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A016826 a(n) = (4n + 2)^2. 10
4, 36, 100, 196, 324, 484, 676, 900, 1156, 1444, 1764, 2116, 2500, 2916, 3364, 3844, 4356, 4900, 5476, 6084, 6724, 7396, 8100, 8836, 9604, 10404, 11236, 12100, 12996, 13924, 14884, 15876, 16900, 17956 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

A bisection of A016742. Sequence arises from reading the line from 4, in the direction 4, 36, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008

LINKS

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

FORMULA

a(n) = a(n-1) + 32*n (with a(0)=4). - Vincenzo Librandi, Dec 15 2010

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(0)=4, a(1)=36, a(2)=100. - Harvey P. Dale, Nov 24 2011

G.f.: -((4*(x^2+6*x+1))/(x-1)^3). - Harvey P. Dale, Nov 24 2011

a(n) = A000290(A016825(n)). - Wesley Ivan Hurt, Feb 24 2014

MAPLE

A016826:=n->(4*n + 2)^2; seq(A016826(n), n=0..40); # Wesley Ivan Hurt, Feb 24 2014

MATHEMATICA

(4*Range[0, 40]+2)^2 (* or *) LinearRecurrence[{3, -3, 1}, {4, 36, 100}, 40] (* Harvey P. Dale, Nov 24 2011 *)

PROG

(PARI) a(n)=(4*n+2)^2 \\ Charles R Greathouse IV, Oct 07 2015

CROSSREFS

Equals A001539 + 1.

Cf. A000290, A016742, A016754, A016802, A016814, A016838.

Sequence in context: A193833 A193183 A152760 * A190318 A193874 A254939

Adjacent sequences:  A016823 A016824 A016825 * A016827 A016828 A016829

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane

STATUS

approved

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Last modified February 24 14:41 EST 2018. Contains 299623 sequences. (Running on oeis4.)