= Solution
Uniform knots give translation invariance $N_{i,k}(x)=N_{0,k}(x-i)$. Put $v_{k,j}=N_{0,k}(j)$ and set missing values to zero. The recurrence in part (a) gives the exact numerical recursion
$$
\boxed{v_{k,j}=\frac{jv_{k-1,j}+(k-j)v_{k-1,j-1}}{k-1}.}
$$
The order-two starting row is $(0,1,0)$. For example, the order-four middle value is $(2\cdot\frac12+2\cdot\frac12)/3=2/3$, and the order-six central value is $(3\cdot\frac{11}{24}+3\cdot\frac{11}{24})/5=66/120$. Iteration gives the requested array, listing $N_{0,k}(0),\ldots,N_{0,k}(k)$:
$$
\boxed{\begin{array}{c|rrrrrrr}
2&0&1&0&&&&\\
3&0&\frac12&\frac12&0&&&\\
4&0&\frac16&\frac46&\frac16&0&&\\
5&0&\frac1{24}&\frac{11}{24}&\frac{11}{24}&\frac1{24}&0&\\
6&0&\frac1{120}&\frac{26}{120}&\frac{66}{120}&\frac{26}{120}&\frac1{120}&0
\end{array}}
$$
The symmetry follows from reflection of the uniform <Cardinal B-spline> about the midpoint of its support. The row sums equal one, consistent with sampling the uniform partition of unity. These are partition-normalized values; normalizing each spline to have peak one would produce a different array.
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