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A230207
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.
3
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, -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, 3, -1, -11, -6, 14, 14, -6, -11, -1, 3, 1, 1, 4, 2, -12, -17, 8
OFFSET
1,2
COMMENTS
The array is trapezoidal rather than triangular because C(n,k) is not uniquely defined for all negative n and negative k.
Row sums are 0.
Coefficients of (x-1)^4 (x+1)^(n-1) for n > 0.
LINKS
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
FORMULA
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.
EXAMPLE
Trapezoid begins:
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, -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;
etc.
MATHEMATICA
Flatten[Table[CoefficientList[(x - 1)^4 (x + 1)^n, x], {n, 0, 7}]] (* T. D. Noe, Oct 25 2013 *)
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 *)
PROG
(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
(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
(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
CROSSREFS
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).
Sequence in context: A155675 A365942 A365947 * A277949 A244081 A279445
KEYWORD
easy,sign,tabf
AUTHOR
Dixon J. Jones, Oct 12 2013
STATUS
approved