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a(n) = 49*n^2 + 2*n.
4

%I #36 Sep 17 2022 14:14:39

%S 51,200,447,792,1235,1776,2415,3152,3987,4920,5951,7080,8307,9632,

%T 11055,12576,14195,15912,17727,19640,21651,23760,25967,28272,30675,

%U 33176,35775,38472,41267,44160,47151,50240,53427,56712,60095,63576,67155

%N a(n) = 49*n^2 + 2*n.

%C The identity (4802*n^2 + 196*n + 1)^2 - (49*n^2 + 2*n)*(686*n + 14)^2 = 1 can be written as A157367(n)^2 - a(n)*A157366(n)^2 = 1.

%C This formula is the case s=7 of the identity (2*s^4*n^2 + 4*s^2*n + 1)^2 - (s^2*n^2 + 2*n)*(2*s^3*n + 2*s)^2 = 1. - _Bruno Berselli_, Feb 11 2012

%C Also, the identity (49*n + 1)^2 - (49*n^2 + 2*n)*7^2 = 1 can be written as A158066(n)^2 - a(n)*7^2 = 1 (see Barbeau's paper in link). - _Vincenzo Librandi_, Feb 11 2012

%C The continued fraction expansion of sqrt(4*a(n)) is [14n; {3, 1, 1, 7n-1, 1, 1, 3, 28n}]. - _Magus K. Chu_, Sep 17 2022

%H Vincenzo Librandi, <a href="/A157365/b157365.txt">Table of n, a(n) for n = 1..10000</a>

%H E. J. Barbeau, <a href="http://www.math.toronto.edu/barbeau/home.html">Polynomial Excursions</a>, Chapter 10: <a href="http://www.math.toronto.edu/barbeau/hxpol10.pdf">Diophantine equations</a> (2010), pages 84-85 (row 15 in the first table at p. 85, case d(t) = t*(7^2*t+2)).

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F G.f.: x*(51+47*x)/(1-x)^3.

%F a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).

%F E.g.f.: (51*x + 49*x^2)*exp(x). - _G. C. Greubel_, Feb 02 2018

%t LinearRecurrence[{3,-3,1},{51,200,447},50]

%o (Magma) I:=[51, 200, 447]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];

%o (PARI) a(n)=49*n^2+2*n \\ _Charles R Greathouse IV_, Dec 23 2011

%Y Cf. A157366, A157367, A158066.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Feb 28 2009