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A016814
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a(n) = (4n+1)^2.
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12
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1, 25, 81, 169, 289, 441, 625, 841, 1089, 1369, 1681, 2025, 2401, 2809, 3249, 3721, 4225, 4761, 5329, 5929, 6561, 7225, 7921, 8649, 9409, 10201, 11025, 11881, 12769, 13689, 14641, 15625, 16641, 17689, 18769, 19881, 21025, 22201, 23409, 24649, 25921, 27225, 28561, 29929
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OFFSET
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0,2
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COMMENTS
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A bisection of A016754. Sequence arises from reading the line from 1, in the direction 1, 25, ..., in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008
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LINKS
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Ivan Panchenko, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
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FORMULA
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a(n) = a(n-1) + 32 * n - 8, n > 0. - Vincenzo Librandi, Dec 15 2010
From George F. Johnson, Sep 28 2012: (Start)
G.f.: (1 + 22*x + 9*x^2)/(1 - x)^3.
a(n+1) = a(n) + 16 + 8*sqrt(a(n)).
a(n+1) = 2*a(n) - a(n-1) + 32 = 3*a(n) - 3*a(n-1) + a(n-2).
a(n-1)*a(n+1) = (a(n) - 16)^2 ; a(n+1) - a(n-1) = 16*sqrt(a(n)).
a(n) = A016754(2*n) = (A016813(n))^2.
(End)
a(0) = 1, a(n) = a(n - 1) + 32(n - 1) - 8. - Alonso del Arte, Aug 19 2017
Sum_{n>=0} 1/a(n) = G/2 + Pi^2/16, where G is the Catalan constant (A006752). - Amiram Eldar, Jun 28 2020
Product_{n>=1} (1 - 1/a(n)) = 2*Gamma(5/4)^2/sqrt(Pi) = 2 * A068467^2 * A087197. - Amiram Eldar, Feb 01 2021
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EXAMPLE
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a(5) = (4 * 5 + 1)^2 = 21^2 = 441.
a(6) = (4 * 6 + 1)^2 = 25^2 = 625.
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MAPLE
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A016814:=n->(4*n+1)^2; seq(A016814(k), k=0..100); # Wesley Ivan Hurt, Nov 02 2013
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MATHEMATICA
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(4 * Range[0, 40] + 1)^2 (* or *) LinearRecurrence[{3, -3, 1}, {1, 25, 81}, 40] (* Harvey P. Dale, Nov 20 2012 *)
Accumulate[32Range[0, 47] - 8] + 9 (* Alonso del Arte, Aug 19 2017 *)
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PROG
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(PARI) a(n)=(4*n+1)^2 \\ Charles R Greathouse IV, Oct 07 2015
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CROSSREFS
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Cf. A000290, A001539, A016742, A016754, A016802, A016826, A016838, A068467, A087197.
Sequence in context: A134153 A251311 A223181 * A280343 A204708 A174623
Adjacent sequences: A016811 A016812 A016813 * A016815 A016816 A016817
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KEYWORD
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nonn,easy
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AUTHOR
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N. J. A. Sloane
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STATUS
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approved
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