Parent and childs of a full d-node tree i have a full d-node tree (by that mean a tree that each node has exactly d nodes as kids). 
My question is, if i get a random k node of this tree, in which position do i get his kids and his parent?
For example, if i have a full binary tree, the positions that i can find the parent,left and right kid of the k node are $\dfrac k2, 2k, 2k+1$ respectively.
Thanks in advance.

It looks like you're starting numbering at 1 for the root, and numbering "left to right" on each level/depth.
If the root has depth $0$, then there are $d^{t}$ nodes with depth $t$ from the root in a full $d$-dimensional tree. Also, the depth of node $k$ is $\ell_{k}=\lceil\log_{d}(k-1)\rceil$.
The number of nodes at depths below the depth of node $k$, then, is
$$n_{k} = \sum_{i = 0}^{\ell_{k-1}}d^{i} = \frac{d^{\ell_{k}}-1}{d-1}.$$
So indexing on the row containing node $k$ starts at $n_{k}+1$.
Children of node $k$:
The position of $k$ in its row is just $p_{k}=k-n_{k}$. The children of node $k$ have depth one more than that of $k$, and in their respective row, the first child $c$ has position $d(p_{k}-1)+1=d(k-n_{k}-1)+1$, so the $j^{th}$ child of $k$ has position
$$j + \sum_{i = 0}^{\ell_{k}+1}d^{i} + d(k-\sum_{i = 0}^{\ell_{k}}d^{i}-1) = j + \frac{d^{\ell_{k}+1}-1}{d-1} + d(k-\frac{d^{\ell_{k}}-1}{d-1}-1)$$
$$=j +dk-d +1$$
Parent of node $k$:
Since this formula applies for the parent $p$ of $k$, for $1 \leq j \leq d$
$$1+dp-d +1\leq k \leq d+dp - d + 1 = dp+1$$
$$\Rightarrow dp - d \leq k-1 \leq dp$$
$$\Rightarrow p-1 \leq \frac{k-1}{d} \leq p$$
so $p = \lceil\frac{k-1}{d}\rceil$.
