Notes on A092671 Jon E. Schoenfield Apr 09 2017 Some examples: 6 is in the sequence because 1/2 + 1/3 + 1/6 = 1. (Note that the prime factorization of 6 is 2*3, and if we start with 1/6, adding 1/3 yields 1/2, which removes the factor 3 from the denominator; then adding 1/2 removes the 2.) 23 cannot be in the sequence because it is a prime: for any positive integer j1 < 23, 1/j1 + 1/23 = (23 + j1)/(23*j1), which cannot be reduced; adding another 1/j2 to the sum (with j2 < 23) will give (23*(j1 + j2) + j1*j2)/(23*j1*j2), from which the factor of 23 in the denominator still cannot be removed by reduction (since 23 does not divide j1*j2, so 23 cannot divide the numerator); and adding any further reciprocals of integers < 23 similarly cannot remove the factor of 23 from the denominator. 25 cannot be in the sequence because it is a prime power: for any positive integer j1 < 25, 1/j1 + 1/25 = (25 + j1)/(25*j1), which cannot be reduced unless 5 divides j1, but even then the denominator would remain divisible by 25 and, as above, this would continue to be the case after the addition any number of reciprocals of other integers < 25). 28 is in the sequence, and a corresponding set of integers can be constructed as follows: 28 = 2^2 * 7, so we will need a set of distinct integers, all divisible by 7, with 28 as the largest, whose sum of reciprocals does not have 7 in the denominator, and the only such sets are {21, 28} and {7, 14, 28}. The latter set gives a simpler sum: 1/7 + 1/14 + 1/28 = 1/4, from which we can reach 1 simply by adding to it 1/4 + 1/2. (We could also use the former set, whose sum is 1/21 + 1/28 = 1/12, to which we could, e.g., add 1/12 to reach 1/6, and then, making use of the above solution for n=6, add 1/3 + 1/2.) 50, although composite, is not in the sequence; 50 = 2*5^2, and the only numbers <= 50 and divisible by 5^2 are 25 and 50, and 1/25 + 1/50 = 3/50, so there is no set of distinct positive integers (having 50 as the largest) whose sum of reciprocals will not have 5^2 remaining in the denominator (thus no way to reach a sum of 1). 75 = 3*5^2 is in the sequence; 1/50 + 1/75 = 1/30 (removing one of the powers of 5 from the denominator); 1/30 + 1/20 = 1/12 (removing the other); and 1/12 + 1/12 = 1/6 (removing one power of 2 from 12 = 2^2 * 3), after which we can reuse the above solution for n=6. Additional notes: For an integer k > 1, let S(k) be the set of fractions that can be obtained by summing 1/k and the reciprocals of one or more distinct positive integers < k. We then have S(2) = {3/2}, S(3) = {5/6, 4/3, 11/6}, S(4) = {7/12, 3/4, 13/12, 5/4, 19/12, 7/4, 25/12}, etc. The prime number 11 is a divisor of one of the numerators in S(3) (it arises from 1/1 + 1/2 + 1/3 = 11/6), but not in S(2). Thus, n = 22 = 11*2 cannot be in A092671, but for n = 33 = 11*3, we can use 1/1 + 1/2 + 1/3 = 11/6 and thus (1/11) * (1/1 + 1/2 + 1/3) = (1/11) * (11/6) = 1/6 i.e., 1/11 + 1/22 + 1/33 = 1/6 to which we can add the reciprocals of two not-yet-used integers, 2 and 3, to reach a sum of 1. Similarly, the prime 19 is not a divisor of any of the numerators in S(2) or S(3), but it does divide one of the numerators in S(4) (the one resulting from 1/1 + 1/3 + 1/4 = 19/12). Thus, neither 19*2 = 38 nor 19*3 = 57 is in A092671, but for 19*4 = 76, we can use 1/1 + 1/3 + 1/4 = 19/12 and thus (1/19) * (1/1 + 1/3 + 1/4) = (1/19) * (19/12) = 1/12 i.e., 1/19 + 1/57 + 1/76 = 1/12 and we can get from the sum 1/12 up to 1 by adding, e.g., the not-yet-used reciprocals 1/6 (to get to 1/4), 1/4 (to get to 1/2), and 1/2 (to get to 1). For numbers n with no relatively large primes (or for proceeding towards a sum of 1 from some partial sum obtained after steps like those in the examples above), a full solution can often be obtained in several steps by adding, at each step, the reciprocal of the largest not-yet-used integer < n that would bring the sum to a unit fraction whose denominator's largest prime factor (LPF) is either less than that of the existing sum or the same as that of the existing sum, but with a smaller multiplicity. For example, for n = 630 = 2 * 3^2 * 5 * 7 (LPF = 7): 1/630 + 1/350 = 1/225 (225 = 3^2 * 5^2; LPF decreased to 5) 1/630 + 1/350 + 1/400 = 1/144 (144 = 2^4 * 3^2; LPF decreased to 3) 1/630 + 1/350 + 1/400 + 1/288 = 1/96 (96 = 2^5 * 3; smaller multiplicity) 1/630 + 1/350 + 1/400 + 1/288 + 1/192 = 1/64 (64 = 2^6; LPF decreased to 2) (and to the partial sum 1/64 we can then add the reciprocals of the unused integers 64, 32, 16, 8, 4, and 2 to reach a final sum of 1).