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a(1) = 1, a(2) = 2; for n >= 3, a(n) = (n-1)^3 - a(n-1) - a(n-2).
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%I #28 Mar 04 2023 08:59:29

%S 1,2,5,20,39,66,111,166,235,328,437,566,725,906,1113,1356,1627,1930,

%T 2275,2654,3071,3536,4041,4590,5193,5842,6541,7300,8111,8978,9911,

%U 10902,11955,13080,14269,15526,16861,18266,19745,21308,22947,24666,26475,28366,30343

%N a(1) = 1, a(2) = 2; for n >= 3, a(n) = (n-1)^3 - a(n-1) - a(n-2).

%C The sum of every three consecutive terms is equal to the cube of the index of the middle one, i.e., a(n-1) + a(n) + a(n+1) = n^3.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,2,-3,3,-1).

%F G.f.: x*(2*x^5 - 7*x^4 + 9*x^3 + 2*x^2 - x + 1)/((x^2 + x + 1)*(x - 1)^4).

%F a(n) = (A242135(n) - 6*cos(2*n*Pi/3) + 2*sin(2*n*Pi/3)/sqrt(3))/3. - _Stefano Spezia_, Mar 04 2023

%e a(5) = (5-1)^3 - a(4) - a(3) = 4^3 - 20 - 5 = 64 - 20 - 5 = 39.

%t a[1] = 1; a[2] = 2; a[n_] := a[n] = (n - 1)^3 - a[n - 1] - a[n - 2]; Array[a, 45] (* _Amiram Eldar_, Mar 03 2023 *)

%o (PARI) lista(nn) = my(va = vector(nn)); va[1] = 1; va[2] = 2; for (n=3, nn, va[n] = (n-1)^3 - va[n-1] - va[n-2];); va; \\ _Michel Marcus_, Mar 03 2023

%Y Cf. A000578, A152728, A242135.

%K nonn,easy

%O 1,2

%A _Tamas Sandor Nagy_, Mar 02 2023