proposed

approved

editing

proposed

The first pair of consecutive integers that are both terms is (728, 729), and the first run of three consecutive integers all appearing as terms begins at 127251 (see Examples). The first run of four consecutive integers appearing as terms begins at 4405832. Do arbitrarily long runs of consecutive integers occur?

The first pair of consecutive integers that are both terms is (728, 729), and the first run of three consecutive integers all appearing as terms begins at 127251 (see Examples). The first run of four consecutive integers appearing as terms begins at 4405832. Do arbitrarily long runs of integers occur?

A run of four consecutive integers that are terms begins at 4405832.

A run of four consecutive integers that are terms begins at 4405832.

allocated for Jon E. Schoenfield

Lesser of two consecutive integers such that one has more prime factors (counted with multiplicity), but the other has more divisors.

495, 728, 729, 975, 1071, 1424, 1616, 1700, 1862, 1875, 2024, 2144, 2223, 2349, 2384, 2415, 2624, 2996, 3104, 3124, 3125, 3159, 3184, 3483, 3663, 4095, 4130, 4292, 4304, 4335, 4779, 4976, 5103, 5312, 5427, 5535, 5589, 5624, 5775, 6224, 6416, 6544, 6560, 6704

1,1

495 = 3*3*5*11 has 4 prime factors and 12 divisors, while 496 = 2*2*2*2*31 has 5 prime factors but only 10 divisors, so 495 is a term.

728 = 2*2*2*7*13, 729 = 3*3*3*3*3*3, and 730 = 2*5*73 have 5, 6, and 3 prime factors and 16, 7, and 8 divisors, respectively, so both 728 and 729 are terms.

127251 = 3*3*3*3*1571, 127252 = 2*2*29*1097, 127253 = 7*7*7*7*53, and 127254 = 2*3*127*167 have 5, 4, 5, and 4 prime factors, and 10, 12, 10, and 16 divisors, respectively, so 127251, 127252, and 127253 are all terms.

Cf. A000005, A001222.

allocated

nonn

Jon E. Schoenfield, May 14 2023

approved

editing

allocated for Jon E. Schoenfield

recycled

allocated

allocated for Jack Braxton

allocated

recycled

allocated for Jack Braxton

allocated

approved