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A171640 a(n) = 10*a(n-1)-a(n-2)-4 with a(1)=1 and a(2)=3.

%I A171640 #27 Jan 21 2023 02:47:59
%S A171640 1,3,25,243,2401,23763,235225,2328483,23049601,228167523,2258625625,
%T A171640 22358088723,221322261601,2190864527283,21687323011225,
%U A171640 214682365584963,2125136332838401,21036680962799043,208241673295152025,2061380051988721203,20405558846592060001
%N A171640 a(n) = 10*a(n-1)-a(n-2)-4 with a(1)=1 and a(2)=3.
%H A171640 Colin Barker, <a href="/A171640/b171640.txt">Table of n, a(n) for n = 1..1000</a>
%H A171640 Giovanni Lucca, <a href="http://forumgeom.fau.edu/FG2018volume18/FG201808index.html">Integer Sequences and Circle Chains Inside a Circular Segment</a>, Forum Geometricorum, Vol. 18 (2018), 47-55.
%H A171640 Giovanni Lucca, <a href="https://ijgeometry.com/product/giovanni-lucca-circle-chains-inside-the-arbelos-and-integer-sequences/">Circle chains inside the arbelos and integer sequences</a>, Int'l J. Geom. (2023) Vol. 12, No. 1, 71-82.
%H A171640 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (11,-11,1).
%F A171640 a(n) = A132596(n-1)+1.
%F A171640 2*a(n) + 2*A132596(n-1) = A087799(n-1).
%F A171640 a(n+1) = [(sqrt(3)-sqrt(2))^n +(sqrt(3)+ sqrt(2))^n]^2 / 4.
%F A171640 From _Colin Barker_, Oct 02 2015: (Start)
%F A171640 a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) for n>3.
%F A171640 G.f.: -x*(3*x^2-8*x+1) / ((x-1)*(x^2-10*x+1)).
%F A171640 (End)
%F A171640 6*a(n)*(a(n)-1) = [A122653(n-1)]^2. - _Jean-Luc Manguin_, Jun 02 2020
%e A171640 a(2+1) = [5-sqrt(24)+5+sqrt(24)]^2/4 = 100/4 = 25.
%t A171640 RecurrenceTable[{a[n] == 10 a[n - 1] - a[n - 2] - 4, a[1] == 1, a[2] == 3}, a, {n, 1, 21}] (* _Michael De Vlieger_, Oct 02 2015 *)
%t A171640 LinearRecurrence[{11,-11,1},{1,3,25},30] (* _Harvey P. Dale_, May 05 2018 *)
%o A171640 (PARI) Vec(-x*(3*x^2-8*x+1)/((x-1)*(x^2-10*x+1)) + O(x^30)) \\ _Colin Barker_, Oct 02 2015
%Y A171640 Cf. A132596, A087799, A001263.
%K A171640 nonn,easy
%O A171640 1,2
%A A171640 _Mark Dols_, Dec 13 2009