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a(n)*a(n+3) - a(n+1)*a(n+2) = 4, given a(0)=a(1)=1, a(2)=5.
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%I #21 Dec 26 2018 09:39:39

%S 1,1,5,9,49,89,485,881,4801,8721,47525,86329,470449,854569,4656965,

%T 8459361,46099201,83739041,456335045,828931049,4517251249,8205571449,

%U 44716177445,81226783441,442644523201,804062262961,4381729054565,7959395846169,43374646022449,78789896198729,429364731169925

%N a(n)*a(n+3) - a(n+1)*a(n+2) = 4, given a(0)=a(1)=1, a(2)=5.

%H Seiichi Manyama, <a href="/A080872/b080872.txt">Table of n, a(n) for n = 0..2000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0, 10, 0, -1).

%F G.f.: (-x^3 - 5*x^2 + x + 1)/(x^4 - 10*x^2 + 1).

%F a(n) = (3+sqrt(3))/12*(sqrt(3)-sqrt(2))^n+(3-sqrt(3))/12*(-sqrt(3)+sqrt(2))^n+(3+sqrt(3))/12*(sqrt(3)+sqrt(2))^n+(3-sqrt(3))/12*(-sqrt(3)-sqrt(2))^n. [_Richard Choulet_, Dec 03 2008]

%F a(n+4) = 10*a(n+2)-a(n). [_Richard Choulet_, Dec 04 2008]

%t CoefficientList[Series[(-x^3-5 x^2+x+1)/(x^4-10 x^2+1),{x,0,30}],x] (* or *) LinearRecurrence[{0,10,0,-1},{1,1,5,9},30] (* _Harvey P. Dale_, May 06 2012 *)

%o (PARI) Vec( (-x^3 - 5*x^2 + x + 1)/(x^4 - 10*x^2 + 1) + O(x^66) ) \\ _Joerg Arndt_, Jan 29 2016

%Y Cf. A080871, A080873, A080874, A080875.

%Y Bisections are A001079 and A072256.

%K nonn,easy

%O 0,3

%A _Paul D. Hanna_, Feb 22 2003