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a(0) = 1, a(1) = 2; for n>1, a(n) = a(n-1) + a(n-2) + 4.
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%I #21 Sep 08 2022 08:45:55

%S 1,2,7,13,24,41,69,114,187,305,496,805,1305,2114,3423,5541,8968,14513,

%T 23485,38002,61491,99497,160992,260493,421489,681986,1103479,1785469,

%U 2888952,4674425,7563381,12237810,19801195,32039009,51840208,83879221,135719433,219598658

%N a(0) = 1, a(1) = 2; for n>1, a(n) = a(n-1) + a(n-2) + 4.

%H G. C. Greubel, <a href="/A182415/b182415.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 (2,0,-1).

%F From _Colin Barker_, May 07 2012: (Start)

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

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

%F a(n) = Fibonacci(n+3) + 3*Fibonacci(n+1) - 4. - _G. C. Greubel_, Jul 22 2019

%e a(3) = 7 + 2 + 4 = 13.

%t With[{f=Fibonacci}, Table[F[n+3]+3*F[n+1]-4, {n,0,40}]] (* _G. C. Greubel_, Jul 22 2019 *)

%t RecurrenceTable[{a[0]==1,a[1]==2,a[n]==a[n-1]+a[n-2]+4},a,{n,40}] (* or *) LinearRecurrence[{2,0,-1},{1,2,7},40] (* _Harvey P. Dale_, Nov 24 2020 *)

%o (PARI) vector(40, n, n--; f=fibonacci; f(n+3) +3*f(n+1) -4 ) \\ _G. C. Greubel_, Jul 22 2019

%o (Magma) F:=Fibonacci; [F(n+3)+3*F(n+1)-4: n in [0..40]]; // _G. C. Greubel_, Jul 22 2019

%o (Sage) f=fibonacci; [f(n+3)+3*f(n+1)-4 for n in (0..40)] # _G. C. Greubel_, Jul 22 2019

%o (GAP) F:=Fibonacci;; List([0..40], n-> F(n+3)+3*F(n+1)-4); # _G. C. Greubel_, Jul 22 2019

%Y Cf. A000045, A014739, A022095 (first differences), A171516.

%K nonn,easy

%O 0,2

%A _Alex Ratushnyak_, Apr 28 2012