Value a(6) in doubt for a =

[Maximilian Hasler]

PS: from my previous post retain only the %N / %F, the other comment
is irrelevant since I read "sets S that achieve n" instead of "achieve
a(n)".
Sorry...
Maximilian

On Jan 15, 2008 8:41 AM, Maximilian Hasler <maximilian.hasler at gmail.com> wrote:
> In http://www.research.att.com/~njas/sequences/A101877 ,
> which could be more concisely be defined by:
>
> %N or
> %F A101877 a(n) = min { max S | S c N*, sum( 1/x, x in S) = n }
>
> njas writes:
>
> Little is known about the number of different sets S that achieve
> a(n). Paul Hanna asks if it is always true that a solution set S for
> n+1 must necessarily contain a solution set for n as a subset. This is
> true for small n, apparently, but seems to me unlikely to hold in
> general. - njas, Dec 31 2005
>
> ME:
> The number of sets S such that sum( 1/x, x in S) = n  is clearly infinite :
> one can always write the last fraction (1/k = 1/a(n) initially) as a
> sum of two different reciprocals with greater denominator :
> 1 / k = 1/[ k+1 ] + 1/[ k (k+1) ]
>
> Maximilian
>
> On Jan 14, 2008 12:46 PM, Rainer Rosenthal <r.rosenthal at web.de> wrote:
> > The values of
>
> > are defined as  a(n) = max(S_n) where the sum of
> > reciprocals of S_n is n. It is claimed that no S_n
> > with smaller maximum could be found.
> >
> > I checked
> >
> >         ReciprocalSum(S_n) = n
> >
> > for n=1..6 and it was correct for n=1..5 but incorrect
> > for n=6:
> >
> >     ReciprocalSum(S_6) is less than 6 (about 5.76).
> >
> > I wasn't able to guess what's wrong with S_6 but following
> > a remark about the inclusion of S_n in S_{n+1} I checked
> > whether S_5 was a subset of the S_6, given there. An no,
> > it is not. S_6 lacks {80, 112, 144, 154, 165, 175, 176},
> > but adding these is not enough.
> >
> > How to repair the description of A101877 and confirm
> > a(6) = 469?
> >
> > Rainer Rosenthal
> > r.rosenthal at web.de
> >
> >
> >
> >
>
>
>
>
> --
> Maximilian F. Hasler (Maximilian.Hasler(AT)gmail.com)
>



-- 
Maximilian F. Hasler (Maximilian.Hasler(AT)gmail.com)