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%I #19 Jul 09 2020 02:58:36
%S 1,1,1,0,1,1,2,1,2,2,0,1,1,3,3,4,6,3,3,3,0,1,1,4,6,8,13,12,10,12,6,4,
%T 4,0,1,1,5,10,15,25,31,30,35,30,20,20,10,5,5,0,1,1,6,15,26,45,66,76,
%U 90,96,80,75,60,35,30,15,6,6,0,1,1,7,21,42,77,126,168,211,252,252,245,231
%N Triangular array read by rows: T(n,m) = number of ways your team can score m points in n rounds of a soccer competition (loss=0 point, draw=1 point, win=3 points), for n >= 0, 0 <= m <= 3n.
%C n-th row has 3n+1 entries.
%H S. R. Finch, P. Sebah and Z.-Q. Bai, <a href="http://arXiv.org/abs/0802.2654">Odd Entries in Pascal's Trinomial Triangle</a>, arXiv:0802.2654 [math.NT], 2008.
%H Szymon Lukaszyk, <a href="https://arxiv.org/abs/2007.03782">Four Cubes</a>, arXiv:2007.03782 [math.GM], 2020.
%F T(n, m) = T(n-1, m) + T(n-1, m-1) + T(n-1, m-3).
%F G.f. Sum T(n, m)*z^n*w^m = 1/(1-z(1+w+w^3)). Hence m-th column is a polynomial in n of degree m given by C(n, m) + C(m-2, 1)*C(n, m-2) + C(m-4, 2)*C(n, m-4) + C(m-6, 3)*C(n, m-6) + ... E.g. column 5 is C(n, 5)+3C(n, 3). - _N. J. A. Sloane_, May 24 2005
%e Triangle begins:
%e 0...1...2...3...4...5...6...7...8...9..10..11..12 points
%e --------------------------------------------------------
%e 1
%e 1...1...0...1
%e 1...2...1...2...2...0...1
%e 1...3...3...4...6...3...3...3...0...1
%e 1...4...6...8..13..12..10..12...6...4...4...0...1
%t T[_, 0] = 1;
%t T[n_, m_] /; 0 <= m <= 3n = T[n, m] = T[n-1, m]+T[n-1, m-1]+T[n-1, m-3];
%t T[_, _] = 0;
%t Table[T[n, m], {n, 0, 7}, {m, 0, 3n}] // Flatten (* _Jean-François Alcover_, Sep 10 2018 *)
%K nonn,tabf
%O 0,7
%A _Floor van Lamoen_, May 02 2000
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