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A105073 Define a(1)=0, a(2)=2 then a(n) = 3*a(n-1) - a(n-2), a(n+1) = 3*a(n)-a(n-1) and a(n+2) = 3*a(n+1) - a(n) + 2.

%I A105073 #36 Sep 08 2022 08:45:17
%S A105073 0,2,6,16,44,116,304,798,2090,5472,14328,37512,98208,257114,673134,
%T A105073 1762288,4613732,12078908,31622992,82790070,216747218,567451584,
%U A105073 1485607536,3889371024,10182505536,26658145586,69791931222,182717648080,478361013020,1252365390980
%N A105073 Define a(1)=0, a(2)=2 then a(n) = 3*a(n-1) - a(n-2), a(n+1) = 3*a(n)-a(n-1) and a(n+2) = 3*a(n+1) - a(n) + 2.
%C A105073 From _Jon E. Schoenfield_, Jan 18 2019: (Start)
%C A105073 Previously, the Name had included the comment, "This sequence is such that 20*(a(n)^2) + 20*a(n) + 1 = j^2 = a square."
%C A105073 However, _Anthony Hernandez_ observed that this statement is not true for all terms; e.g., at a(4)=16, 20*16^2 + 20*16 + 1 = 5441, a nonsquare.
%C A105073 It is true that 20*a(n)^2 + 20*a(n) + 1 = A305315(n/3)^2 when n == 0 (mod 3) and A305316((n-2)/3)^2 when n == 2 (mod 3); however, for n == 1 (mod 3) with n > 1, sqrt(20*a(n)^2 + 20*a(n) + 1) is a noninteger number whose fractional part apparently approaches 3 - sqrt(5) as n increases, and _Andrey Zabolotskiy_ observes that round(sqrt(20*a(n)^2 + 20*a(n) + 1) + sqrt(5)) appears to be equal to A002878(n). (End)
%H A105073 Vincenzo Librandi, <a href="/A105073/b105073.txt">Table of n, a(n) for n = 1..1000</a>
%H A105073 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3,-1,1,-3,1).
%F A105073 a(n) = (1/6)*(Fibonacci(2n+4) - 2*Fibonacci(2n) - 2*cos((n+2)(2*Pi/3)) - 4). - _Ralf Stephan_, May 20 2007
%F A105073 From _R. J. Mathar_, Nov 13 2009: (Start)
%F A105073 a(n) = 3*a(n-1) - a(n-2) + a(n-3) - 3*a(n-4) + a(n-5).
%F A105073 G.f.: 2*x^2/((1-x) * (1+x+x^2) * (1-3*x+x^2)).
%F A105073 a(n) = A061347(n+2)/6 + A001519(n+2)/2 - 2/3. (End)
%F A105073 a(n) = floor(A027941(n)/2). - _Anthony Hernandez_, Jan 03 2019
%t A105073 a[n_]:=(1/6)*(Fibonacci[2*n+4] - 2*Fibonacci[2*n] - 2*Cos[(n+2)*(2*Pi/3)] - 4 ); Array[a,50] (* _Stefano Spezia_, Jan 11 2019 *)
%t A105073 RecurrenceTable[{a[1]==0, a[2]==2, a[3]==6, a[4]==16, a[5]==44, a[n]== 3 a[n-1] - a[n-2] + a[n-3] - 3 a[n-4] + a[n-5]}, a, {n, 35}] (* _Vincenzo Librandi_, Jan 13 2019 *)
%o A105073 (Magma) I:=[0,2,6,16,44]; [n le 5 select I[n] else 3*Self(n-1) - Self(n-2) + Self(n-3) - 3*Self(n-4) + Self(n-5): n in [1..35]]; // _Vincenzo Librandi_, Jan 13 2019
%Y A105073 Cf. A001519, A061347, A027941.
%Y A105073 Cf. A305315, A305316.
%K A105073 nonn,easy
%O A105073 1,2
%A A105073 _Pierre CAMI_, Apr 06 2005
%E A105073 Extended by _R. J. Mathar_, Nov 13 2009