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a(n)*a(n+3) - a(n+1)*a(n+2) = 2^n, given a(0)=1, a(1)=1, a(2)=3.
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%I #13 Jun 13 2015 00:50:52

%S 1,1,3,4,14,20,72,104,376,544,1968,2848,10304,14912,53952,78080,

%T 282496,408832,1479168,2140672,7745024,11208704,40553472,58689536,

%U 212340736,307302400,1111830528,1609056256,5821620224,8425127936,30482399232

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

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0, 6, 0, -4).

%F G.f.: (1 + x - 3*x^2 - 2*x^3) / (1 - 6*x^2 + 4*x^4). a(n) = 6*a(n-2) - 4*a(n-4). - _Michael Somos_, Mar 05 2003

%F a(2n) = A080877(2n+1), a(2n+1) = A080877(2n+2)/2.

%F a(n) = (1/20*10^(1/2) + 1/4)*(sqrt(3 + sqrt(5)))^n + (1/20*10^(1/2) + 1/4)*(sqrt(3 - sqrt(5)))^n + ( - 1/20*10^(1/2) + 1/4)*( - (sqrt(3 + sqrt(5))))^n + ( - 1/20*10^(1/2) + 1/4)*( - (sqrt(3 - sqrt(5))))^n. - _Richard Choulet_, Dec 07 2008

%F a(-n) = a(n) / 2^n. a(2*n) = A098648(n). a(2*n + 1) = A082761(n). - _Michael Somos_, May 25 2014

%F 0 = a(n)*(+2*a(n+2)) + a(n+1)*(+2*a(n+1) - 7*a(n+2) + a(n+3)) + a(n+2)*(+a(n+2)) for all n in Z. - _Michael Somos_, May 25 2014

%e G.f. = 1 + x + 3*x^2 + 4*x^3 + 14*x^4 + 20*x^5 + 72*x^6 + 104*x^7 + 376*x^8 + ...

%t a[ n_] := If[ n < 0, 2^n, 1] SeriesCoefficient[ (1 + x - 3*x^2 - 2*x^3)/(1 - 6*x^2 + 4*x^4), {x, 0, Abs@n}]; (* _Michael Somos_, May 25 2014 *)

%t a[ n_] := 2^Quotient[ n - 1, 2] If[ OddQ@n, Fibonacci@n, LucasL@n]; (* _Michael Somos_, May 25 2014 *)

%t LinearRecurrence[{0,6,0,-4},{1,1,3,4},40] (* _Harvey P. Dale_, Dec 07 2014 *)

%o (PARI) {a(n) = if( n<0, 2^n, 1) * polcoeff( (1 + x - 3*x^2 - 2*x^3) / (1 - 6*x^2 + 4*x^4) + x * O(x^abs(n)), abs(n))}; /* _Michael Somos_, May 25 2014 */

%o (PARI) {a(n) = 2^((n - 1)\2) * if( n%2, fibonacci(n), fibonacci(n-1) + fibonacci(n+1))}; /* _Michael Somos_, May 25 2014 */

%Y Cf. A080876, A080877, A080879, A080880, A080881, A080882.

%Y Cf. A000079, A082761, A098648.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Feb 22 2003