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a(n) = (n - 1)*(2*floor(n/2) + 1).
4

%I #27 Jan 18 2023 14:44:53

%S -1,0,3,6,15,20,35,42,63,72,99,110,143,156,195,210,255,272,323,342,

%T 399,420,483,506,575,600,675,702,783,812,899,930,1023,1056,1155,1190,

%U 1295,1332,1443,1482,1599,1640,1763,1806,1935,1980,2115,2162,2303,2352,2499,2550,2703,2756,2915

%N a(n) = (n - 1)*(2*floor(n/2) + 1).

%C Summing a(n) by pairs, one gets -1, 9, 35, 77, 135, ... = A033566.

%C A198442(k) is a member of this sequence if k == 0 or 1 (mod 4). - _Bruno Berselli_, Jul 04 2017

%H Colin Barker, <a href="/A289296/b289296.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).

%F a(n) = A023443(n) * A109613(n).

%F a(n) = n^2-1 if n is even and n^2-n if n is odd.

%F n^2 - a(n) = A093178(n).

%F From _Colin Barker_, Jul 02 2017: (Start)

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

%F a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>4. (End)

%t Table[(n - 1) (2 Floor[n/2] + 1), {n, 0, 60}] (* or *) LinearRecurrence[{1, 2, -2, -1, 1}, {-1, 0, 3, 6, 15}, 61]

%o (PARI) a(n)=(n\2*2+1)*(n-1) \\ _Charles R Greathouse IV_, Jul 02 2017

%o (PARI) Vec(-(1 - x - 5*x^2 - x^3 - 2*x^4) / ((1 - x)^3*(1 + x)^2) + O(x^60)) \\ _Colin Barker_, Jul 02 2017

%o (Python)

%o def A289296(n): return (n-1)*(n|1) # _Chai Wah Wu_, Jan 18 2023

%Y Subsequence of A214297.

%Y Cf. A023443, A033566, A093178, A109613.

%K sign,easy

%O 0,3

%A _Jean-François Alcover_ and _Paul Curtz_, Jul 02 2017