The smallest sequence is 1 ... 10 which has lowest common multiple of 2,520 and we note that
2520 = (2^3)(3^2)(5^1)(7^1).
Taking the exponents (3, 2, 1, 1), adding 1 to each gives (4, 3, 2, 2) which when multiplied gives 48 as the number of divisors of 2520. We now look to add to the exponents so that the same product gives 144.
Adding a new prime the product of (6,3,2,2,2) gives 144. Reducing each by 1 gives (5,2,1,1,1) which when attached to the primes gives (2^5)(3^2)(5^1)(7^1)(11^1) = 110,880 as the smallest n. |