%I
%S 13,55,125,223,349,503,685,895,1133,1399,1693,2015,2365,2743,3149,
%T 3583,4045,4535,5053,5599,6173,6775,7405,8063,8749,9463,10205,10975,
%U 11773,12599,13453,14335,15245,16183,17149,18143,19165,20215,21293,22399
%N 14n^2  1.
%C The identity (14*n^21)^2(49*n^27)*(2*n)^2=1 can be written as a(n)^2A158484(n)*A005843(n)^2=1.
%C Sequence found by reading the line from 13, in the direction 13, 55,..., in the square spiral whose vertices are the generalized enneagonal numbers numbers A118277. Also sequence found by reading the same line in the square spiral whose edges have length A195019 and whose vertices are the numbers A195020.  Omar E. Pol, Sep 13 2011
%H Vincenzo Librandi, <a href="/A158485/b158485.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,3,1).
%F a(n) = 3*a(n1) 3*a(n2) +a(n3).
%F G.f: x*(1316*x+x^2)/(x1)^3.
%t LinearRecurrence[{3,3,1},{13,55,125},50]
%o (MAGMA) I:=[13, 55, 125]; [n le 3 select I[n] else 3*Self(n1)3*Self(n2)+1*Self(n3): n in [1..40]];
%o (PARI) a(n) = 14*n^21
%Y Cf. A005843, A158484.
%K nonn,easy
%O 1,1
%A _Vincenzo Librandi_, Mar 20 2009
