Value a(6) in doubt for a =

[franktaw at netscape.net]

There is a shortcut available in the calculation of
http://www.research.att.com/~njas/sequences/A101877

For each prime p, calculate a value t(p), which is the minimal
value of the maximum of the non-empty set T(p), where
T(p) is the set whose maximal element is minimized,
and whose reciprocal sum has p in the numerator.  For
example, T(5) = {2,3}, since 1/2 + 1/3 = 5/6.

Now, inclusion of a prime power p^k in S_n forces the
maximum of S_n to be at least p^k*t(p).  This means to
calculate a(n), you can look at, for increasing m, S(m) =
{j s.t. j <= m and
     for all prime powers p^k s.t. p^k|j, p^k*t(p) <= m}.
(This can be done in an incremental manner.)  Until
sum_{j in S, 1/j} >= n, m is not large enough.

T(p) for p>2 can be calculated by first building a table of
reciprocals modulo p, and looking for the first non-empty
sum that sums to 0 modulo p.  (T(2) = {1,3}, the only case
where t(p) > p.)  For p = 5, this table is
1 2 3 4
1 3 2 4
and the first zero sum is 3 + 2, corresponding to {2,3}, as
specified above.

Further, once a candidate S(m) is found, determine whether
(sum_{j in S(m)} 1/j) - n can be represented as a sum of
reciprocals of elements in S(m).  If it can, then S_n is S(m)
with those elements removed.  Note that for most m,
S(m) = S(m-1).  Obviously, it is not necessary to repeat the
calculation for these m.

For example, for n = 3, we have:
S(20) = S(23) = {1,2,3,4,5,6,9,10,12,15,18,20}, whose
reciprocal sum is 2 11/12.  For S(24), we add 8 and 24,
since 8*t(2) = 24, making the sum 3 1/12.  Since 1/12
is the reciprocal sum of {12},
S_3 = S(24) - {12} = {1,2,3,4,5,6,8,9,10,15,18,20,24}.

Franklin T. Adams-Watters

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