a(8) = 212 because 212 is the smallest deficient number that begins a run of 8 consecutive deficient numbers (212, 213, 214, 215, 217, 218, 219, 221) with the same number of distinct prime factors, i.e. 2.
a(9) = 295 because 295 is the smallest deficient number that begins a run of 9 consecutive deficient numbers (295, 296, 297, 298, 299, 301, 302, 303, 305) with the same number of distinct prime factors, i.e. 2.
