Permutation Its Own Inverse?

[franktaw at netscape.net]

Yes, it is (its own inverse - it isn't all that easily answered).

It is easiest to subtract out the triangular numbers, making each row 
its own permutation:

1
2,1
3,2,1
4,2,3,1
5,3,2,4,1
6,3,2,4,5,1
7,4,3,2,5,6,1
...

We can now see the rule: if ceiling(n/ceiling(n/k)) = k, then T(n,k) = 
ceiling(n/k); otherwise T(n,k) = k.  Equivalently, for each pair i, j 
with i*j >= n, but (i-1)*j and i*(j-1) < n, we have T(n,i) = j, T(n,j) 
= i.  For any other k, T(n,k) = k.

Franklin T. Adams-Watters


-----Original Message-----
From: Leroy Quet <qq-quet at mindspring.com>

[By the way, I am sending to old email address. Does new seq.fan 
address
work yet?]

Form the triangle where a(n,m) is the (ceiling(n/m))th integer from 
among
those positive integers not occurring earlier in the (concatenated)
sequence.

So we have the triangle beginning:

1
3,2
6,5,4
10,8,9,7
15,13,12,14,11
21,18,17,19,20,16
28,25,24,23,26,27,22

Now, this may be an easily answered question, but, is this sequence 
(the
triangle read by rows: 1,3,2,6,5,4,10,8,9,7,...) its own inverse
permutation of the positive integers?

thanks,
Leroy Quet