The sequence of larger least prime divisors of Smarandache numbers Sm(n)

Let Sm(n)=$1234567891011...$n.Numbers from concatenating the first n natural numbers ($123456789...$n)is also called a Smarandache number. And now consider this sequence: numbers n greater than one, such that Sm(n) create a new record for larger least prime divisor for all $1<m<=n$. The sequence begins, n=$2$,$3$,$7$,$61$,$121$,$133$...,with the corresponding least prime divisors of $2$,$3$,$127$,$10386763$,$278240783$,$8223519074965787731$. What is the next term after $133$ ?. I've checked Fleuren(?)'s factorizations of Smarandache numbers Sm(n), and other sites but I couldn't find the next term after $133$. I couldn't find the next term myself because of factoring problem,but I'm quite sure that the next term is greater than $319$. Can you find the next term(s) after $133$ ?

• Link. – Lucian Nov 17 '15 at 18:16
• @Lucian, that site has a limit for the number of digits. – Takuma Sannz Nov 17 '15 at 18:31

The next term is $133$ with the least prime divisor of $8 223 519 074 965 787 731$. And the next term after $133$ is greater than 319