A137928 The even principal diagonal of a 2n X 2n square spiral.

2, 4, 10, 16, 26, 36, 50, 64, 82, 100, 122, 144, 170, 196, 226, 256, 290, 324, 362, 400, 442, 484, 530, 576, 626, 676, 730, 784, 842, 900, 962, 1024, 1090, 1156, 1226, 1296, 1370, 1444, 1522, 1600, 1682, 1764, 1850, 1936, 2026, 2116, 2210, 2304, 2402, 2500, 2602, 2704, 2810

Offset

1,1

Comments

This is concerned with 2n X 2n square spirals of the form illustrated in the Example section.

Links

Table of n, a(n) for n=1..53.

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

Formula

a(n) = 2*n + 4*floor((n-1)^2/4) = 2*n + 4*A002620(n-1).

a(n) = A171218(n) - A171218(n-1). - Reinhard Zumkeller, Dec 05 2009

From R. J. Mathar, Jun 27 2011: (Start)

G.f.: 2*x*(1 + x^2) / ( (1 + x)*(1 - x)^3 ).

a(n) = 2*A000982(n). (End)

a(n+1) = (3 + 4*n + 2*n^2 + (-1)^n)/2 = A080335(n) + (-1)^n. - Philippe Deléham, Feb 17 2012

a(n) = 2 * ceiling(n^2/2). - Wesley Ivan Hurt, Jun 15 2013

a(n) = n^2 + (n mod 2). - Bruno Berselli, Oct 03 2017

Sum_{n>=1} 1/a(n) = Pi*tanh(Pi/2)/4 + Pi^2/24. - Amiram Eldar, Jul 07 2022

Example

Example with n = 2:
.
   7---8---9--10
   |           |
   6   1---2  11
   |       |   |
   5---4---3  12
               |
  16--15--14--13
.
a(1) = 2(1) + 4*floor((1-1)/4) = 2;
a(2) = 2(2) + 4*floor((2-1)/4) = 4.

Maple

A137928:=n->2*ceil(n^2/2): seq(A137928(n), n=1..100); # Wesley Ivan Hurt, Jul 25 2017

Mathematica

LinearRecurrence[{2, 0, -2, 1}, {2, 4, 10, 16}, 60] (* Harvey P. Dale, Aug 28 2017 *)

Prog

(Python) a = lambda n: 2*n + 4*floor((n-1)**2/4)
(PARI) a(n)=2*n+(n-1)^2\4*4 \\ Charles R Greathouse IV, May 21 2015

Crossrefs

Cf. A000982, A002061 (odd diagonal), A002620, A080335, A171218.

Sequence in context: A189558 A111149 A123689 * A293154 A144834 A006584

Adjacent sequences: A137925 A137926 A137927 * A137929 A137930 A137931

Keyword

nonn,easy

Author

William A. Tedeschi, Feb 29 2008

Status

approved