All the numbers $1, 6, 3, 10$ are the triangular numbers but just rearranged. We can make a pattern like so to generate these numbers:

$$0 + 1 = 1\tag*{$T_1$}$$ $$0 + 1 + 2 = 3\tag*{$T_2$}$$ $$0 + 1 + 2 + 3 = 6\tag*{$T_3$}$$ $$0 + 1 + 2 + 3 + 4 = 10\tag*{$T_4$}$$

By this reason, I believe that

the next number is also a triangular number.

Now the sequence $1, 6, 3, 10$ is ordered as $T_1, T_3, T_2, T_4$ which has the pattern

$T_{(n)}, T_{(n+2)}, T_{(n)+1}, T_{(n+2)+1}$ for $n = 1$.

Therefore, the next answer is

$T_{(n)+2}$ which when $n = 1$, we have that $T_{(n)+2} = T_{1+2} = T_3 = 6$.

If this pattern continued, the next triangular number would be $T_{(n+2)+2} = 21$, however my reasoning does not explain why we have to add certain numbers to form this pattern.

All the numbers $1, 6, 3, 10$ are the triangular numbers but just rearranged. We can make a pattern like so to generate these numbers:

$$0 + 1 = 1\tag*{$T_1$}$$ $$0 + 1 + 2 = 3\tag*{$T_2$}$$ $$0 + 1 + 2 + 3 = 6\tag*{$T_3$}$$ $$0 + 1 + 2 + 3 + 4 = 10\tag*{$T_4$}$$

By this reason, I believe that

the next number is also a triangular number.

Now the sequence $1, 6, 3, 10$ is ordered as $T_1, T_3, T_2, T_4$ which has the pattern

$T_{(n)}, T_{(n+2)}, T_{(n)+1}, T_{(n+2)+1}$ for $n = 1$.

Therefore, the next answer is

$T_{(n)+2}$ which when $n = 1$, we have that $T_{(n)+2} = T_{1+2} = T_3 = 6$ which has exactly four factors, namely $1$, $2$, $3$ and $6$.

If this pattern continued, the next triangular number would be $T_{(n+2)+2} = 21$, however my reasoning does not explain why we have to add certain numbers to form this pattern.