On Mon, 15 Jan 2024, Mikael Sundqvist wrote:

> Hi,
>
> you can try something like
>
> \sum_{\mstack{k=0, k\equiv p + 1 (\mtext{mod }2)}}^{p -1}
>
> but it will not be too pretty with such a large sub-index to the sum.

There is also

\sum_{\startsubstack \NC a \NR \NC b \NR \stopsubstack}

which imitates the \substack command from latex.

OT but it is better to use one of \mod, \pmod, \bmod, rather than explicit \mtext{mod }.

Aditya
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