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%I #27 Sep 08 2022 08:46:06
%S 1,-6,15,-20,15,-6,1,1,-5,9,-5,-5,9,-5,1,1,-4,4,4,-10,4,4,-4,1,1,-3,0,
%T 8,-6,-6,8,0,-3,1,1,-2,-3,8,2,-12,2,8,-3,-2,1,1,-1,-5,5,10,-10,-10,10,
%U 5,-5,-1,1,1,0,-6,0,15,0,-20,0,15,0,-6,0,1,1,1,-6
%N Trapezoid of dot products of row 6 (signs alternating) with sequential 7-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 7-tuples (C(6,0), -C(6,1), ..., -C(6,5), C(6,6)) and (C(n-1,k-6), C(n-1,k-5), ..., C(n-1,k)), n >= 1, 0 <= k <= n+5.
%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)^6 (x+1)^(n-1).
%H G. C. Greubel, <a href="/A230209/b230209.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+m)*C(m,i)*C(n-1,k-i), n >= 1, with T(n,0) = (-1)^m and m=6.
%e Trapezoid begins:
%e 1, -6, 15, -20, 15, -6, 1;
%e 1, -5, 9, -5, -5, 9, -5, 1;
%e 1, -4, 4, 4, -10, 4, 4, -4, 1;
%e 1, -3, 0, 8, -6, -6, 8, 0, -3, 1;
%e 1, -2, -3, 8, 2, -12, 2, 8, -3, -2, 1;
%e 1, -1, -5, 5, 10, -10, -10, 10, 5, -5, -1, 1;
%e 1, 0, -6, 0, 15, 0, -20, 0, 15, 0, -6, 0, 1;
%e etc.
%t Flatten[Table[CoefficientList[(x - 1)^6 (x + 1)^n, x], {n, 0, 7}]] (* _T. D. Noe_, Oct 25 2013 *)
%t m=6; 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=6; 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:=6; [[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=6; [[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-A230208 (j=3 to j=5), A230210-A230212 (j=7 to j=9).
%K easy,sign,tabf
%O 1,2
%A _Dixon J. Jones_, Oct 12 2013
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