Alternatively: sum = \( (1 + 2^2)(1 + 3^2)(1 + 5^0) \) but note: the general formula for sum of square divisors of \( n = \prod p^e_i \) is \( \prod \left( \sum_k=0^\lfloor e_i/2 \rfloor p^2k \right) \). Here, only up to \( c=0 \), so: - Crosslake