Algorithms >Divisor Problem?

You can use reverse approach. Since n-th triangle number can be found as (n^2 + n)/2, you can just iterate n and for each number count its divisors. Some optimizations: (n^2+n)/2 = n(n+1)/2.

N and n+1 do not have any common divisors (except for 1), and only one of them is even. Therefore the number of divisors is either multiplied number of divisors of n/2 and n+1, or multiplied number of divisors of n and (n+1)/2 number of divisors can be found by the formula you mentioned, therefore you only need a list of prime numbers (get it here for example) This approach seems to be a bit more straightforward and optimal. Moreover, it guarantees that you will find the first triangle number.

You can use reverse approach. Since n-th triangle number can be found as (n^2 + n)/2, you can just iterate n and for each number count its divisors. Some optimizations: (n^2+n)/2 = n(n+1)/2.

N and n+1 do not have any common divisors (except for 1), and only one of them is even. Therefore the number of divisors is either multiplied number of divisors of n/2 and n+1, or multiplied number of divisors of n and (n+1)/2. Number of divisors can be found by the formula you mentioned, therefore you only need a list of prime numbers (get it here, for example) This approach seems to be a bit more straightforward and optimal.

Moreover, it guarantees that you will find the first triangle number.

You can use reverse approach. Since n-th triangle number can be found as (n^2 + n)/2, you can just iterate n and for each number count its divisors. Some optimizations.

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