Explicit formula for orthonormalized vectors (using Gram-Schmidt) For $n\in\mathbb{N}$, let
$$
B(n):=\{b_1,b_2,\ldots,b_n\}:=\left\{\begin{pmatrix}1\\2\\3\\4\\\vdots\\n\end{pmatrix},\begin{pmatrix}2\\-1\\0\\0\\\vdots\\0\end{pmatrix},\begin{pmatrix}3\\0\\-1\\0\\\vdots\\0\end{pmatrix},\begin{pmatrix}4\\0\\0\\-1\\\vdots\\0\end{pmatrix},\ldots,\begin{pmatrix}n\\0\\0\\0\\\vdots\\-1\end{pmatrix}\right\}.
$$
Now I would like to orthogonalize (by Gram-Schmidt) and then normalize $B(n)$. 
Question: Is it possible to determine the orthonormalized vectors explicitly?

I write the vectors $b_k\in B(n)~(k=1,2,\ldots,n)$ as 
$$
b_k=(b_{k,1},b_{k,2},\ldots,b_{k,k},b_{k,k+1},\ldots,b_{k,n})^T. 
$$
where $b_{k,1}=k$ and $b_{k,k}=-1$ while $b_{k,i}=0$ otherwise. Hence, for $k,j\geq 2$ and $k\neq j$, I note that $b_k\cdot b_j= kj$ and $b_k\cdot b_k=k^2+1$.
Moreoever, since $b_k\cdot b_1=0$ for all $k\geq 2$, the orthogonalized vectors, which I denote by $C(n):=\{c_1,c_2,\ldots,c_n\}$, are
$$
\begin{align*}
c_1&:=b_1,\\
c_2&:=b_2,\\
c_k&:=b_k-\frac{2k}{5}b_2-\sum_{i=3}^{k-1}\frac{c_i\cdot b_k}{c_i\cdot c_i}c_i,\quad k\geq 3
\end{align*}
$$
And if we normalize, $d_k:=\frac{c_k}{\sqrt{c_k\cdot c_k}}$ one has the desired set $D(n):=\{d_1,d_2,\ldots,d_n\}$ of orthonormalized vectors.
For the first three vectors, what I get is:
$$
\begin{align*}
d_1&=\frac{b_1}{\sqrt{b_1\cdot b_1}}=\frac{b_1}{\sqrt{\frac{1}{6}(n(n+1)(2n+1))}}=\frac{1}{\sqrt{\frac{n(n+1)(2n+1)}{6}}}\begin{pmatrix}1\\2\\3\\\vdots\\n\end{pmatrix}\\
d_2&=\frac{b_2}{\sqrt{b_2\cdot b_2}}=\frac{b_2}{\sqrt{5}}=\begin{pmatrix}\frac{2}{\sqrt{5}}\\-\frac{1}{\sqrt{5}}\\0\\\vdots\\0\end{pmatrix}\\
d_3&=\frac{5}{\sqrt{70}}c_3,\quad c_3=b_3-\frac{6}{5}b_2=\begin{pmatrix}\frac{3}{5}\\\frac{6}{5}\\-1\\0\\\vdots\\0\end{pmatrix}\\
d_4&=\frac{c_4}{\sqrt{c_4\cdot c_4}},\quad c_4=b_4-\frac{8}{5}b_2-\frac{c_3\cdot b_4}{c_3\cdot c_3}c_3=?
\end{align*}
$$
But I don't see a way to reach explicit formulae for $d_j~(j=1,2,3,\ldots,n)$ in general. 

It seems that 
$$v_k=(1,2,3,\ldots,k-1,-x_k,0,\ldots,0)$$
works for constants $x_k$ that are easy to work out by dotting with $b_1$.  Then normalize each $v_k$.  If $v_k$ is normal to $v_1$ then it is normal to $v_h$ for all $h\gt k$
