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A137928
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The even principal diagonal of a 2n X 2n square spiral.
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14
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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
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OFFSET
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1,1
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COMMENTS
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This is concerned with 2n X 2n square spirals of the form illustrated in the Example section.
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LINKS
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Table of n, a(n) for n=1..53.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
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FORMULA
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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
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EXAMPLE
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Example with n = 2:
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7---8---9--10
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6 1---2 11
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5---4---3 12
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16--15--14--13
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a(1) = 2(1) + 4*floor((1-1)/4) = 2;
a(2) = 2(2) + 4*floor((2-1)/4) = 4.
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MAPLE
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A137928:=n->2*ceil(n^2/2): seq(A137928(n), n=1..100); # Wesley Ivan Hurt, Jul 25 2017
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MATHEMATICA
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LinearRecurrence[{2, 0, -2, 1}, {2, 4, 10, 16}, 60] (* Harvey P. Dale, Aug 28 2017 *)
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PROG
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(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
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CROSSREFS
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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
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KEYWORD
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nonn,easy
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AUTHOR
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William A. Tedeschi, Feb 29 2008
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STATUS
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approved
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