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%I #30 Sep 08 2022 08:46:06
%S 1,-4,6,-4,1,1,-3,2,2,-3,1,1,-2,-1,4,-1,-2,1,1,-1,-3,3,3,-3,-1,1,1,0,
%T -4,0,6,0,-4,0,1,1,1,-4,-4,6,6,-4,-4,1,1,1,2,-3,-8,2,12,2,-8,-3,2,1,1,
%U 3,-1,-11,-6,14,14,-6,-11,-1,3,1,1,4,2,-12,-17,8
%N Trapezoid of dot products of row 4 (signs alternating) with sequential 5-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 5-tuples (C(4,0), -C(4,1), C(4,2), -C(4,3), C(4,4)) and (C(n-1,k-4), C(n-1,k-3), C(n-1,k-2), C(n-1,k-1), C(n-1,k)), n >= 1, 0 <= k <= n+3.
%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)^4 (x+1)^(n-1) for n > 0.
%H G. C. Greubel, <a href="/A230207/b230207.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=4.
%e Trapezoid begins:
%e 1, -4, 6, -4, 1;
%e 1, -3, 2, 2, -3, 1;
%e 1, -2, -1, 4, -1, -2, 1;
%e 1, -1, -3, 3, 3, -3, -1, 1;
%e 1, 0, -4, 0, 6, 0, -4, 0, 1;
%e 1, 1, -4, -4, 6, 6, -4, -4, 1, 1;
%e 1, 2, -3, -8, 2, 12, 2, -8, -3, 2, 1;
%e etc.
%t Flatten[Table[CoefficientList[(x - 1)^4 (x + 1)^n, x], {n, 0, 7}]] (* _T. D. Noe_, Oct 25 2013 *)
%t m=4; 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 29 2018 *)
%o (PARI) m=4; 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 29 2018
%o (Magma) m:=4; [[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 29 2018
%o (Sage) m=4; [[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 29 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 (j=3), A230208-A230212 (j=5 to j=9).
%K easy,sign,tabf
%O 1,2
%A _Dixon J. Jones_, Oct 12 2013
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