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A158485 a(n) = 14*n^2 - 1.

%I A158485 #32 Sep 08 2022 08:45:43
%S A158485 13,55,125,223,349,503,685,895,1133,1399,1693,2015,2365,2743,3149,
%T A158485 3583,4045,4535,5053,5599,6173,6775,7405,8063,8749,9463,10205,10975,
%U A158485 11773,12599,13453,14335,15245,16183,17149,18143,19165,20215,21293,22399
%N A158485 a(n) = 14*n^2 - 1.
%C A158485 The identity (14*n^2-1)^2-(49*n^2-7)*(2*n)^2=1 can be written as a(n)^2-A158484(n)*A005843(n)^2=1.
%C A158485 Sequence found by reading the line from 13, in the direction 13, 55,..., in the square spiral whose vertices are the generalized enneagonal 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 A158485 Vincenzo Librandi, <a href="/A158485/b158485.txt">Table of n, a(n) for n = 1..10000</a>
%H A158485 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158485 a(n) = 3*a(n-1) -3*a(n-2) +a(n-3).
%F A158485 G.f: x*(-13-16*x+x^2)/(x-1)^3.
%F A158485 From _Amiram Eldar_, Feb 04 2021: (Start)
%F A158485 Sum_{n>=1} 1/a(n) = (1 - (Pi/sqrt(14))*cot(Pi/sqrt(14)))/2.
%F A158485 Sum_{n>=1} (-1)^(n+1)/a(n) = ((Pi/sqrt(14))*csc(Pi/sqrt(14)) - 1)/2.
%F A158485 Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(14))*csc(Pi/sqrt(14)).
%F A158485 Product_{n>=1} (1 - 1/a(n)) = csc(Pi/sqrt(14))*sin(Pi/sqrt(7))/sqrt(2). (End)
%t A158485 LinearRecurrence[{3,-3,1},{13,55,125},50]
%o A158485 (Magma) I:=[13, 55, 125]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]];
%o A158485 (PARI) a(n) = 14*n^2-1
%Y A158485 Cf. A005843, A158484.
%K A158485 nonn,easy
%O A158485 1,1
%A A158485 _Vincenzo Librandi_, Mar 20 2009