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%I #27 Sep 08 2022 08:46:06
%S -1,9,-36,84,-126,126,-84,36,-9,1,-1,8,-27,48,-42,0,42,-48,27,-8,1,-1,
%T 7,-19,21,6,-42,42,-6,-21,19,-7,1,-1,6,-12,2,27,-36,0,36,-27,-2,12,-6,
%U 1,-1,5,-6,-10,29,-9,-36,36,9,-29,10,6,-5,1,-1,4,-1,-16
%N Trapezoid of dot products of row 9 (signs alternating) with sequential 10-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 10-tuples (C(9,0), -C(9,1), ..., C(9,8), -C(9,9)) and (C(n-1,k-9), C(n-1,k-8), ..., C(n-1,k)), n >= 1, 0 <= k <= n+8.
%C The array is trapezoidal rather than triangular because C(n,k) is not uniquely defined for all negative n and negative k.
%C Row sums are 0.
%C Coefficients of (x-1)^9 (x+1)^(n-1), n > 0.
%H G. C. Greubel, <a href="/A230212/b230212.txt">Rows n=1..50 of trapezoid, flattened</a>
%H Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, Graça Tomaz, <a href="https://www.emis.de/journals/JIS/VOL21/Falcao/falcao2.html">Combinatorial Identities Associated with a Multidimensional Polynomial Sequence</a>, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
%F T(n,k) = Sum_(i=0)^(n+m-1) [((-1)^i)*C(m,i)*C(n-1,k-i))], n>=1, with T(n,0) = (-1)^m and m=9.
%e Trapezoid begins:
%e -1, 9, -36, 84, -126, 126, -84, 36, -9, 1;
%e -1, 8, -27, 48, -42, 0, 42, -48, 27, -8, 1;
%e -1, 7, -19, 21, 6, -42, 42, -6, -21, 19, -7, 1;
%e -1, 6, -12, 2, 27, -36, 0, 36, -27, -2, 12, -6, 1;
%e -1, 5, -6, -10, 29, -9, -36, 36, 9, -29, 10, 6, -5, 1;
%e -1, 4, -1, -16, 19, 20, -45, 0, 45, -20, -19, 16, 1, -4, 1;
%e -1, 3, 3, -17, 3, 39, -25, -45, 45, 25, -39, -3, 17, -3, -3, 1;
%e etc.
%t Flatten[Table[CoefficientList[(x - 1)^9 (x + 1)^n, x], {n, 0, 7}]] (* _T. D. Noe_, Oct 25 2013 *)
%t m=9; Table[If[k == 0, (-1)^m, Sum[(-1)^(j+m)*Binomial[m, j]*Binomial[n-1, k-j], {j, 0, n+m-1}]], {n, 1, 10}, {k, 0, n+m-1}]//Flatten (* _G. C. Greubel_, Nov 28 2018 *)
%o (PARI) m=9; for(n=1, 10, for(k=0, n+m-1, print1(if(k==0, (-1)^m, sum(j=0, n+m-1, (-1)^(j+m)*binomial(m,j)*binomial(n-1,k-j))), ", "))) \\ _G. C. Greubel_, Nov 28 2018
%o (Magma) m:=9; [[k le 0 select (-1 )^m else (&+[(-1)^(j+m)* Binomial(m,j) *Binomial(n-1,k-j): j in [0..(n+m-1)]]): k in [0..(n+m-1)]]: n in [1..10]]; // _G. C. Greubel_, Nov 28 2018
%o (Sage) m=9; [[sum((-1)^(j+m)*binomial(m,j)*binomial(n-1,k-j) for j in range(n+m)) for k in range(n+m)] for n in (1..10)] # _G. C. Greubel_, Nov 28 2018
%Y Using row j of the alternating Pascal triangle as generator: A007318 (j=0), A008482 and A112467 (j=1 after the first term in each), A182533 (j=2 after the first two rows), A230206-A230211 (j=3 to j=8).
%K easy,sign,tabf
%O 1,2
%A _Dixon J. Jones_, Oct 12 2013
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