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a(n) = (n-3)*A006918(n-2)/2 for n >= 2, with a(0) = a(1) = 0.
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%I #24 Sep 19 2024 14:40:53

%S 0,0,0,0,1,5,12,28,50,90,140,220,315,455,616,840,1092,1428,1800,2280,

%T 2805,3465,4180,5060,6006,7150,8372,9828,11375,13195,15120,17360,

%U 19720,22440,25296,28560,31977,35853,39900,44460,49210,54530,60060,66220,72611

%N a(n) = (n-3)*A006918(n-2)/2 for n >= 2, with a(0) = a(1) = 0.

%D K. B. Reid and L. W. Beineke "Tournaments", pp. 169-204 in L. W. Beineke and R. J. Wilson, editors, Selected Topics in Graph Theory, Academic Press, NY, 1978, p. 186 Theorem 6.11.

%H Colin Barker, <a href="/A038376/b038376.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2,2,-6,0,6,-2,-2,1).

%F From _Colin Barker_, Nov 19 2016: (Start)

%F a(n) = (n^4-3*n^3-4*n^2+12*n)/48 for n even.

%F a(n) = (n^4-3*n^3-n^2+3*n)/48 for n odd.

%F a(n) = 2*a(n-1) + 2*a(n-2) - 6*a(n-3) + 6*a(n-5) - 2*a(n-6) - 2*a(n-7) + a(n-8) for n>7.

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

%F a(n) = n*(n - 3)*(2*n^2 - 3*(-1)^n - 5)/96. - _Paolo Xausa_, Sep 17 2024 (derived from _Bruno Berselli_ formula in A006918)

%t LinearRecurrence[{2, 2, -6, 0, 6, -2, -2, 1}, {0, 0, 0, 0, 1, 5, 12, 28}, 100]

%t (* or *)

%t A038376[n_] := n*(n - 3)*(2*n^2 - 3*(-1)^n - 5)/96;

%t Array[A038376, 100, 0] (* _Paolo Xausa_, Sep 16 2024 *)

%o (PARI) concat(vector(4), Vec(x^4*(1+3*x) / ((1-x)^5 * (1+x)^3) + O(x^100))) \\ _Colin Barker_, Nov 19 2016

%Y Cf. A006918.

%K nonn,easy

%O 0,6

%A _N. J. A. Sloane_

%E Name corrected by _Paolo Xausa_, Sep 16 2024