Thanks a lot, Aditya.
On Mon, Jan 15, 2024 at 8:41 PM Aditya Mahajan <adit...@umich.edu> wrote:
> 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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--
Respect,
Shiv Shankar Dayal
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