A153265 - OEIS

%I #20 Mar 17 2024 12:54:54

%S 4,1,-9,-2,31,1,-126,1,511,-2,-2049,1,8194,1,-32769,-2,131071,1,

%T -524286,1,2097151,-2,-8388609,1,33554434,1,-134217729,-2,536870911,1,

%U -2147483646,1,8589934591,-2,-34359738369,1

%N a(n) = (-2*I)^n + (2*I)^n + (1/2 + 1/2*I*sqrt(3))^n + (1/2 - 1/2*I*sqrt(3))^n.

%C For all n there is an m such that: ||a(n)| - 2^m| <= 2. In the Python program which will be provided, the sequence (a(n)) is given by 4tesseq(X) where X = 1.5'i + .25(ii + jj + kk + ee) is the generating floretion.

%H Vincenzo Librandi, <a href="/A153265/b153265.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,-5,4,-4).

%F a(n) = a(n-1) - 5a(n-2) + 4a(n-3) - 4a(n-4).

%F G.f.: 4 + x*(1-10*x+12*x^2-16*x^3)/((x^2-x+1)*(4*x^2+1)). - corrected by _Vaclav Kotesovec_, Jun 25 2014

%e a(4) = 32 + (1/2 + 1/2*I*sqrt(3))^4 + (1/2 - 1/2*I*sqrt(3))^4 = 31 -or- a(4) = a(n-1) - 5a(n-2) + 4a(n-3) - 4a(n-4) = -2 - 5*(-9) + 4*(1) - 4*4 = 31

%p a := n-> (2*I)^n+(-2*I)^n+(1/2+1/2*I*sqrt(3))^n+(1/2-1/2*I*sqrt(3))^n;

%t CoefficientList[Series[4 + x (1 - 10 x + 12 x^2 - 16 x^3)/((x^2 - x + 1) (4 x^2 + 1)), {x, 0, 100}], x] (* _Vincenzo Librandi_, Jun 26 2014 *)

%o (PARI) a(n)=2*(n%2<1)*(-4)^(n\2)+3*(n%3<1)*(-1)^(n\3)-(-1)^n \\ _Tani Akinari_, Jun 25 2014

%o (Magma) I:=[4,1,-9,-2]; [n le 4 select I[n] else Self(n-1)-5*Self(n-2)+4*Self(n-3) -4*Self(n-4): n in [1..50]]; // _Vincenzo Librandi_, Jun 26 2014

%Y Cf. A153266, A153267, A153268.

%K easy,sign

%O 0,1

%A _Creighton Dement_, Dec 22 2008, Dec 31 2008