cp's OEIS Frontend

A230208 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.

%I A230208 #33 Sep 08 2022 08:46:06
%S A230208 -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 A230208 -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 A230208 5,-10,-10,10,10,-5,-5,1,1,-1,-2,4,10,-5,-20,0,20,5
%N A230208 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 A230208 The array is trapezoidal rather than triangular because C(n,k) is not uniquely defined for all negative n and negative k.
%C A230208 Row sums are 0.
%C A230208 Coefficients of (x-1)^5 (x-1)^(n-1), n > 0.
%H A230208 G. C. Greubel, <a href="/A230208/b230208.txt">Rows n=1..50 of trapezoid, flattened</a>
%H A230208 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 A230208 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 A230208 Trapezoid begins:
%e A230208   -1,  5, -10, 10,  -5,   1;
%e A230208   -1,  4,  -5,  0,   5,  -4,  1;
%e A230208   -1,  3,  -1, -5,   5,   1, -3,  1;
%e A230208   -1,  2,   2, -6,   0,   6, -2, -2,  1;
%e A230208   -1,  1,   4, -4,  -6,   6,  4, -4, -1,  1;
%e A230208   -1,  0,   5,  0, -10,   0, 10,  0, -5,  0, 1;
%e A230208   -1, -1,   5,  5, -10, -10, 10, 10, -5, -5, 1, 1;
%e A230208 etc.
%t A230208 Flatten[Table[CoefficientList[(x - 1)^5 (x + 1)^n, x], {n, 0, 7}]] (* _T. D. Noe_, Oct 25 2013 *)
%t A230208 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 A230208 (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 A230208 (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 A230208 (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 A230208 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 A230208 easy,sign,tabf
%O A230208 1,2
%A A230208 _Dixon J. Jones_, Oct 12 2013