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a(n) = (11^n + 5^n)/2.
1

%I #13 Sep 08 2022 08:45:39

%S 1,8,73,728,7633,82088,893593,9782648,107374753,1179950408,

%T 12973595113,142680249368,1569336258673,17261966423528,

%U 189877968549433,2088639343496888,22974941225731393,252723895719373448

%N a(n) = (11^n + 5^n)/2.

%C Binomial transform of A081343.

%C Inverse binomial transform of A143079.

%H Vincenzo Librandi, <a href="/A152429/b152429.txt">Table of n, a(n) for n = 0..200</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (16,-55).

%F a(n) = 16*a(n-1) - 55*a(n-2), with a(0)=1, a(1)=8.

%F G.f.: (1-8*x)/(1 - 16*x + 55*x^2).

%F a(n) = Sum_{k=0..n} A098158(n,k)*8^(2k-n)*9^(n-k).

%F a(n) = ((8 + sqrt(9))^n + (8 - sqrt(9))^n)/2. - Al Hakanson (hawkuu(AT)gmail.com), Dec 08 2008

%F E.g.f.: (exp(11*x) + exp(5*x))/2. - _G. C. Greubel_, Jan 08 2020

%p seq( (11^n+5^n)/2, n=0..20); # _G. C. Greubel_, Jan 08 2020

%t LinearRecurrence[{16,-55}, {1,8}, 20] (* _G. C. Greubel_, Jan 08 2020 *)

%o (Magma) [(11^n+5^n)/2: n in [0..20]]; // _Vincenzo Librandi_, Jun 01 2011

%o (PARI) vector(21, n, (11^(n-1) + 5^(n-1))/2 ) \\ _G. C. Greubel_, Jan 08 2020

%o (Sage) [(11^n+5^n)/2 for n in (0..20)] # _G. C. Greubel_, Jan 08 2020

%o (GAP) List([0..20], n-> (11^n+5^n)/2); # _G. C. Greubel_, Jan 08 2020

%Y Cf. A162516.

%K nonn,less

%O 0,2

%A _Philippe Deléham_, Dec 03 2008