15 Jan
2024
15 Jan
'24
4:08 p.m.
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