - 1 + ceiling((-3 + n^2)/2) if n > 1.

a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - Joerg Arndt, Apr 02 2011

From Guenther Schrack, Jun 06 2018: (Start)

a(n) = (2*n^2 + 4*n - 9 + (-1)^n)/4 for n > 1.

a(n) = a(n-2) + 2*n for n > 3.

a(-n) = a(n-2) for n > 1.

a(n) = n - 1 + A047838(n) for n > 1. (End)

G.f.: x * (1 + x^2 + 2*x^3 - 2*x^4) / (1 - 2*x + 2*x^3 - x^4). - Michael Somos, Oct 28 2018

Sum_{n>=1} 1/a(n) = 8/3 + tan(sqrt(5)*Pi/2)*Pi/(2*sqrt(5)) - cot(sqrt(3/2)*Pi)*Pi/(2*sqrt(6)). - Amiram Eldar, Sep 16 2022

Maple

a:=n->n-1+ceil((-3+n^2)/2): 1, seq(a(n), n=2..60); # Muniru A Asiru, Aug 05 2018

Mathematica

Table[n-1+Ceiling[(n*n-3)/2], {n, 60}] (* Vladimir Joseph Stephan Orlovsky, Apr 02 2011 *)
Join[{1}, LinearRecurrence[{2, 0, -2, 1}, {2, 5, 10, 15}, 52]] (* Ray Chandler, Jul 15 2015 *)

Prog

(GAP) a:=[2, 5, 10, 15];; for n in [5..60] do a[n]:=2*a[n-1]-2*a[n-3]+a[n-4]; od; a:=Concatenation([1], a); # Muniru A Asiru, Aug 05 2018

Crossrefs

Cf. A182835, A047838.

First differences: A109613(n) for n > 2. - Guenther Schrack, Jun 06 2018

Sequence in context: A024390 A125622 A080551 * A008822 A267454 A013927

Adjacent sequences: A179204 A179205 A179206 * A179208 A179209 A179210

Keyword

nonn,easy

Author

Clark Kimberling, Jan 07 2011

Status

approved