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a(n) = floor(3*n/2)*floor(n/3).
1

%I #15 Feb 20 2024 03:30:24

%S 0,0,0,4,6,7,18,21,24,40,45,49,72,78,84,112,120,127,162,171,180,220,

%T 231,241,288,300,312,364,378,391,450,465,480,544,561,577,648,666,684,

%U 760,780,799,882,903,924,1012,1035,1057,1152,1176,1200,1300,1326

%N a(n) = floor(3*n/2)*floor(n/3).

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,1,-1,0,1,-1,0,-1,1).

%F a(n) = a(n-1)+a(n-3)-a(n-4)+a(n-6)-a(n-7)-a(n-9)+a(n-10).

%F G.f.: (4*x^3 + 2*x^4 + x^5 + 7*x^6 + x^7 + 2*x^8 + x^9)/(1 - x - x^3 + x^4 - x^6 + x^7 + x^9 - x^10).

%t a[n_] := Floor[(3 n/2)*Floor[n/3]]

%t Table[a[n], {n, 0, 90}] (* A213034 *)

%t LinearRecurrence[{1, 0, 1, -1, 0, 1, -1, 0, -1, 1}, {0, 0, 0, 4, 6, 7, 18, 21, 24, 40}, 90]

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Jun 05 2012