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%I #16 Aug 22 2015 22:45:04
%S 1,1,1,1,1,1,1,1,2,2,3,3,3,2,2,1,1,1,1,2,3,4,5,7,7,8,8,8,7,7,5,4,3,2,
%T 1,1,1,1,2,3,5,6,9,11,14,16,19,20,23,23,24,23,23,20,19,16,14,11,9,6,5,
%U 3,2,1,1,1,1,2,3,5,7,10,13,18,22,28,33,40,45,52,57,63,66,70,71
%N Triangle of coefficients of polynomials p(n), with p(3)=1, p(n) = (1 - t^(2*n - 4))*(1 - t^(2*n - 3))*p(n - 1)/((1 - t^(n - 3))*(1 - t^n)).
%H J. Riordan, <a href="http://dx.doi.org/10.1016/S0021-9800(68)80032-8">The number of score sequences in tournaments</a>, J. Combin. Theory, 5 (1968), 87-89. [But beware errors.]
%e 1;
%e 1 + t + t^2 + t^3 + t^4 + t^5;
%e t^10 + t^9 + 2*t^8 + 2*t^7 + 3*t^6 + 3*t^5 + 3*t^4 + 2*t^3 + 2*t^2 + t + 1;
%e ...
%p p := proc(n) option remember; if n = 3 then 1 else (1-t^(2*n-4))*(1-t^(2*n-3))*p(n-1)/((1-t^(n-3))*(1-t^n)); fi; end;
%t p[3] = 1; p[n_] := p[n] = (1 - t^(2*n-4))*(1 - t^(2*n-3))*p[n-1]/((1 - t^(n-3))*(1 - t^n)) // Simplify; Table[ CoefficientList[ Series[p[n], {t, 0, n^3}], t], {n, 3, 8}] // Flatten (* _Jean-François Alcover_, Oct 25 2013 *)
%Y Cf. A046919.
%K nonn,easy,nice,tabf
%O 3,9
%A _N. J. A. Sloane_
%E Keyword tabf by _Michel Marcus_, Dec 05 2014
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