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a(n) = 1^n + 4^n + 6^n.
0

%I #16 Jun 19 2024 18:42:12

%S 3,11,53,281,1553,8801,50753,296321,1745153,10339841,61514753,

%T 366991361,2193559553,13127802881,78632599553,471258726401,

%U 2825404874753,16943839313921,101628676145153,609634617917441

%N a(n) = 1^n + 4^n + 6^n.

%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (11,-34,24).

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

%F G.f.: 1/(1-x) + 1/(1-4*x) + 1/(1-6*x).

%F E.g.f.: e^x + e^(4*x) + e^(6*x). (End)

%F a(n) = 10*a(n-1) - 24*a(n-2) + 15 with a(0)=3, a(1)=11. - _Vincenzo Librandi_, Jul 21 2010

%t Table[1^n + 4^n + 6^n, {n, 0, 20}]

%t LinearRecurrence[{11,-34,24},{3,11,53},30] (* _Harvey P. Dale_, Jun 19 2024 *)

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

%K easy,nonn

%O 0,1

%A _Robert G. Wilson v_, Aug 23 2002