A230208 - OEIS

%I #33 Sep 08 2022 08:46:06

%S -1,5,-10,10,-5,1,-1,4,-5,0,5,-4,1,-1,3,-1,-5,5,1,-3,1,-1,2,2,-6,0,6,

%T -2,-2,1,-1,1,4,-4,-6,6,4,-4,-1,1,-1,0,5,0,-10,0,10,0,-5,0,1,-1,-1,5,

%U 5,-10,-10,10,10,-5,-5,1,1,-1,-2,4,10,-5,-20,0,20,5

%N Trapezoid of dot products of row 5 (signs alternating) with sequential 6-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 6-tuples (C(5,0), -C(5,1), ..., -C(5,5)) and (C(n-1,k-5), C(n-1,k-4), ..., C(n-1,k)), n >= 1, 0 <= k <= n+4.

%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)^5 (x-1)^(n-1), n > 0.

%H G. C. Greubel, <a href="/A230208/b230208.txt">Rows n=1..50 of trapezoid, 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=5.

%e Trapezoid begins:

%e   -1,  5, -10, 10,  -5,   1;

%e   -1,  4,  -5,  0,   5,  -4,  1;

%e   -1,  3,  -1, -5,   5,   1, -3,  1;

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

%e   -1,  1,   4, -4,  -6,   6,  4, -4, -1,  1;

%e   -1,  0,   5,  0, -10,   0, 10,  0, -5,  0, 1;

%e   -1, -1,   5,  5, -10, -10, 10, 10, -5, -5, 1, 1;

%e etc.

%t Flatten[Table[CoefficientList[(x - 1)^5 (x + 1)^n, x], {n, 0, 7}]] (* _T. D. Noe_, Oct 25 2013 *)

%t m=5; 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=5; 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:=5; [[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=5; [[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-A230207 (j=3 and j=4), A230209-A230212 (j=6 to j=9).

%K easy,sign,tabf

%O 1,2

%A _Dixon J. Jones_, Oct 12 2013