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a(n)*a(n+3) - a(n+1)*a(n+2) = 2^n, given a(0)=1, a(1)=2, a(2)=10.
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%I #13 Jun 10 2024 16:31:14

%S 1,2,10,21,106,223,1126,2369,11962,25167,127078,267361,1350010,

%T 2840303,14341798,30173889,152359738,320551567,1618589926,3405371681,

%U 17195050234,36176882223,182671192870,384324217729,1940602920634

%N a(n)*a(n+3) - a(n+1)*a(n+2) = 2^n, given a(0)=1, a(1)=2, a(2)=10.

%H Harvey P. Dale, <a href="/A080881/b080881.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 (0,11,0,-4).

%F G.f.: (-x^3 - x^2 + 2*x + 1)/(4*x^4 - 11*x^2 + 1)

%F a(n + 4) = 11*a(n + 2) - 4*a(n) [From _Richard Choulet_, Dec 06 2008]

%F a(n) = (3/140*15^(1/2)*7^(1/2) + 1/4 + 3/56*7^(1/2) + 1/24*15^(1/2))*sqrt((11 + sqrt(105))/2)^n + ( - 3/140*15^(1/2)*7^(1/2) + 1/4 - 3/56*7^(1/2) + 1/24*15^(1/2))*sqrt((11 - sqrt(105))/2)^n + (3/140*15^(1/2)*7^(1/2) - 1/24*15^(1/2) - 3/56*7^(1/2) + 1/4)*( - sqrt((11 + sqrt(105))/2))^n + (1/4 + 3/56*7^(1/2) - 1/24*15^(1/2) - 3/140*15^(1/2)*7^(1/2))*( - sqrt((11 - sqrt(105))/2))^n [From _Richard Choulet_, Dec 07 2008]

%t CoefficientList[Series[(-x^3-x^2+2x+1)/(4x^4-11x^2+1),{x,0,30}],x] (* or *) LinearRecurrence[ {0,11,0,-4},{1,2,10,21},30] (* _Harvey P. Dale_, Jun 10 2024 *)

%Y Cf. A080876, A080877, A080878, A080879, A080880, A080882.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Feb 22 2003