shoelaces and aglets

[Jack Brennen]

Neil Fernandez wrote:
> I have submitted the sequence as:
> 
> %I A000001
> %S A000001 1,2,2,3,3,4,5,6,6,7,8,9,10,11,11,12,13,14,15
> %N A000001 Aglet pairs to be picked up, given n shoelaces, so that
> probably none remain
> %C A000001 Assistance in extending sequence given by Gerald McGarvey
> <Gerald.McGarvey at comcast.net>
> %e A000001 a(2)=2 because given 2 shoelaces, p=1/3 that the first two
> aglets picked up will be on a single shoelace, requiring another pick-
> up, and p=2/3 that they won't, so the mean no. of pick-ups is (1/3)*2 +
> (2/3)*1 = 4/3, for which the ceiling is 2.

I think that you may need to either change the example or the
description.  As originally stated, you proposed that the sequence is
the minimum number of aglet pairs to pull to have at least a 50% chance
of pulling all shoelaces.  But then you calculate the mean number of
aglet pairs to pull.  It seems to me that you've described the median,
but then calculated the mean...

  Jack