Matrices of Paul D Hanna

[Roland Bacher]

Here a small remark on matrices considered by Paul D Hanna
one week ago. Sorry for the Latex-style:



Let $M_b$ be a matrix as considered by Hanna. One has hence
$$\tilde m_{i,j}=m_{i,j}$$ 
where $\tilde M=M_b^b$ with coefficients
$\tilde m_{i,j}$ is the $b-$th power of the matrix $M_b$ with
coefficients $m_{i,j}$. 

Let $L_b=-\sum_{k=1} (-1)^k\frac{(M-I)^k}{k}$ be the matrix logarithm of 
$M_b$. One has then clearly 
$$l_{i+k,j+k}=b^kl_{i,j}$$
and $l_{i,j}=0$ if $i\leq j$.

The matrix-logarithm $L_b$ of $M_b$
is hence completely determined by its first column-vector
$\gamma_i=l_{i+1,1}, \ i=1,2,\dots$.

The matrix $L_{-1}$ seems particularly nice:

Consider the recursive sequence 
$$\beta_1=1,\ \beta_{2n}=\frac{-1}{2}\beta_{2n-1},\ \beta_{2n+1}=
\frac{-n}{2n+1}\beta_{2n}$$
starting as
$$1,\ \frac{-1}{2},\ \frac{1}{6},\ \frac{-1}{12},\ \frac{1}{30},\
\frac{-1}{60},\ \frac{1}{140},\dots$$
For $b=-1$ one has then seemingly
$$\gamma_n=\beta_n-2\sum_{k=1}^{n-1}(-1)^{n+k}\gamma_k$$
with first terms
$$1,\ \frac{3}{2},\ \frac{7}{6},\ \frac{5}{4},\ \frac{73}{60},\
\frac{169}{140},\ \frac{1}{140},\dots$$
(by the way, this series seems to be convergent).

The associated Hanna-matrix $M_{-1}$ seems also to be very simple:
$m_{i,j}=1$ if $j$ is odd, $m_{2i,2i}=1,\ m_{2i+1,2i}=-1,\ m_{2i+2,2i}=2$
and  $m_{2i+k,2i}=-1$ for $k>2$.

Another particularly simple case (where all can actually be proven)
is $L_1$ where $\gamma_i=l_{i+1,1}=1/i$ and $m_{i,j}=1$ for
$i\geq j$ (this is obvious). 

In the original case $b=2$, one seems to have $\gamma_{2i}=0$ for
all $i$.

In general, the sequence $\gamma_i(b)$ is a polynomial in $b$ with degree
$\leq {i\choose 2}$.


Best whishes    Roland Bacher


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Original message of Paul D Hanna

   Consider the set of lower triangular matrices that satisfy 
the peculiar exponential property: 
   M^b = M excluding the first row and column;
then by construction,
   M^(b^n) = M excluding the first n rows and n columns,
where integer b>1 and the first column and main diagonal consist of all
1's. 

   These matrices are interesting in themselves, but there may be other 
relations hidden in them, as the following conjecture would indicate.

Conjecture:
   The sum of the n-th row of the above matrix M is equal to 
the partitions of b^n into powers of b.  

   Can anyone prove this?  If true, it would be remarkable.  
Listed here are some examples at b=2 and b=3.

(b=2)  Sequence A002577 (pasted below) lists the "Partitions of 2^n 
into powers of 2". 
   Benoit Cloitre pointed out to me that it appears that this sequence 
A002577 forms the sums of the rows of the lower triangular matrix M, 
b=2, (A078121 - pasted below), which satisfies:
   M^2 = M excluding first row and column. 

   Benoit also prompted me to look for a similar occurrence for b=3 and
the 
partitions of 3^n into powers of 3; it seems that Benoit's hunch was
right. 

(b=3)  It appears that the sequence A078125 (pasted below) is equal to 
the "partitions of 3^n into powers of 3" (Not in EIS), which begin 
1,2,5,23,239,... as in 
a(0)=1: {1}, 
a(1)=2: {3; 1+1+1}, 
a(2)=5: {9; 3+3+3; 3+3+1+1+1; 3+1+1+1+1+1+1; 1+1+1+1+1+1+1+1+1}, 
... 
Of course, these are the coefficients of x^(3^n) in the power series 
expansion of 1/[(1-x)(1-x^3)(1-x^9)...(1-x^(3^k))...].

   And the sequence A078125 is the sum of the rows of the lower 
triangular matrix M, b=3, (A078122 - pasted below), which satisfies: 
   M^3 = M excluding first row and column. 

   Could someone verify that the sequence A078125 is indeed equal to 
the partitions of 3^n into powers of 3?  

 Thanks, 
  Paul D Hanna (pauldhanna at juno.com) 
-------------------------------------------------------------------

ID Number: A078121
Sequence:  1,1,1,1,2,1,1,4,4,1,1,10,16,8,1,1,36,84,64,16,1,1,202,656,
           680,256,32,1,1,1828,8148,10816,5456,1024,64,1,1,27338,
           167568,274856,174336,43680,4096,128,1,1,692004,5866452,
           11622976,8909648,2794496,349504
Name:      Infinite lower triangular matrix, M, that satisfies 
              [M^2](i,j) = M(i+1,j+1) for all i,j>=0 
              where [M^n](i,j) denotes the element at row i, column j, 
              of the n-th power of matrix M, with M(0,k)=1 and M(k,k)=1
              for all k>=0.
Comments:  M also satisfies: [M^(2k)](i,j) = [M^k](i+1,j+1) for all
              i,j,k>=0; thus [M^(2^n)](i,j) = M(i+n,j+n) for all n>=0.
Example:   The square of the matrix is the same matrix excluding the 
              first row and column:
           [1,_0,_0,0,0]^2=[_1,_0,_0,_0,0]
           [1,_1,_0,0,0]___[_2,_1,_0,_0,0]
           [1,_2,_1,0,0]___[_4,_4,_1,_0,0]
           [1,_4,_4,1,0]___[10,16,_8,_1,0]
           [1,10,16,8,1]___[36,84,64,16,1]
-------------------------------------------------------------------

ID Number: A002577 (Formerly M1239 and N0473)
Sequence:  1,2,4,10,36,202,1828,27338,692004,30251722,2320518948,
           316359580362,77477180493604,34394869942983370,
           27893897106768940836,41603705003444309596874,
           114788185359199234852802340,588880400923055731115178072778
Name:      Partitions of 2^n into powers of 2.
-------------------------------------------------------------------

ID Number: A078125
Sequence:  1,2,5,23,239,5828,342383,50110484,18757984046,
           18318289003448
Name:      First column of matrix A078123, which is the square of the 
       infinite lower triangular matrix A078122 that shifts 
       left and up when cubed.
Example:   Square of A078122 = A078123 as can be seen by 4x4 submatrix:
           [1,_0,_0,0]^2=[_1,_0,_0,_0]
           [1,_1,_0,0]___[_2,_1,_0,_0]
           [1,_3,_1,0]___[_5,_6,_1,_0]
           [1,12,_9,1]___[23,51,18,_1]
-------------------------------------------------------------------

ID Number: A078122
Sequence:  1,1,1,1,3,1,1,12,9,1,1,93,117,27,1,1,1632,3033,1080,81,1,1,
           68457,177507,86373,9801,243,1,1,7112055,24975171,15562314,
           2371761,88452,729,1,1,1879090014,8786827629,6734916423,
           1291958181,64392813,796797
Name:      Infinite lower triangular matrix, M, that satisfies 
       [M^3](i,j) = M(i+1,j+1) for all i,j>=0 
       where [M^n](i,j) denotes the element at row i, column j, 
       of the n-th power of matrix M, with M(0,k)=1 and M(k,k)=1
              for all k>=0.
Comments:  M also satisfies: [M^(3k)](i,j) = [M^k](i+1,j+1) for all
              i,j,k>=0; thus [M^(3^n)](i,j) = M(i+n,j+n) for all n>=0.

Example:   [1,_0,_0,0]^3=[_1,__0,_0,_0]
           [1,_1,_0,0]___[_3,__1,_0,_0]
           [1,_3,_1,0]___[12,__9,_1,_0]
           [1,12,_9,1]___[93,117,27,_1]

Further examples of powers of triangular matrix M of A078122:

M=
1,
1,1,
1,3,1,
1,12,9,1,
1,93,117,27,1,
1,1632,3033,1080,81,1,
1,68457,177507,86373,9801,243,1,
1,7112055,24975171,15562314,2371761,88452,729,1,
...

M^2=
1,
2,1,
5,6,1,
23,51,18,1,
239,861,477,54,1,
5828,32856,25263,4347,162,1,
342383,3013980,3016107,699813,39285,486,1,
50110484,690729981,865184724,253656252,19053063,354051,1458,1,
...

M^3 = M excluding first row and column =
1,
3,1,
12,9,1,
93,117,27,1,
1632,3033,1080,81,1,
68457,177507,86373,9801,243,1,
7112055,24975171,15562314,2371761,88452,729,1,
...

M^6 = M^2 excluding first row and column =
1,
6,1,
51,18,1,
861,477,54,1,
32856,25263,4347,162,1,
3013980,3016107,699813,39285,486,1,
690729981,865184724,253656252,19053063,354051,1458,1,
...

M^9 = M^3 excluding first row and column =
1,
9,1,
117,27,1,
3033,1080,81,1,
177507,86373,9801,243,1,
24975171,15562314,2371761,88452,729,1,
...
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----- End of forwarded message from Paul D Hanna -----