silly but fun (recreational purposes only) - Matrix Inverse and Moebius Function?

[wouter meeussen]

Hi Paul,

have a happy (transition to) 2006! --- other readers: same!
('though I myself don't believe in calenders vs. happiness..)

Yes, up to n=65, the first column of the transpose *equals* the MoebiusMu!
Keen eyes, you have!
And the matrix inverse *does* look too regular, you're right.

So, 't might be worthwhile after all?

W.

----- Original Message ----- 
From: "Paul D Hanna" <pauldhanna at juno.com>
To: <wouter.meeussen at pandora.be>
Cc: <seqfan at ext.jussieu.fr>
Sent: Friday, December 30, 2005 10:35 PM
Subject: Re: silly but fun (recreational purposes only) - Matrix Inverse and Moebius Function?


Wouter,
       I find the matrix inverse of your triangle interesting ...
It appears to have the Moebius function A008683 in column 1 ...
at least for the few initial terms in column 1:
[1,-1,-1,0,-1,1,-1,0,0,1,-1,0,-1,1,1,0,-1].

Coincidence?   Given the definition, I tend to think column 1 is indeed A008683.
Would you like to verify that the trend continues?

Other patterns seem to emerge in the matrix inverse as well.
Paul

1;
-1, 1;
-1, -1, 1;
0, -2, -1, 1;
-1, 0, -2, -1, 1;
1, -1, -1, -2, -1, 1;
-1, 1, -1, -1, -2, -1, 1;
0, 0, 2, -2, -1, -2, -1, 1;
0, 1, -1, 2, -2, -1, -2, -1, 1;
1, -1, 3, 0, 1, -2, -1, -2, -1, 1;
-1, 1, -1, 3, 0, 1, -2, -1, -2, -1, 1;
0, 2, 2, 0, 4, -1, 1, -2, -1, -2, -1, 1;
-1, 0, 2, 2, 0, 4, -1, 1, -2, -1, -2, -1, 1;
1, -3, 1, 3, 4, 1, 3, -1, 1, -2, -1, -2, -1, 1;
1, 2, -2, 2, 2, 4, 1, 3, -1, 1, -2, -1, -2, -1, 1;
0, 2, 1, -1, 3, 4, 5, 0, 3, -1, 1, -2, -1, -2, -1, 1;
-1, 0, 2, 1, -1, 3, 4, 5, 0, 3, -1, 1, -2, -1, -2, -1, 1;


-- "wouter meeussen" <wouter.meeussen at pandora.be> wrote:
...
what is the sequence that remains unchanged under the operation: "replace every integer k by the
(sorted) sequence of divisors of k+1"? What is the length of this list at step n?
----------------------------------------------------------
...
invariant of Flatten[# /. k_ :>Divisors[1+k] ]

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

length of list after n steps: (ziltch from SuperSeeker)
{1, 2, 4, 9, 19, 43, 94, 210, 464, 1035, 2295, 5111, 11352,
25259, 56145, 124888, 277669}

it even produces a triangular table if you're in the mood for digit-counting: how many ones, two's,
three's .. at each step?
{1},
{1,1},
{2,1,1},
{4,3,1,1},
{9,5,3,1,1},
{19,13,6,3,1,1},
{43,26,14,6,3,1,1},
{94,61,29,15,6,3,1,1},
{210,130,68,30,15,6,3,1,1},
{464,297,146,71,31,15,6,3,1,1},
{1035,648,331,152,72,31,15,6,3,1,1},
{2295,1457,727,347,155,73,31,15,6,3,1,1},
{5111,3215,1628,759,353,156,73,31,15,6,3,1,1},
{11352,7184,3602,1704,774,356,157,73,31,15,6,3,1,1},
{25259,15923,8038,3765,1736,780,357,157,73,31,15,6,3,1,1},
{56145,35482,17832,8411,3839,1751,783,358,157,73,31,15,6,3,1,1},
{124888,78794,39713,18647,8571,3870,1757,784,358,157,73,31,15,6,3,1,1}

with T[n+1,1]= row-sum of n-th row (of course).
Tail end backwards: 1,1,3,6,15,31,73,157,358,.. does SuperSeek something,
and had me screaming for help earlier-on.
Would it then be worth its weight in bits after all?