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a(n) = 2^n + 3^n + 4^n.
4

%I #19 Sep 08 2022 08:45:07

%S 3,9,29,99,353,1299,4889,18699,72353,282339,1108649,4373499,17312753,

%T 68711379,273234809,1088123499,4338079553,17309140419,69107159369,

%U 276040692699,1102999460753,4408508961459,17623571298329

%N a(n) = 2^n + 3^n + 4^n.

%H Vincenzo Librandi, <a href="/A074526/b074526.txt">Table of n, a(n) for n = 0..200</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (9,-26,24).

%F From _Mohammad K. Azarian_, Dec 26 2008: (Start)

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

%F E.g.f.: exp(2*x)+exp(3*x)+exp(4*x). (End)

%t Table[2^n + 3^n + 4^n, {n, 0, 23}]

%t LinearRecurrence[{9,-26,24},{3,9,29},30] (* _Harvey P. Dale_, Jun 14 2022 *)

%o (Magma) [2^n + 3^n + 4^n: n in [0..25]]; // _Vincenzo Librandi_, Jun 11 2011

%Y Cf. A001550, A001576, A034513, A001579, A074501 - A074580.

%K easy,nonn

%O 0,1

%A _Robert G. Wilson v_, Aug 23 2002