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a(n) = (8/9)*(2+7*(-8)^(n-1)).
2

%I #14 Apr 07 2016 16:10:43

%S 1,8,-48,400,-3184,25488,-203888,1631120,-13048944,104391568,

%T -835132528,6681060240,-53448481904,427587855248,-3420702841968,

%U 27365622735760,-218924981886064,1751399855088528,-14011198840708208

%N a(n) = (8/9)*(2+7*(-8)^(n-1)).

%H G. C. Greubel, <a href="/A165748/b165748.txt">Table of n, a(n) for n = 0..500</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (-7,8).

%F a(n) = (-8)*a(n-1) + 16 for n>=1, with a(0) = 1.

%F a(n) = 8*a(n-2) - 7*a(n-1), a(0)=1, a(1)=8.

%F G.f.: (1+15x)/(1+7x-8x^2).

%F a(n) = Sum_{0<=k<=n} A112555(n,k)*7^(n-k).

%F From _G. C. Greubel_, Apr 07 2016: (Start)

%F a(n) = -7*a(n-1) + 8*a(n-2).

%F E.g.f.: (1/9)*(16*exp(x) - 7*exp(-8*x)). (End)

%t Table[(8/9)*(2 + 7*(-8)^(n - 1)), {n, 0, 100}] or

%t LinearRecurrence[{-7,8}, {1,8}, 100] (* _G. C. Greubel_, Apr 07 2016 *)

%o (PARI) x='x+O('x^99); Vec((1+15*x)/(1+7*x-8*x^2)) \\ _Altug Alkan_, Apr 07 2016

%K easy,sign

%O 0,2

%A _Philippe Deléham_, Sep 26 2009