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 A157446 a(n) = 16*n^2 - n. 4
 15, 62, 141, 252, 395, 570, 777, 1016, 1287, 1590, 1925, 2292, 2691, 3122, 3585, 4080, 4607, 5166, 5757, 6380, 7035, 7722, 8441, 9192, 9975, 10790, 11637, 12516, 13427, 14370, 15345, 16352, 17391, 18462, 19565, 20700, 21867, 23066, 24297, 25560 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS The identity (2048*n^2 - 128*n + 1)^2 - (16*n^2 - n)*(512*n - 16)^2 = 1 can be written as A157448(n)^2 - a(n)*A157447(n)^2 = 1. - Vincenzo Librandi, Jan 26 2012 This is the case s=4 of the identity (8*n^2*s^4 - 8*n*s^2 + 1)^2 - (n^2*s^2 - n)*(8*n*s^3 - 4*s)^2 = 1. - Bruno Berselli, Jan 26 2012 Sequence found by reading the line from 15, in the direction 15, 62, ... in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Nov 02 2012 The continued fraction expansion of sqrt(a(n)) is [4n-1; {1, 6, 1, 8n-2}]. For n=1, this collapses to [3; {1, 6}]. - Magus K. Chu, Sep 22 2022 LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..10000 Vincenzo Librandi, X^2-AY^2=1 Index entries for linear recurrences with constant coefficients, signature (3,-3,1). FORMULA G.f.: x*(15 + 17*x)/(1-x)^3. - Vincenzo Librandi, Jan 26 2012 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jan 26 2012 MATHEMATICA LinearRecurrence[{3, -3, 1}, {15, 62, 141}, 40] (* Vincenzo Librandi, Jan 26 2012 *) PROG (Magma) I:=[15, 62, 141]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jan 26 2012 (PARI) for(n=1, 22, print1(16*n^2 - n", ")); \\ Vincenzo Librandi, Jan 26 2012 CROSSREFS Cf. A157447, A157448. Sequence in context: A072201 A218811 A219819 * A220084 A240711 A212055 Adjacent sequences: A157443 A157444 A157445 * A157447 A157448 A157449 KEYWORD nonn,easy AUTHOR Vincenzo Librandi, Mar 01 2009 STATUS approved

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Last modified December 3 23:46 EST 2022. Contains 358544 sequences. (Running on oeis4.)