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a(n+3) = 3*(a(n+2) - a(n+1)) + 2*a(n).
3

%I #22 Dec 16 2023 17:00:35

%S 1,2,2,2,4,10,22,44,86,170,340,682,1366,2732,5462,10922,21844,43690,

%T 87382,174764,349526,699050,1398100,2796202,5592406,11184812,22369622,

%U 44739242,89478484,178956970,357913942,715827884,1431655766,2863311530

%N a(n+3) = 3*(a(n+2) - a(n+1)) + 2*a(n).

%C Binomial transform of period-3 sequence with period 1 1 -1.

%H Harvey P. Dale, <a href="/A130707/b130707.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,2).

%F a(n) = 2^n/3 + 4*(-1)^n*(1/3)*cos((2n+1)*Pi/3). - _Emeric Deutsch_, Jul 27 2007

%F From _R. J. Mathar_, Nov 18 2007: (Start)

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

%F a(n) = (2*A057079(n) + 2^n)/3. (End)

%p a:=proc(n) options operator, arrow: (1/3)*2^n+(4/3)*(-1)^n*cos((1/3)*(2*n+1)*Pi) end proc: seq(a(n), n = 0 .. 33); # _Emeric Deutsch_, Jul 27 2007

%t RecurrenceTable[{a[0]==1,a[1]==a[2]==2,a[n]==3(a[n-1]-a[n-2])+2a[n-3]},a,{n,40}] (* or *) LinearRecurrence[{3,-3,2},{1,2,2},40] (* _Harvey P. Dale_, Jan 18 2015 *)

%K nonn

%O 0,2

%A _Paul Curtz_, Jul 01 2007

%E More terms from _Emeric Deutsch_, Jul 27 2007

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Last modified September 20 02:41 EDT 2024. Contains 376016 sequences. (Running on oeis4.)