$47$-smooth Triangular Numbers

A number is $p$-smooth if it has no prime factors larger than $p$.
Let $T$ be the sequence of triangular numbers, i.e. $T(n)=n(n+1)/2$.
Find the sum of all indices $n$ such that $T(n)$ is $47$-smooth.

This problem is quite challenging as it requires a good understanding of number theory.

A triangular number $T(n)$ is the sum of the positive integers up to $n$ described by the formula $T(n)=n(n+1)/2$.

A number is called $p$-smooth if it does not include any prime factors larger than $p$.

So, to solve this problem, we need to find all numbers $n$ for which the triangular number $T(n)$ is $47$-smooth, i.e., its prime factorization contains only primes that are less than or equal to $47$ and then sum up all these $n$’s.

Firstly, for a number to be 47-smooth, take note that the prime factorization of $n(n+1)/2$ would not contain any prime factors greater than 47. This considerably narrows down the search as we know the primes less than or equal to 47 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47.

Notice that 48 is $T(9)$ and $T(10)$, and we need $n(n+1)/2 < 48$ in order for $n(n+1)/2$ to be $47$-smooth. Thus, $n\leq9$. Manually checking the numbers 1 to 9, we find that $T(1),$ $T(2),$ $T(3),$ $T(4),$ $T(6),$ and $T(7)$ are $47$-smooth. Summing these indices we get $1 + 2 + 3 + 4 + 6 + 7 = 23$. So, the sum of all indices $n$ such that $T(n)$ is $47$-smooth is $23$.

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