Four three-digit numbers are in arithmetic progression, and the number of factors of each is also in arithmetic progression. In fact that second arithmetic progression has a constant difference of 1, so that each successive number in the original arithmetic progression has one more factor than the number before it.
Note: These are not the prime factors of the numbers, but rather any factor, including the number itself and 1, so, for example, 46 has four factors (1, 2, 23 and 46), as does 8 (1, 2, 4 and 8).
What is the original arithmetic progression of three-digit numbers?
The sequence 229, 361, 493, and 625 differ by 132; their respective factor counts are 2, 3, 4, and 5 (i.e. increasing by one).
Staightforward: I generated a table of the factor counts of all numbers between 1 and 999. I then considered each starting number N1 from 1 to 996, and for each gap G from 1 to 332 (if N+(G*3) < 1000); then for each of those calculated the other numbers from that start, i.e. N2=N1+G, N3=N2+G, and N4=N3+G. I then looked up the factor counts F(x) for N1, N2, N3, and N4, and selected if F(N2) = F(N1)+1; F(N3)=F(N2)+1, and F(N4)=F(N3)+1. This yielded the single solution above.
Clearly a task for a computer program.