sum of unit fractions
hv at crypt.org
hv at crypt.org
Wed Dec 8 01:08:06 CET 2004
Define a(n) as the least integer k such that there is a sum of distinct unit
fractions equal to _n_ of which the greatest denominator is k.
Alternatively:
a(n) = k implies that there exists a set of positive integers S such that
sum{1/s_i} = n, and max(S) = k, and no set S' exists with the same sum
but a smaller maximal element.
eg a(1) = 1 (S = { 1 })
a(2) = 6 (S = { 1 2 3 6 })
a(3) = 24 (S = { 1 2 3 4 5 6 8 9 10 15 18 20 24 })
Proof of a(3): for k = 24, we have a candidate set of 1 .. 24, of which the
prime powers greater than 12 can immediately be discarded as unusable;
the multiples of 11 are unavailable since no partial sum of [ 2/1, 2/2 ]
is divisible by 11; the multiples of 7 are unavailable since no partial
sum of [ 6/1, 6/2, 6/3 ] is divisible by 7.
That leaves the candidate set as S U { 12 }, with a sum of 3 1/12. It
immediately follows that S is a candidate set with the right sum; further,
since the greatest denominator is 24, no two fractions from this candidate
set can be <= 1/12, so no candidate set with a lower maximal element can
give the required sum.
Working by hand, I believe I've established a(4) = 65, with the set {
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 18 20 22 24
26 27 28 30 33 35 36 40 42 45 48 52 54 56 60 63 65
} from which the only optional discards are { 21 39 44 55 }.
I'd appreciate if someone could help verify this set is a) correct and
b) minimal, and possibly further extend the sequence; I can't think of
any easy way to search efficiently for these minimal sets by program,
but that's the direction I'll be aiming.
There is no sequence currently in the database matching 1, 6, 24, 65.
Hugo van der Sanden
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