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%I #44 Dec 12 2023 09:36:57
%S 3,2,4,1,2,0,3,2,4,1,2,0,3,2,4,1,2,0,3,2,4,1,2,0,3,2,4,1,2,0,3,2,4,1,
%T 2,0,3,2,4,1,2,0,3,2,4,1,2,0,3,2,4,1,2,0,3,2,4,1,2,0,3,2,4,1,2,0,3,2,
%U 4,1,2,0,3,2,4,1,2,0,3,2,4,1,2,0,3,2
%N Period 6: repeat [3, 2, 4, 1, 2, 0].
%C Associate to sequence identical to half its p-th differences.
%C Corresponding n-th differences table:
%C 3, 2, 4, 1, 2, 0, 3;
%C -1, 2, -3, 1, -2, 3, -1;
%C 3, -5, 4, -3, 5, -4, 3;
%C -8, 9, -7, 8, -9, 7, -8;
%C 17, -16, 15, -17, 16, -15, 17;
%C -33, 31, -32, 33, -31, 32, -33;
%C 64, -63, 65, -64, 63, -65, 64;
%C Note that the main diagonal is 3 followed by A000079(n+1).
%C Note also the southeast diagonal 4, 1, 5, 7, 17 is 4 followed by A014551(n+1).
%C Note also 3*A001045(n+1), one signed and one unsigned, in two southeast diagonals.
%C Starting from second line, the first column is A130750 signed.
%C Starting from second line, the second column is A130752 signed.
%C Starting from second line, the third column is A130755 signed.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,-1,1).
%F From _Wesley Ivan Hurt_, Aug 29 2014: (Start)
%F G.f.: (3-x+2*x^2)/((1-x)*(1+x^3)).
%F a(n) = a(n-1)-a(n-3)+a(n-4);
%F a(n) = 2 + ((-n-2) mod 3) * (-1)^n. (End)
%F a(n) = (6 + 3*cos(n*Pi) + 2*sqrt(3)*sin(n*Pi/3))/3. - _Wesley Ivan Hurt_, Jun 20 2016
%p A140430:=n->2+((-n-2) mod 3)*(-1)^n: seq(A140430(n), n=0..100); # _Wesley Ivan Hurt_, Aug 29 2014
%t CoefficientList[Series[(3 - x + 2 x^2)/((1 - x)*(1 + x^3)), {x, 0, 30}], x] (* _Wesley Ivan Hurt_, Aug 29 2014 *)
%t PadRight[{},120,{3,2,4,1,2,0}] (* _Harvey P. Dale_, Jan 21 2023 *)
%o (PARI) a(n)=[3,2,4,1,2,0][n%6+1] \\ _Charles R Greathouse IV_, Jun 02 2011
%o (Magma) [2 + ((-n-2) mod 3)*(-1)^n : n in [0..100]]; // _Wesley Ivan Hurt_, Aug 29 2014
%Y Cf. A000079 (2^n), A001045 (Jacobsthal), A014551 (Jacobsthal-Lucas).
%Y Cf. A130750, A130752, A130755.
%K nonn,easy
%O 0,1
%A _Paul Curtz_, Jun 25 2008
%E More terms from _Wesley Ivan Hurt_, Aug 29 2014
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