Thanks a lot, Aditya.
On Mon, Jan 15, 2024 at 8:41 PM Aditya Mahajan
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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