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a(n) = 1^n + 3^n + 6^n.
1

%I #15 Jan 26 2023 10:05:36

%S 3,10,46,244,1378,8020,47386,282124,1686178,10097380,60525226,

%T 362974204,2177313778,13062288340,78368947066,470199333484,

%U 2821152954178,16926788584900,101560344088906,609360902271964

%N a(n) = 1^n + 3^n + 6^n.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (10,-27,18).

%F G.f.: 1/(1-x)+1/(1-3*x)+1/(1-6*x). E.g.f.: e^x+e^(3*x)+e^(6*x). [_Mohammad K. Azarian_, Dec 26 2008]

%F a(n) = 9*a(n-1) - 18*a(n-2) + 10, n>1. [_Gary Detlefs_, Jun 21 2010]

%t Table[1^n + 3^n + 6^n, {n, 0, 20}]

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

%K easy,nonn

%O 0,1

%A _Robert G. Wilson v_, Aug 23 2002