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a(n) = 1^n + 4^n + 9^n.
2

%I #23 Mar 14 2021 12:07:26

%S 3,14,98,794,6818,60074,535538,4799354,43112258,387682634,3487832978,

%T 31385253914,282446313698,2541932937194,22877060890418,

%U 205892205836474,1853024483819138,16677198879535754,150094704016475858

%N a(n) = 1^n + 4^n + 9^n.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (14,-49,36).

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

%F G.f.: 1/(1-x) + 1/(1-4*x) + 1/(1-9*x).

%F E.g.f.: e^x + e^(4*x) + e^(9*x). (End)

%F a(n) = 13*a(n-1) - 36*a(n-2) + 24 with a(0)=3, a(1)=14. - _Vincenzo Librandi_, Jul 21 2010

%t Table[1^n + 4^n + 9^n, {n, 0, 20}]

%t LinearRecurrence[{14,-49,36},{3,14,98},30] (* _Harvey P. Dale_, Aug 06 2013 *)

%o (Python)

%o def a(n): return 1 + 4**n + 9**n

%o print([a(n) for n in range(19)]) # _Michael S. Branicky_, Mar 14 2021

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

%Y Cf. A052539.

%K easy,nonn

%O 0,1

%A _Robert G. Wilson v_, Aug 23 2002