Thanks a lot, Aditya.
On Mon, Jan 15, 2024 at 8:41 PM Aditya Mahajan 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
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}
It works nicely. You are a lifesaver. Thanks a lot.
On Mon, Jan 15, 2024 at 7:30 PM 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.
>
> /Mikael
>
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.
/Mikael
On Mon, Jan 15, 2024 at 2:41 PM Shiv Shankar Dayal
wrote:
>
> I have following formula
> \sum_{k=0\\ k\equiv p +
