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A230206
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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.
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7
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-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, 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, -14, -8, 3, 4, 1, -1, -5, -7, 5, 22, 14, -14, -22, -5, 7
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OFFSET
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1,2
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COMMENTS
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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)^3 (x+1)^(n-1) for n > 0.
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LINKS
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FORMULA
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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.
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EXAMPLE
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Trapezoid begins
-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, 1, 1;
-1, -2, 2, 6, 0, -6, -2, 2, 1;
-1, -3, 0, 8, 6, -6, -8, 0, 3, 1;
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MATHEMATICA
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Flatten[Table[CoefficientList[(x - 1)^3 (x + 1)^n, x], {n, 0, 7}]] (* T. D. Noe, Oct 25 2013 *)
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 *)
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PROG
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(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
(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
(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
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CROSSREFS
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KEYWORD
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easy,sign,tabf
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AUTHOR
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STATUS
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approved
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