Permutation Its Own Inverse?
franktaw at netscape.net
franktaw at netscape.net
Thu Jul 13 01:22:56 CEST 2006
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
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