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a(n) = 4*a(n-1) + 5*a(n-2) for n > 1, with a(0) = 2 and a(1) = 4.
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%I #26 Aug 26 2024 09:47:12

%S 2,4,26,124,626,3124,15626,78124,390626,1953124,9765626,48828124,

%T 244140626,1220703124,6103515626,30517578124,152587890626,

%U 762939453124,3814697265626,19073486328124,95367431640626,476837158203124,2384185791015626,11920928955078124,59604644775390626

%N a(n) = 4*a(n-1) + 5*a(n-2) for n > 1, with a(0) = 2 and a(1) = 4.

%C Let F(x) = Product_{n>=0} (1 - x^(3*n+1))/(1 - x^(3*n+2)). This sequence is the simple continued fraction expansion of the real number 1 + F(-1/5) = 2.24761 97788 60361 46849 ... = 2 + 1/(4 + 1/(26 + 1/(124 + 1/(626 + ...)))). See A111317. - _Peter Bala_, Dec 26 2012

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (4,5).

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

%F a(n) = 5^n + (-1)^n.

%F From _Elmo R. Oliveira_, Aug 23 2024: (Start)

%F E.g.f.: exp(-x)*(exp(6*x) + 1).

%F a(n) = 2*A081340(n). (End)

%t CoefficientList[Series[(2 - 4x)/(1 - 4x - 5x^2), {x, 0, 25}], x]

%t LinearRecurrence[{4,5},{2,4},30] (* _Harvey P. Dale_, May 13 2022 *)

%o (Sage) [lucas_number2(n,4,-5) for n in range(0, 22)] # _Zerinvary Lajos_, May 14 2009

%Y Cf. A081340, A111317.

%K easy,nonn

%O 0,1

%A Mario Catalani (mario.catalani(AT)unito.it), Sep 01 2003

%E a(22)-a(24) from _Elmo R. Oliveira_, Aug 23 2024