A230206 - OEIS

%I #44 Sep 08 2022 08:46:06

%S -1,3,-3,1,-1,2,0,-2,1,-1,1,2,-2,-1,1,-1,0,3,0,-3,0,1,-1,-1,3,3,-3,-3,

%T 1,1,-1,-2,2,6,0,-6,-2,2,1,-1,-3,0,8,6,-6,-8,0,3,1,-1,-4,-3,8,14,0,

%U -14,-8,3,4,1,-1,-5,-7,5,22,14,-14,-22,-5,7

%N Trapezoid of dot products of row 3 (signs alternating) with sequential 4-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 4-tuples (C(3,0), -C(3,1), C(3,2), -C(3,3)) and (C(n-1,k-3), C(n-1,k-2), C(n-1,k-1), C(n-1,k)), n >= 1, 0 <= k <= n+2.

%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)^3 (x+1)^(n-1) for n > 0.

%H Dixon J. Jones, <a href="/A230206/b230206.txt">Rows n = 1..100 for irregular triangle, flattened</a>

%H Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and 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=3.

%e Trapezoid begins

%e   -1,  3, -3,  1;

%e   -1,  2,  0, -2,  1;

%e   -1,  1,  2, -2, -1,  1;

%e   -1,  0,  3,  0, -3,  0,  1;

%e   -1, -1,  3,  3, -3, -3,  1, 1;

%e   -1, -2,  2,  6,  0, -6, -2, 2, 1;

%e   -1, -3,  0,  8,  6, -6, -8, 0, 3, 1;

%t Flatten[Table[CoefficientList[(x - 1)^3 (x + 1)^n, x], {n, 0, 7}]] (* _T. D. Noe_, Oct 25 2013 *)

%t m=3; 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=3; 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:=3; [[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=3; [[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), A230207-A230212 (j=4 to j=9).

%K easy,sign,tabf

%O 1,2

%A _Dixon J. Jones_, Oct 11 2013