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%I #26 Sep 08 2022 08:46:19
%S 0,1,1,3,5,8,12,19,31,51,83,134,216,349,565,915,1481,2396,3876,6271,
%T 10147,16419,26567,42986,69552,112537,182089,294627,476717,771344,
%U 1248060,2019403,3267463,5286867,8554331,13841198,22395528,36236725,58632253,94868979
%N a(n) = 2*a(n-1) - a(n-2) + a(n-4), n>3, a(0)=0, a(1)=a(2)=1, a(3)=3.
%C Difference table for a(n):
%C 0, 1, 1, 3, 5, 8, 12, 19, 31, 51, 83, 134, 216, ...
%C 1, 0, 2, 2, 3, 4, 7, 12, 20, 32, 51, 82, 133, ...
%C -1, 2, 0, 1, 1, 3, 5, 8, 12, 19, 31, 51, 83, ...
%C 3, -2, 1, 0, 2, 2, 3, 4, 7, 12, 20, 32, 51, ...
%C etc.
%C The pair a(n) = 0, 1, 1, 3, 5, 8, 12, 19, 31, 51, ...
%C and b(n) = 0, 2, 2, 3, 4, 7, 12, 20, 32, 51, ...
%C is interesting. a(n) and b(n) are autosequences of the first kind (see Link). a(n) and b(n) have the same first trisection: 3*A001076(n).
%C a(n) + b(n) = A022086(n) = 3*A000045(n) (Fibonacci).
%C b(n) - a(n) = 0, 1, 1, 0, -1, -1, 0, ... = A128834(n).
%C a(n+6) - a(n) = b(n+6) - b(n) = 6*Fib(n+3).
%C a(n) - a(n) mod 9 = 9*A004699(n) = b(n) - b(n) mod 9.
%H Colin Barker, <a href="/A286311/b286311.txt">Table of n, a(n) for n = 0..1000</a>
%H OEIS Wiki, <a href="https://oeis.org/wiki/Autosequence">Autosequence</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,0,1).
%F a(n) = 2*a(n-1) - a(n-2) + a(n-4). Valid for b(n).
%F G.f.: x*(1 - x + 2*x^2) / ((1 - x + x^2)*(1 - x - x^2)). - _Colin Barker_, May 06 2017
%t LinearRecurrence[{2, -1, 0, 1}, {0, 1, 1, 3}, 40] (* or *)
%t CoefficientList[Series[x (1 - x + 2 x^2)/((1 - x + x^2) (1 - x - x^2)), {x, 0, 39}], x] (* _Michael De Vlieger_, May 07 2017 *)
%o (PARI) concat(0, Vec(x*(1 - x + 2*x^2) / ((1 - x + x^2)*(1 - x - x^2)) + O(x^60))) \\ _Colin Barker_, May 06 2017
%o (Magma) I:=[0,1,1,3]; [n le 4 select I[n] else 2*Self(n-1) - Self(n-2) + Self(n-4): n in [1..30]]; // _G. C. Greubel_, Jan 15 2018
%Y Cf. A000034, A000045, A001076, A004699, A022086, A128834.
%K nonn,easy
%O 0,4
%A _Paul Curtz_, May 06 2017
%E More terms from _Colin Barker_, May 06 2017
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