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 A016814 a(n) = (4n+1)^2. 12
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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 LINKS Ivan Panchenko, Table of n, a(n) for n = 0..200 Index entries for linear recurrences with constant coefficients, signature (3,-3,1). FORMULA 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 EXAMPLE a(5) = (4 * 5 + 1)^2 = 21^2 = 441. a(6) = (4 * 6 + 1)^2 = 25^2 = 625. MAPLE A016814:=n->(4*n+1)^2; seq(A016814(k), k=0..100); # Wesley Ivan Hurt, Nov 02 2013 MATHEMATICA (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 *) PROG (PARI) a(n)=(4*n+1)^2 \\ Charles R Greathouse IV, Oct 07 2015 CROSSREFS Cf. A000290, A001539, A016742, A016754, A016802, A016826, A016838. Sequence in context: A134153 A251311 A223181 * A280343 A204708 A174623 Adjacent sequences:  A016811 A016812 A016813 * A016815 A016816 A016817 KEYWORD nonn,easy AUTHOR STATUS approved

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Last modified November 27 23:47 EST 2020. Contains 338685 sequences. (Running on oeis4.)