[seqfan] Re: Sharing no digit business ((ADD)END(UM))

[franktaw at netscape.net]

This is no coincidence; it is easily established by induction.  Let n 
be the smallest integer such that a(a(n))  != n; call a(n) m.  
Manifestly n and m have no digit in common, so there are two cases:

1) n is the value of some a(k) with k < m.  But then k would have been 
a preferred choice for a(n).

2) a(m) = k, with k < n.  But then by induction, we must have a(a(k)) = 
k, and so a(k) = m.  But this contradicts a(n) = m.

Note that the only property of the "shares no digit" relation used in 
this proof is that it is symmetric, so this result also applies in a 
number of similar cases.

Franklin T. Adams-Watters

-----Original Message-----
From: David Wilson <davidwwilson at comcast.net>

....

1. a(n) = smallest unseen positive integer such that a(n) shares no 
digit with n.
....

I also found something rather amusing. As far as I calculated (1), I 
found that a(a(n)) = n.