$g_1 \geq \dots \geq g_{m+s}$. We say that ${\mathbf g}$ is majorized by ${\mathbf d}$ and ${\mathbf a}$ $({\mathbf g}\prec' ({\mathbf d},{\mathbf a}))$ if $$\label{gmaj1}
d_i\geq g_{i+s}, \quad 1\leq i\leq m,$$ $$\label{gmaj2}
\sum_{i=1}^{h_j}g_i-\sum_{i=1}^{h_j-j}d_i\leq \sum_{i=1}^j a_i, \quad 1\leq j\leq s,$$ where $h_j=\min\{i\; : \; d_{i-j+1}<g_i\}, \   1\leq j\leq s$   $(d_{m+1}=-\infty)$, $$\label{gmaj3}
\sum_{i=1}^{m+s}g_i=\sum_{i=1}^md_i+\sum_{i=1}^sa_i.$$

In the case that $s=0$, condition (\[gmaj2\]) disappears, and conditions (\[gmaj1\]) and (\[gmaj3\]) are equivalent to ${\mathbf d}={\mathbf g}$.

In the case that $s=1$, from condition (\[gmaj3\]) we observe that $a_1$ is completely determined by ${\mathbf d}$ and ${\mathbf g}$ ($a_1=\sum_{i=1}^{m+1}g_i-\sum_{i=1}^{m}d_i$), therefore we will write ${\mathbf g}\prec'({\mathbf d}, a_1)$ as ${\mathbf g}\prec'{\mathbf d}$ and we will refer to it as [*$1$step-generalized majorization*]{}. Moreover, it is easy to see that ${\mathbf g}\prec'{\mathbf d}$ if and only if $$d_{i}=g_{i+1}, \quad h \leq i \leq m,$$ where $h=\min\{i: d_i<g_i\}$.

\[aux\]

1.  \[aux1\] If ${\mathbf g}$ and ${\mathbf d}$ satisfy that $g_i\leq d_i$ for $1\leq i \leq m$, then $h=m+1$ and ${\mathbf g}\prec'{\mathbf d}$.

2.  \[aux2\] Notice that if ${\mathbf g}\prec'{\mathbf d}$ and for some index $1\leq i \leq m$ we have $d_i>g_{i+1}$, then $i<h$.

3.  \[aux3\] In [@DoStEJC10], the 1step-generalized majorization is called [*elementary generalized majorization*]{} and it is denoted by ${\mathbf 