Date: Fri, 22 Mar 2002 10:09:38 +0000
Comment from: Claude Lenormand, <hlne.lenormand(AT)voonoo.net>
Subject: A055881

Using left mirrors operating on the word 12345,
you get 21345, 32145, 43215, 54321.

Beginning with 12345, you want to built all the 120 permutations,
proeding from one to the next by only one mirror operating on a left factor
(as an hamiltonian cycle)

Here is a bijection f, of integers written in factorial numeration, on permutations:

f(1)=12.345, f(01)=213.45, f(11)=31.245, f(02)=132.45, f(12)=23.145,
f(001)=3214.5, 
f(101)=41.235, f(011)=142.35, f(111)=24.135, f(021)=421.35,
f(121)=12.435, f(002)=2143.5,
f(102)=34.125, f(012)=431.25, f(112)=13.425, f(022)=314.25,
f(122)=41.325, f(003)=1432.5,
f(103)=23.415, f(013)=324.15, f(113)=42.315, f(023)=243.15,
f(123)=34.215, f(0001)=43215.,
f(1001)=51.234, f(0101)=152.34,.... and so on.

The sequences A055881 gives the number of zeros, plus one, on the left,
when integers n>0 are written according to the factorial numeration system,
n=c(1).1!+c(2).2!....+c(k).k!+....
(=c(0)c(1)...c(k)....)

So the permutation following f(021)=421.35 is f(121)=12.435
because 021 beginning by one zeros,
we operate a mirror on the left factor of length 3=1+2.