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A266505 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = -1, a(1) = 1, a(2) = 3, a(3) = 5.

%I A266505 #44 Mar 05 2025 06:53:43
%S A266505 -1,1,3,5,5,11,13,27,31,65,75,157,181,379,437,915,1055,2209,2547,5333,
%T A266505 6149,12875,14845,31083,35839,75041,86523,181165,208885,437371,504293,
%U A266505 1055907,1217471,2549185,2939235,6154277,7095941,14857739,17131117,35869755,41358175,86597249,99847467
%N A266505 a(n) = 2*a(n - 2) + a(n - 4) with a(0) = -1, a(1) = 1, a(2) = 3, a(3) = 5.
%C A266505 a(n)/A266504(n) converges to sqrt(2).
%C A266505 Alternatively, bisection of A266506.
%C A266505 Alternatively, A135532(n) and A048655(n) interlaced.
%C A266505 Alternatively, A255236(n-1), A054490(n), A038762(n) and A101386(n) interlaced.
%C A266505 Let b(n) = (a(n) - (a(n) mod 2))/2, that is b(n) = {-1, 0, 1, 2, 2, 5, 6, 13, 15, 32, 37, 78, 90, ...}. Then:
%C A266505 A006451(n) = {b(4n+0) U b(4n+1)} gives n in N such that triangular(n) + 1 is square;
%C A266505 A216134(n) = {b(4n+2) U b(4n+3)} gives n in N such that triangular(n) follows form n^2 + n + 1 (twice a triangular number + 1).
%H A266505 G. C. Greubel, <a href="/A266505/b266505.txt">Table of n, a(n) for n = 0..1000</a>
%H A266505 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,0,1).
%F A266505 G.f.: (-1 + 3*x)*(1 + x)^2/(1 - 2*x^2 - x^4).
%F A266505 a(n) = (-(1+sqrt(2))^floor(n/2)*(-1)^n - sqrt(8)*(1-sqrt(2))^floor(n/2) - (1-sqrt(2))^floor(n/2)*(-1)^n + sqrt(8)*(1+sqrt(2))^floor(n/2))/2.
%F A266505 a(n) = 3*(((1+sqrt(2))^floor(n/2)-(1-sqrt(2))^floor(n/2))/sqrt(8)) - (-1)^n*(((1+sqrt(2))^(floor(n/2)-(-1)^n)-(1-sqrt(2))^(floor(n/2)-(-1)^n))/sqrt(8)).
%F A266505 a(n) = (3*A000129(floor(n/2)) - A000129(n-(-1)^n)), where A000129 gives the Pell numbers.
%F A266505 a(n) = sqrt(2*A266504(n)^2 - 7*(-1)^A266504(n))*sgn(2*n-1), where A266504 gives all x in N such that 2*x^2 - 7*(-1)^x = y^2. This sequence gives associated y values.
%F A266505 a(2n) = (-(1 + sqrt(2))^n - sqrt(8)*(1 - sqrt(2))^n - (1 - sqrt(2))^n + sqrt(8)*(1 + sqrt(2))^n)/2 = a(2n) = A135532(n).
%F A266505 a(2n) = 3*(((1+sqrt(2))^n-(1-sqrt(2))^n)/sqrt(8)) - (((1+sqrt(2))^(n-1)-(1-sqrt(2))^(n-1))/sqrt(8)) = A135532(n).
%F A266505 a(2n+1) = (+(1 + sqrt(2))^n - sqrt(8)*(1 - sqrt(2))^n + (1 - sqrt(2))^n + sqrt(8)*(1 + sqrt(2))^n)/2 = a(2n + 1) = A048655(n).
%F A266505 a(2n+1) = 3*(((1+sqrt(2))^n-(1-sqrt(2))^n)/sqrt(8)) + (((1+sqrt(2))^(n+1)-(1-sqrt(2))^(n+1))/sqrt(8)) = A048655(n).
%F A266505 a(4n + 0) = 6*a(4n - 4) - a(4n - 8) = A255236(n-1).
%F A266505 a(4n + 1) = 6*a(4n - 3) - a(4n - 7) = A054490(n).
%F A266505 a(4n + 2) = 6*a(4n - 2) - a(4n - 6) = A038762(n).
%F A266505 a(4n + 3) = 6*a(4n - 1) - a(4n - 5) = A101386(n).
%F A266505 (sqrt(2*(a(2n + 1) )^2 + 14*(-1)^floor(n/2)))/2 = A266504(n).
%F A266505 (a(2n + 1) + a(2n))/8 = A000129(n), where A000129 gives the Pell numbers.
%F A266505 a(2n + 1) - a(2n) = A002203(n), where A002203 gives the companion Pell numbers.
%F A266505 (a(2n + 2) + a(2n + 1))/2 = A000129(n+2).
%F A266505 (a(2n + 2) - a(2n + 1))/2 = A000129(n-1).
%p A266505 a:=proc(n) option remember; if n=0 then -1 elif n=1 then 1 elif n=2 then 3 elif n=3 then 5 else 2*a(n-2)+a(n-4); fi; end:  seq(a(n), n=0..50); # _Wesley Ivan Hurt_, Jan 01 2016
%t A266505 LinearRecurrence[{0, 2, 0, 1}, {-1, 1, 3, 5}, 70] (* _Vincenzo Librandi_, Dec 31 2015 *)
%t A266505 Table[SeriesCoefficient[(-1 + 3 x) (1 + x)^2/(1 - 2 x^2 - x^4), {x, 0, n}], {n, 0, 42}] (* _Michael De Vlieger_, Dec 31 2015 *)
%o A266505 (Magma) I:=[-1,1,3,5]; [n le 4 select I[n] else 2*Self(n-2)+Self(n-4): n in [1..70]]; // _Vincenzo Librandi_, Dec 31 2015
%o A266505 (PARI) my(x='x+O('x^40)); Vec((-1+3*x)*(1+x)^2/(1-2*x^2-x^4)) \\ _G. C. Greubel_, Jul 26 2018
%Y A266505 Cf. A000129, A001333, A002203, A002965, A006451, A006452, A002965, A038761, A038762, A048654, A048655, A054490, A078343, A098586, A098790, A100525, A101386, A135532, A216134, A216162, A253811, A255236, A266504, A266505, A266507.
%K A266505 sign,easy
%O A266505 0,3
%A A266505 _Raphie Frank_, Dec 30 2015