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Is it true that if is a finite set with then there is a partition such that
for ?
Source: erdosproblems.com/316
An accepted solution exists. The statement is false.
Disproved. The answer is no: the divisors of other than
and , the set the site attributes to Sándor's 1997 paper, have
reciprocal sum and no admissible split, and the eleven-element
set of the site's commentary (sum
) has none either; both are finite computations rechecked
here with exact arithmetic. Sándor's paper (J. Number Theory 63 (1997),
refereed) is not held, so his statements are recorded second-hand from the
site. The site's label is DISPROVED (LEAN); its Lean qualifier is a catalog
label whose referent is the formal-conjectures file, which proves the disproof by
a decide +kernel step on the eleven-element set; nothing was built
here, and no local kernel credit is claimed (see Formalization). The
accepted claim is recorded on
Sándor's claim page (1997).