A320085 Triangle read by rows, 0 <= k <= n: T(n,k) is the numerator of the derivative of the k-th Bernstein basis polynomial of degree n evaluated at the interval midpoint t = 1/2; denominator is A320086.
0, -1, 1, -1, 0, 1, -3, -3, 3, 3, -1, -1, 0, 1, 1, -5, -15, -5, 5, 15, 5, -3, -3, -15, 0, 15, 3, 3, -7, -35, -63, -35, 35, 63, 35, 7, -1, -3, -7, -7, 0, 7, 7, 3, 1, -9, -63, -45, -63, -63, 63, 63, 45, 63, 9, -5, -5, -135, -15, -105, 0, 105, 15, 135, 5, 5
Offset: 0
Examples
Triangle begins: 0; -1, 1; -1, 0, 1; -3, -3, 3, 3; -1, -1, 0, 1, 1; -5, -15, -5, 5, 15, 5; -3, -3, -15, 0, 15, 3, 3; -7, -35, -63, -35, 35, 63, 35, 7; -1, -3, -7, -7, 0, 7, 7, 3, 1; -9, -63, -45, -63, -63, 63, 63, 45, 63, 9; -5, -5, -135, -15, -105, 0, 105, 15, 135, 5, 5; ...
Links
- G. C. Greubel, Rows n = 0..50 of the triangle, flattened
- Rita T. Farouki, The Bernstein polynomial basis: A centennial retrospective, Computer Aided Geometric Design Vol. 29 (2012), 379-419.
- Ron Goldman, Pyramid Algorithms. A Dynamic Programming Approach to Curves and Surfaces for Geometric Modeling, Morgan Kaufmann Publishers, 2002, Chap. 5.
- Eric Weisstein's World of Mathematics, Bernstein Polynomial
- Wikipedia, Bernstein polynomial
Programs
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Maple
T:=proc(n,k) n*(binomial(n-1,k-1)-binomial(n-1,k))/gcd(n*(binomial(n-1,k-1)-binomial(n-1,k)),2^(n-1)); end proc: seq(seq(T(n,k),k=0..n),n=0..11); # Muniru A Asiru, Oct 06 2018
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Mathematica
Table[Numerator[n*(Binomial[n-1, k-1] - Binomial[n-1, k])/2^(n-1)], {n, 0, 12}, {k, 0, n}]//Flatten
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Maxima
T(n, k) := n*(binomial(n - 1, k - 1) - binomial(n - 1, k))/gcd(n*(binomial(n - 1, k - 1) - binomial(n - 1, k)), 2^(n - 1))$ tabl(nn) := for n:0 thru nn do print(makelist(T(n, k), k, 0, n))$
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Sage
def A320085(n,k): return numerator(n*(binomial(n-1, k-1) - binomial(n-1, k))/2^(n-1)) flatten([[A320085(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jul 19 2021
Formula
T(n, k) = n*(binomial(n-1, k-1) - binomial(n-1, k))/gcd(n*(binomial(n-1, k-1) - binomial(n-1, k)), 2^(n-1)).
T(n, n-k) = -T(n,k).
T(n, 0) = -n.
T(2*n+1, 1) = -A000466(n).
T(2*n, 1) = -A069834(n-1), n > 1.
Sum_{k=0..n} k*T(n,k)/A320086(n,k) = n.
Sum_{k=0..n} k^2*T(n,k)/A320086(n,k) = n^2.
Sum_{k=0..n} k*(k-1)*T(n,k)/A320086(n,k) = n*(n - 1).
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