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%I #21 Mar 19 2022 06:38:47
%S 1,3,6,2,-3,-9,-2,6,15,5,-6,-18,-5,9,24,8,-9,-27,-8,12,33,11,-12,-36,
%T -11,15,42,14,-15,-45,-14,18,51,17,-18,-54,-17,21,60,20,-21,-63,-20,
%U 24,69,23,-24,-72,-23,27,78,26,-27,-81,-26,30,87,29,-30,-90,-29
%N a(n) = 1 + 2 + 3 - 4 - 5 - 6 + 7 + 8 + 9 - 10 - 11 - 12 + ... - (up to n).
%C In general, for sequences that add the first k natural numbers and then subtract the next k natural numbers, and continue to alternate in this way up to n, we have a(n) = Sum_{i=1..n} i*(-1)^floor((i-1)/k). Here, k=3.
%H Colin Barker, <a href="/A319674/b319674.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,-2,2,0,-1,1).
%F a(n) = Sum_{i=1..n} i*(-1)^floor((i-1)/3).
%F From _Colin Barker_, Sep 26 2018: (Start)
%F G.f.: x*(1 + 2*x + 3*x^2 - 2*x^3 - x^4) / ((1 - x)*(1 + x)^2*(1 - x + x^2)^2).
%F a(n) = a(n-1) - 2*a(n-3) + 2*a(n-4) - a(n-6) + a(n-7) for n>7.
%F (End)
%e a(1) = 1;
%e a(2) = 1 + 2 = 3;
%e a(3) = 1 + 2 + 3 = 6;
%e a(4) = 1 + 2 + 3 - 4 = 2;
%e a(5) = 1 + 2 + 3 - 4 - 5 = -3;
%e a(6) = 1 + 2 + 3 - 4 - 5 - 6 = -9;
%e a(7) = 1 + 2 + 3 - 4 - 5 - 6 + 7 = -2;
%e a(8) = 1 + 2 + 3 - 4 - 5 - 6 + 7 + 8 = 6;
%e a(9) = 1 + 2 + 3 - 4 - 5 - 6 + 7 + 8 + 9 = 15;
%e a(10) = 1 + 2 + 3 - 4 - 5 - 6 + 7 + 8 + 9 - 10 = 5; etc.
%t Table[Sum[i (-1)^Floor[(i - 1)/3], {i, n}], {n, 60}]
%t Accumulate[Flatten[If[EvenQ[#[[1]]],-#,#]&/@Partition[Range[70],3]]] (* or *) LinearRecurrence[{1,0,-2,2,0,-1,1},{1,3,6,2,-3,-9,-2},70] (* _Harvey P. Dale_, Sep 15 2021 *)
%o (PARI) Vec(x*(1 + 2*x + 3*x^2 - 2*x^3 - x^4) / ((1 - x)*(1 + x)^2*(1 - x + x^2)^2) + O(x^60)) \\ _Colin Barker_, Sep 26 2018
%Y Cf. A001057 (k=1), A077140 (k=2), this sequence (k=3).
%K sign,look,easy
%O 1,2
%A _Wesley Ivan Hurt_, Sep 25 2018
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