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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A319861 Triangle read by rows, 0 <= k <= n: T(n,k) is the numerator of the k-th Bernstein basis polynomial of degree n evaluated at the interval midpoint t = 1/2; denominator is A319862.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 3, 1, 1, 1, 3, 1, 1, 1, 5, 5, 5, 5, 1, 1, 3, 15, 5, 15, 3, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 1, 7, 7, 35, 7, 7, 1, 1, 1, 9, 9, 21, 63, 63, 21, 9, 9, 1, 1, 5, 45, 15, 105, 63, 105, 15, 45, 5, 1, 1, 11, 55, 165, 165, 231, 231, 165, 165, 55, 11, 1
Offset: 0

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In Computer-Aided Geometric Design, the affine combination Sum_{k=0..n} (T(n,k)/A319862(n,k))*P_k is the halfway point for the Bézier curve of degree n defined by the control points P_k, k = 0, 1, ..., n.

Examples

			Triangle begins:
  1;
  1, 1;
  1, 1,  1;
  1, 3,  3,  1;
  1, 1,  3,  1,   1;
  1, 5,  5,  5,   5,  1;
  1, 3, 15,  5,  15,  3,   1;
  1, 7, 21, 35,  35, 21,   7,  1;
  1, 1,  7,  7,  35,  7,   7,  1,  1;
  1, 9,  9, 21,  63, 63,  21,  9,  9, 1;
  1, 5, 45, 15, 105, 63, 105, 15, 45, 5, 1;
  ...
		

Crossrefs

Programs

  • GAP
    Flat(List([0..11],n->List([0..n],k->Binomial(n,k)/Gcd(Binomial(n,k),2^n)))); # Muniru A Asiru, Sep 30 2018
    
  • Maple
    a:=(n,k)->binomial(n,k)/gcd(binomial(n,k),2^n): seq(seq(a(n,k),k=0..n),n=0..11); # Muniru A Asiru, Sep 30 2018
  • Mathematica
    T[n_, k_] = Binomial[n, k]/GCD[Binomial[n, k], 2^n];
    tabl[nn_] = TableForm[Table[T[n, k], {n, 0, nn}, {k, 0, n}]];
  • Maxima
    T(n,k) := binomial(n, k)/gcd(binomial(n, k), 2^n)$
    tabl(nn) := for n:0 thru nn do print(makelist(T(n, k), k, 0, n))$
    
  • Sage
    flatten([[numerator(binomial(n,k)/2^n) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jul 19 2021

Formula

T(n, k) = numerator of binomial(n,k)/2^n.
T(n, k) = binomial(n,k)/A082907(n,k).
T(n, k)/A319862(n,k) = binomial(n,k)/2^n.
T(n, n-k) = T(n,k).
T(n, 0) = 1.
T(n, 1) = A000265(n) (with offset 0, following Peter Luschny's formula).
T(n, 2) = A069834(n-1), n > 1.
Sum_{k=0..n} 2*k*T(n,k)/A319862(n,k) = n.
Sum_{k=0..n} 2*k^2*T(n,k)/A319862(n,k) = A000217(n).