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A158555 a(n) = 196*n^2 + 14.

%I A158555 #29 Jan 15 2025 17:48:12
%S A158555 14,210,798,1778,3150,4914,7070,9618,12558,15890,19614,23730,28238,
%T A158555 33138,38430,44114,50190,56658,63518,70770,78414,86450,94878,103698,
%U A158555 112910,122514,132510,142898,153678,164850,176414,188370,200718,213458,226590,240114,254030
%N A158555 a(n) = 196*n^2 + 14.
%C A158555 The identity (28*n^2 + 1)^2 -(196*n^2 + 14)*(2*n)^2 = 1 can be written as A158556(n)^2 - a(n)*A005843(n)^2 = 1.
%H A158555 Vincenzo Librandi, <a href="/A158555/b158555.txt">Table of n, a(n) for n = 0..10000</a>
%H A158555 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A158555 G.f.: 14*(1 + 12*x + 15*x^2)/(1-x)^3.
%F A158555 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
%F A158555 From _Amiram Eldar_, Mar 09 2023: (Start)
%F A158555 Sum_{n>=0} 1/a(n) = (coth(Pi/sqrt(14))*Pi/sqrt(14) + 1)/28.
%F A158555 Sum_{n>=0} (-1)^n/a(n) = (cosech(Pi/sqrt(14))*Pi/sqrt(14) + 1)/28. (End)
%F A158555 E.g.f.: 14*exp(x)*(1 + 14*x + 14*x^2). - _Elmo R. Oliveira_, Jan 15 2025
%t A158555 LinearRecurrence[{3, -3, 1}, {14, 210, 798}, 50] (* _Vincenzo Librandi_, Feb 05 2012 *)
%o A158555 (Magma) I:=[14, 210, 798]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]]; // _Vincenzo Librandi_, Feb 14 2012
%o A158555 (PARI) for(n=0, 40, print1(196*n^2 + 14", ")); \\ _Vincenzo Librandi_, Feb 14 2012
%Y A158555 Cf. A005843, A158556.
%K A158555 nonn,easy
%O A158555 0,1
%A A158555 _Vincenzo Librandi_, Mar 21 2009
%E A158555 Comment rewritten, a(0) added by _R. J. Mathar_, Oct 16 2009