1,4

The row sums are zero.

If you divide these by n a combinatorial like triangle sequence results:

{Indeterminate},

{-1, 1},

{-1, 0, 1},

{-1, -1, 1, 1},

{-1, -2, 0, 2, 1},

{-1, -3, -2, 2, 3, 1},

{-1, -4, -5,0, 5, 4, 1},

{-1, -5, -9, -5, 5, 9, 5, 1},

{-1, -6, -14, -14, 0, 14, 14, 6, 1},

{-1, -7, -20, -28, -14, 14, 28, 20, 7, 1},

{-1, -8, -27, -48, -42, 0, 42, 48, 27,8, 1}

Weisstein, Eric W. "Bernstein Polynomial." http : // mathworld.wolfram.com/BernsteinPolynomial.html

p(x,n,m)=Binomial[n,m]*x^m*(1-x)^(n-m); t(n,m)=d/dx*p(x,n,m) such that x=1/2.

{0},

{-1, 1},

{-2, 0, 2},

{-3, -3, 3, 3},

{-4, -8, 0, 8, 4},

{-5, -15, -10, 10, 15, 5},

{-6, -24, -30, 0, 30, 24, 6},

{-7, -35, -63, -35, 35, 63, 35, 7},

{-8, -48, -112, -112, 0, 112, 112, 48, 8},

{-9, -63, -180, -252, -126, 126, 252, 180, 63, 9},

{-10, -80, -270, -480, -420, 0, 420, 480, 270, 80, 10}

Clear[BernstenB, n, i, t]; BernsteinB[0, n_, 0] := 1; BernsteinB[n_, n_, t_] := t^n; BernsteinB[i_, n_, t_] := Binomial[n, i]t^i(1 - t)^(n - i); Table[Table[2^(n - 1)*(D[BernsteinB[i, n, t], t] /. t -> 1/2), {i, 0, n}], {n, 0, 10}]; Flatten[%]

Cf. A123948.

Sequence in context: A002125 A135356 A003987 this_sequence A063180 A141693 A028376

Adjacent sequences: A141689 A141690 A141691 this_sequence A141693 A141694 A141695

uned,sign

Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Sep 09 2008