Ternary Egyptian fractions with prime denominator
Abstract
For a prime number p, let A3(p)= | \ m ∈ N: ∃ m1,m2,m3 ∈ N, mp=1m1+1m2+1m3 \ |. In 2019 Luca and Pappalardi proved that x ( x)3 Σp x A3(p) x ( x)5. We improve the upper bound, showing Σp x A3(p) x ( x)3 ( x)2.
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