The Mobius function Blog

Mahonian numbers modulo 2

Posted on February 25, 2011 by Mats Granvik

mats.granvik(AT)abo.fi

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Recurrence 1:
T(1,1)=1, n>1: T(n,1)=0, k>1: T(n,k) = (0 + (sum from i = 1 to k-1 of T(n-i,k-1))) mod 2.

Recurrence 2:
T(1,1)=1, n>1: T(n,1)=0, k>1: T(n,k) = (1 + (sum from i = 1 to k-1 of T(n-i,k-0))) mod 2.

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Eigensequences and the definition of matrix inversion

Posted on February 19, 2011 by Mats Granvik

Step 1. We want to find the eigentriangle for this lower triangular Toeplitz triangle:

${A=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 2&1&0&0&0&0&0&0\\ 3&2&1&0&0&0&0&0\\ 4&3&2&1&0&0&0&0\\ 5&4&3&2&1&0&0&0\\ 6&5&4&3&2&1&0&0\\ 7&6&5&4&3&2&1&0\\ 8&7&6&5&4&3&2&1\end{array}\right)$

Step 2. We begin by shifting down matrix ${A}$ one step and adding ones to the main diagonal:

${A_{down shifted}=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0\\ 2&1&1&0&0&0&0&0\\ 3&2&1&1&0&0&0&0\\ 4&3&2&1&1&0&0&0\\ 5&4&3&2&1&1&0&0\\ 6&5&4&3&2&1&1&0\\ 7&6&5&4&3&2&1&1\end{array}\right)$

Step 3. We multiply matrix ${A_{down shifted}}$ elementwise with matrix ${B}$

${B=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ -1&1&0&0&0&0&0&0\\ -1&-1&1&0&0&0&0&0\\ -1&-1&-1&1&0&0&0&0\\ -1&-1&-1&-1&1&0&0&0\\ -1&-1&-1&-1&-1&1&0&0\\ -1&-1&-1&-1&-1&-1&1&0\\ -1&-1&-1&-1&-1&-1&-1&1\end{array}\right)$

Step 4. We now get matrix ${C}$ as a result:

${C=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ -1&1&0&0&0&0&0&0\\ -2&-1&1&0&0&0&0&0\\ -3&-2&-1&1&0&0&0&0\\ -4&-3&-2&-1&1&0&0&0\\ -5&-4&-3&-2&-1&1&0&0\\ -6&-5&-4&-3&-2&-1&1&0\\ -7&-6&-5&-4&-3&-2&-1&1\end{array}\right)$

Step 5. Calculate the matrix inverse of matrix ${C}$ to get matrix ${C^-1}$ :

${C^-1=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0\\ 3&1&1&0&0&0&0&0\\ 8&3&1&1&0&0&0&0\\ 21&8&3&1&1&0&0&0\\ 55&21&8&3&1&1&0&0\\ 144&55&21&8&3&1&1&0\\ 377&144&55&21&8&3&1&1\end{array}\right)$

Step 6. Now by the definition of matrix inversion discussed here: The inverse of a triangular matrix using forward substitution, we have: ${C*C^-1=I}$ where ${I}$ is the identity matrix. Written as matrices we then can write:

$\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ -1&1&0&0&0&0&0&0\\ -2&-1&1&0&0&0&0&0\\ -3&-2&-1&1&0&0&0&0\\ -4&-3&-2&-1&1&0&0&0\\ -5&-4&-3&-2&-1&1&0&0\\ -6&-5&-4&-3&-2&-1&1&0\\ -7&-6&-5&-4&-3&-2&-1&1\end{array}\right){*}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0\\ 3&1&1&0&0&0&0&0\\ 8&3&1&1&0&0&0&0\\ 21&8&3&1&1&0&0&0\\ 55&21&8&3&1&1&0&0\\ 144&55&21&8&3&1&1&0\\ 377&144&55&21&8&3&1&1\end{array}\right){=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 0&1&0&0&0&0&0&0\\ 0&0&1&0&0&0&0&0\\ 0&0&0&1&0&0&0&0\\ 0&0&0&0&1&0&0&0\\ 0&0&0&0&0&1&0&0\\ 0&0&0&0&0&0&1&0\\ 0&0&0&0&0&0&0&1\end{array}\right)$

Step 7. Write matrix ${C}$ again and look at the identity for the first column of ${C^-1}$

Step 8. Multiplying the expression on the left of the equal sign we get:

$\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ -1&1&0&0&0&0&0&0\\ -2&-1&3&0&0&0&0&0\\ -3&-2&-3&8&0&0&0&0\\ -4&-3&-6&-8&21&0&0&0\\ -5&-4&-9&-16&-21&55&0&0\\ -6&-5&-12&-24&-42&-55&144&0\\ -7&-6&-15&-32&-63&-110&-144&377\\\end{array}\right){=}\left(\begin{array}{cccccccc}1\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right)$

Step 9. Sum the terms on the left side of the equal sign and notice that they equal the right hand side of the equal sign.

Step 10. Remove the elements in the main diagonal to get matrix ${D}$ :

${D=}\left(\begin{array}{cccccccc}0&0&0&0&0&0&0&0\\ -1&0&0&0&0&0&0&0\\ -2&-1&0&0&0&0&0&0\\ -3&-2&-3&0&0&0&0&0\\ -4&-3&-6&-8&0&0&0&0\\ -5&-4&-9&-16&-21&0&0&0\\ -6&-5&-12&-24&-42&-55&0&0\\ -7&-6&-15&-32&-63&-110&-144&0\end{array}\right)$

Step 11. Change the signs of the elements in matrix ${D}$ to get matrix ${E}$ :

${E=}\left(\begin{array}{ccccccc}1&0&0&0&0&0&0\\ 2&1&0&0&0&0&0\\ 3&2&3&0&0&0&0\\ 4&3&6&8&0&0&0\\ 5&4&9&16&21&0&0\\ 6&5&12&24&42&55&0\\ 7&6&15&32&63&110&144\end{array}\right)$

Matrix E is called an eigentriangle because the last elements in each row equals the row sums of the previous row. This property of the eigentriangle is because of the definition of matrix inversion which was illustrated above. $1, 1, 3, 8, 21, 55, 144, 377$ is called an eigensequence.

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The invert transform, Bell numbers, Pascal triangle, permanents, matrix multiplication and matrix inversion, Catalan numbers.

Posted on February 18, 2011 by Mats Granvik

Mats Granvik mats.granvik(AT)abo.fi and Gary W. Adamson INVERT transform

Consider the Pascal triangle:

${C=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0\\ 1&2&1&0&0&0&0&0\\ 1&3&3&1&0&0&0&0\\ 1&4&6&4&1&0&0&0\\ 1&5&10&10&5&1&0&0\\ 1&6&15&20&15&6&1&0\\ 1&7&21&35&35&21&7&1\end{array}\right)$

Shift down the elements one step and add a diagonal of ones in the main diagonal.

${X=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0\\ 1&1&1&0&0&0&0&0\\ 1&2&1&1&0&0&0&0\\ 1&3&3&1&1&0&0&0\\ 1&4&6&4&1&1&0&0\\ 1&5&10&10&5&1&1&0\\ 1&6&15&20&15&6&1&1\end{array}\right)$

Change the sign of all the elements below the main diagonal. ${X_{signed}}$ is found as a comment here: Oeis table A121207.

${X_{signed}=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ -1&1&0&0&0&0&0&0\\ -1&-1&1&0&0&0&0&0\\ -1&-2&-1&1&0&0&0&0\\ -1&-3&-3&-1&1&0&0&0\\ -1&-4&-6&-4&-1&1&0&0\\ -1&-5&-10&-10&-5&-1&1&0\\ -1&-6&-15&-20&-15&-6&-1&1\end{array}\right)$

Calculate the matrix inverse and you will get:

${A=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0\\ 2&1&1&0&0&0&0&0\\ 5&3&1&1&0&0&0&0\\ 15&9&4&1&1&0&0&0\\ 52&31&14&5&1&1&0&0\\ 203&121&54&20&6&1&1&0\\ 877&523&233&85&27&7&1&1\end{array}\right)$

As we see we have the Bell numbers in the first column. Oeis, Bell numbers. The INVERT transform of any number triangle is equivalent to the steps above, here with the INVERT transform of the Pascal triangle as example.

Repeating the procedure or algorithm (above), infinitely many times, will produce the Catalan numbers as a convergent Oeis Catalan numbers in all columns, regardless of starting triangle (or starting sequence). That is you input matrix ${A}$ into the first step instead of matrix ${C}$ .

${Cat=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0\\ 2&1&1&0&0&0&0&0\\ 5&2&1&1&0&0&0&0\\ 14&5&2&1&1&0&0&0\\ 42&14&5&2&1&1&0&0\\ 132&42&14&5&2&1&1&0\\ 429&132&42&14&5&2&1&1\end{array}\right)$

Another way to calculate the first column in matrix A is by taking matrix powers of this matrix M. Here the elements in the main diagonal have been deleted except for the first element which is equal to 1. Taking matrix powers is simply matrix multiplication repeated, ${M^2=M*M}$ , ${M^3=M*M*M}$ and so on.

${M=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 1&1&0&0&0&0&0&0\\ 1&2&1&0&0&0&0&0\\ 1&3&3&1&0&0&0&0\\ 1&4&6&4&1&0&0&0\\ 1&5&10&10&5&1&0&0\\ 1&6&15&20&15&6&1&0\end{array}\right)$

${M^2=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 2&0&0&0&0&0&0&0\\ 4&1&0&0&0&0&0&0\\ 8&5&1&0&0&0&0&0\\ 16&17&7&1&0&0&0&0\\ 32&49&31&9&1&0&0&0\\ 64&129&111&49&11&1&0&0\end{array}\right)$

${M^3=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 2&0&0&0&0&0&0&0\\ 5&0&0&0&0&0&0&0\\ 14&1&0&0&0&0&0&0\\ 41&9&1&0&0&0&0&0\\ 122&52&12&1&0&0&0&0\\ 365&246&88&15&1&0&0&0\end{array}\right)$

${M^4=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 2&0&0&0&0&0&0&0\\ 5&0&0&0&0&0&0&0\\ 15&0&0&0&0&0&0&0\\ 51&1&0&0&0&0&0&0\\ 187&14&1&0&0&0&0&0\\ 715&121&18&1&0&0&0&0\end{array}\right)$

${M^5=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 2&0&0&0&0&0&0&0\\ 5&0&0&0&0&0&0&0\\ 15&0&0&0&0&0&0&0\\ 52&0&0&0&0&0&0&0\\ 202&1&0&0&0&0&0&0\\ 855&20&1&0&0&0&0&0\end{array}\right)$

${M^6=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 2&0&0&0&0&0&0&0\\ 5&0&0&0&0&0&0&0\\ 15&0&0&0&0&0&0&0\\ 52&0&0&0&0&0&0&0\\ 203&0&0&0&0&0&0&0\\ 876&1&0&0&0&0&0&0\end{array}\right)$

${M^7=}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&0\\ 1&0&0&0&0&0&0&0\\ 2&0&0&0&0&0&0&0\\ 5&0&0&0&0&0&0&0\\ 15&0&0&0&0&0&0&0\\ 52&0&0&0&0&0&0&0\\ 203&0&0&0&0&0&0&0\\ 877&0&0&0&0&0&0&0\end{array}\right)$

As promised we have calculated the Bell numbers in the first column.

Yet another way to calculate the first column in matrix ${A}$ , is to calculate permanents of a modified version of matrix ${X}$ . Here the element in the lower right corner has been swapped with the element in the lower right corner.

${permanent}\left(\begin{array}{cccccccc}1&0&0&0&0&0&0&1\\ 1&1&0&0&0&0&0&0\\ 1&1&1&0&0&0&0&0\\ 1&2&1&1&0&0&0&0\\ 1&3&3&1&1&0&0&0\\ 1&4&6&4&1&1&0&0\\ 1&5&10&10&5&1&1&0\\ 1&6&15&20&15&6&1&0\end{array}\right){=}\begin{array}{c}877\end{array}$

i.e. the last element in the first column of matrix ${A}$ .

Similarly for submatrices:

${permanent}\left(\begin{array}{ccccccc}1&0&0&0&0&0&1\\ 1&1&0&0&0&0&0\\ 1&1&1&0&0&0&0\\ 1&2&1&1&0&0&0\\ 1&3&3&1&1&0&0\\ 1&4&6&4&1&1&0\\ 1&5&10&10&5&1&0\end{array}\right){=}\begin{array}{c}203\end{array}$

i.e. the second last element in the first column of matrix ${A}$ .

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Mobius matrix

Posted on January 12, 2011 by Mats Granvik

According to a contributor in the oeis this could perhaps be called a Mobius matrix:
Mobius matrix

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The inverse of a triangular matrix using forward substitution

Posted on December 21, 2010 by Mats Granvik

This post is about how to invert a lower triangular matrix ${A}$ by forward substitution.

Consider a matrix ${X}$ which is the matrix inverse of another ${n*n}$ sized triangular matrix ${A}$ . We take the definition of matrix inversion as a starting point:

${A*X=I}$

Here ${A*X}$ is the matrix multiplication of ${X}$ times ${A}$ and ${I}$ is the identity matrix, kind of the equivalent of the number 1 of the natural numbers 1,2,3… but as a matrix.
By writing out elements this could for example look like:
$\left(\begin{array}{cccccc}a_{\mathrm{11}}&0&0&0&0&0\\ a_{\mathrm{21}}&a_{\mathrm{22}}&0&0&0&0\\ a_{\mathrm{31}}&a_{\mathrm{32}}&a_{\mathrm{33}}&0&0&0\\ a_{\mathrm{41}}&a_{\mathrm{42}}&a_{\mathrm{43}}&a_{\mathrm{44}}&0&0\\ a_{\mathrm{51}}&a_{\mathrm{52}}&a_{\mathrm{53}}&a_{\mathrm{54}}&a_{\mathrm{55}}&0\\ a_{\mathrm{61}}&a_{\mathrm{62}}&a_{\mathrm{63}}&a_{\mathrm{64}}&a_{\mathrm{65}}&a_{\mathrm{66}}\end{array}\right){*}\left(\begin{array}{cccccc}x_{\mathrm{11}}&x_{\mathrm{12}}&x_{\mathrm{13}}&x_{\mathrm{14}}&x_{\mathrm{15}}&x_{\mathrm{16}}\\ x_{\mathrm{21}}&x_{\mathrm{22}}&x_{\mathrm{23}}&x_{\mathrm{24}}&x_{\mathrm{25}}&x_{\mathrm{26}}\\ x_{\mathrm{31}}&x_{\mathrm{32}}&x_{\mathrm{33}}&x_{\mathrm{34}}&x_{\mathrm{35}}&x_{\mathrm{36}}\\ x_{\mathrm{41}}&x_{\mathrm{42}}&a_{\mathrm{43}}&x_{\mathrm{44}}&x_{\mathrm{45}}&x_{\mathrm{46}}\\ x_{\mathrm{51}}&x_{\mathrm{52}}&x_{\mathrm{53}}&x_{\mathrm{54}}&x_{\mathrm{55}}&x_{\mathrm{51}}\\ x_{\mathrm{61}}&x_{\mathrm{62}}&x_{\mathrm{63}}&x_{\mathrm{64}}&x_{\mathrm{65}}&x_{\mathrm{66}}\end{array}\right){=}\left(\begin{array}{cccccc}1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1\end{array}\right)$

However it is a fact that the inverse of a triangular matrix is also a triangular matrix and therefore we can write the equation as follows:
$\left(\begin{array}{cccccc}a_{\mathrm{11}}&0&0&0&0&0\\ a_{\mathrm{21}}&a_{\mathrm{22}}&0&0&0&0\\ a_{\mathrm{31}}&a_{\mathrm{32}}&a_{\mathrm{33}}&0&0&0\\ a_{\mathrm{41}}&a_{\mathrm{42}}&a_{\mathrm{43}}&a_{\mathrm{44}}&0&0\\ a_{\mathrm{51}}&a_{\mathrm{52}}&a_{\mathrm{53}}&a_{\mathrm{54}}&a_{\mathrm{55}}&0\\ a_{\mathrm{61}}&a_{\mathrm{62}}&a_{\mathrm{63}}&a_{\mathrm{64}}&a_{\mathrm{65}}&a_{\mathrm{66}}\end{array}\right){*}\left(\begin{array}{cccccc}x_{\mathrm{11}}&0&0&0&0&0\\ x_{\mathrm{21}}&x_{\mathrm{22}}&0&0&0&0\\ x_{\mathrm{31}}&x_{\mathrm{32}}&x_{\mathrm{33}}&0&0&0\\ x_{\mathrm{41}}&x_{\mathrm{42}}&a_{\mathrm{43}}&x_{\mathrm{44}}&0&0\\ x_{\mathrm{51}}&x_{\mathrm{52}}&x_{\mathrm{53}}&x_{\mathrm{54}}&x_{\mathrm{55}}&0\\ x_{\mathrm{61}}&x_{\mathrm{62}}&x_{\mathrm{63}}&x_{\mathrm{64}}&x_{\mathrm{65}}&x_{\mathrm{66}}\end{array}\right){=}\left(\begin{array}{cccccc}1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1\end{array}\right)$

We then proceed by solving n different equation systems for each column of ${X}$ . Here ${x_{\mathrm{n}}}$ denotes the n:th column of ${X}$ and the somewhat unorthodox ${i_{\mathrm{n}}}$ denotes the n:th column of the identity matrix ${I}$ .

Equation system 1: ${A*x_{\mathrm{1}}=i_{\mathrm{1}}}$ :

$\left(\begin{array}{cccccc}a_{\mathrm{11}}&0&0&0&0&0\\ a_{\mathrm{21}}&a_{\mathrm{22}}&0&0&0&0\\ a_{\mathrm{31}}&a_{\mathrm{32}}&a_{\mathrm{33}}&0&0&0\\ a_{\mathrm{41}}&a_{\mathrm{42}}&a_{\mathrm{43}}&a_{\mathrm{44}}&0&0\\ a_{\mathrm{51}}&a_{\mathrm{52}}&a_{\mathrm{53}}&a_{\mathrm{54}}&a_{\mathrm{55}}&0\\ a_{\mathrm{61}}&a_{\mathrm{62}}&a_{\mathrm{63}}&a_{\mathrm{64}}&a_{\mathrm{65}}&a_{\mathrm{66}}\end{array}\right){*}\left(\begin{array}{cccccc}x_{\mathrm{11}}\\ x_{\mathrm{21}}\\ x_{\mathrm{31}}\\ x_{\mathrm{41}}\\ x_{\mathrm{51}}\\ x_{\mathrm{61}}\end{array}\right){=}\left(\begin{array}{cccccc}1\\ 0\\ 0\\ 0\\ 0\\ 0\end{array}\right)$

Equation system 2: ${A*x_{\mathrm{2}}=i_{\mathrm{2}}}$ :

$\left(\begin{array}{cccccc}a_{\mathrm{11}}&0&0&0&0&0\\ a_{\mathrm{21}}&a_{\mathrm{22}}&0&0&0&0\\ a_{\mathrm{31}}&a_{\mathrm{32}}&a_{\mathrm{33}}&0&0&0\\ a_{\mathrm{41}}&a_{\mathrm{42}}&a_{\mathrm{43}}&a_{\mathrm{44}}&0&0\\ a_{\mathrm{51}}&a_{\mathrm{52}}&a_{\mathrm{53}}&a_{\mathrm{54}}&a_{\mathrm{55}}&0\\ a_{\mathrm{61}}&a_{\mathrm{62}}&a_{\mathrm{63}}&a_{\mathrm{64}}&a_{\mathrm{65}}&a_{\mathrm{66}}\end{array}\right){*}\left(\begin{array}{cccccc}0\\ x_{\mathrm{22}}\\ x_{\mathrm{32}}\\ x_{\mathrm{42}}\\ x_{\mathrm{52}}\\ x_{\mathrm{62}}\end{array}\right){=}\left(\begin{array}{cccccc}0\\ 1\\ 0\\ 0\\ 0\\ 0\end{array}\right)$

Equation system 3: ${A*x_{\mathrm{3}}=i_{\mathrm{3}}}$ :

$\left(\begin{array}{cccccc}a_{\mathrm{11}}&0&0&0&0&0\\ a_{\mathrm{21}}&a_{\mathrm{22}}&0&0&0&0\\ a_{\mathrm{31}}&a_{\mathrm{32}}&a_{\mathrm{33}}&0&0&0\\ a_{\mathrm{41}}&a_{\mathrm{42}}&a_{\mathrm{43}}&a_{\mathrm{44}}&0&0\\ a_{\mathrm{51}}&a_{\mathrm{52}}&a_{\mathrm{53}}&a_{\mathrm{54}}&a_{\mathrm{55}}&0\\ a_{\mathrm{61}}&a_{\mathrm{62}}&a_{\mathrm{63}}&a_{\mathrm{64}}&a_{\mathrm{65}}&a_{\mathrm{66}}\end{array}\right){*}\left(\begin{array}{cccccc}0\\ 0\\ x_{\mathrm{33}}\\ x_{\mathrm{43}}\\ x_{\mathrm{53}}\\ x_{\mathrm{63}}\end{array}\right){=}\left(\begin{array}{cccccc}0\\ 0\\ 1\\ 0\\ 0\\ 0\end{array}\right)$

Equation system 4: ${A*x_{\mathrm{4}}=i_{\mathrm{4}}}$ :

$\left(\begin{array}{cccccc}a_{\mathrm{11}}&0&0&0&0&0\\ a_{\mathrm{21}}&a_{\mathrm{22}}&0&0&0&0\\ a_{\mathrm{31}}&a_{\mathrm{32}}&a_{\mathrm{33}}&0&0&0\\ a_{\mathrm{41}}&a_{\mathrm{42}}&a_{\mathrm{43}}&a_{\mathrm{44}}&0&0\\ a_{\mathrm{51}}&a_{\mathrm{52}}&a_{\mathrm{53}}&a_{\mathrm{54}}&a_{\mathrm{55}}&0\\ a_{\mathrm{61}}&a_{\mathrm{62}}&a_{\mathrm{63}}&a_{\mathrm{64}}&a_{\mathrm{65}}&a_{\mathrm{66}}\end{array}\right){*}\left(\begin{array}{cccccc}0\\ 0\\ 0\\ x_{\mathrm{44}}\\ x_{\mathrm{54}}\\ x_{\mathrm{64}}\end{array}\right){=}\left(\begin{array}{cccccc}0\\ 0\\ 0\\ 1\\ 0\\ 0\end{array}\right)$

Equation system 5: ${A*x_{\mathrm{5}}=i_{\mathrm{5}}}$ :

$\left(\begin{array}{cccccc}a_{\mathrm{11}}&0&0&0&0&0\\ a_{\mathrm{21}}&a_{\mathrm{22}}&0&0&0&0\\ a_{\mathrm{31}}&a_{\mathrm{32}}&a_{\mathrm{33}}&0&0&0\\ a_{\mathrm{41}}&a_{\mathrm{42}}&a_{\mathrm{43}}&a_{\mathrm{44}}&0&0\\ a_{\mathrm{51}}&a_{\mathrm{52}}&a_{\mathrm{53}}&a_{\mathrm{54}}&a_{\mathrm{55}}&0\\ a_{\mathrm{61}}&a_{\mathrm{62}}&a_{\mathrm{63}}&a_{\mathrm{64}}&a_{\mathrm{65}}&a_{\mathrm{66}}\end{array}\right){*}\left(\begin{array}{cccccc}0\\ 0\\ 0\\ 0\\ x_{\mathrm{55}}\\ x_{\mathrm{65}}\end{array}\right){=}\left(\begin{array}{cccccc}0\\ 0\\ 0\\ 0\\ 1\\ 0\end{array}\right)$

Equation system 6: ${A*x_{\mathrm{6}}=i_{\mathrm{6}}}$ :

$\left(\begin{array}{cccccc}a_{\mathrm{11}}&0&0&0&0&0\\ a_{\mathrm{21}}&a_{\mathrm{22}}&0&0&0&0\\ a_{\mathrm{31}}&a_{\mathrm{32}}&a_{\mathrm{33}}&0&0&0\\ a_{\mathrm{41}}&a_{\mathrm{42}}&a_{\mathrm{43}}&a_{\mathrm{44}}&0&0\\ a_{\mathrm{51}}&a_{\mathrm{52}}&a_{\mathrm{53}}&a_{\mathrm{54}}&a_{\mathrm{55}}&0\\ a_{\mathrm{61}}&a_{\mathrm{62}}&a_{\mathrm{63}}&a_{\mathrm{64}}&a_{\mathrm{65}}&a_{\mathrm{66}}\end{array}\right){*}\left(\begin{array}{cccccc}0\\ 0\\ 0\\ 0\\ 0\\ x_{\mathrm{66}}\end{array}\right){=}\left(\begin{array}{cccccc}0\\ 0\\ 0\\ 0\\ 0\\ 1\end{array}\right)$

Now recall matrix multiplication with this example:
$\left(\begin{array}{cccccc}1&0&0&0&0&0\\ 1&2&0&0&0&0\\ 1&2&3&0&0&0\\ 1&2&3&4&0&0\\ 1&2&3&4&5&0\\ 1&2&3&4&5&6\end{array}\right){*}\left(\begin{array}{cccccc}1\\ 2\\ 3\\ 4\\ 5\\ 6\end{array}\right)$

Which has the intermediary step:
$\begin{array}{ccccccccccccc}1&+&0&+&0&+&0&+&0&+&0&=&1\\ 1&+&4&+&0&+&0&+&0&+&0&=&5\\ 1&+&4&+&9&+&0&+&0&+&0&=&14\\ 1&+&4&+&9&+&16&+&0&+&0&=&30\\ 1&+&4&+&9&+&16&+&25&+&0&=&55\\ 1&+&4&+&9&+&16&+&25&+&36&=&91\end{array}$

That is also how we will rewrite the equations systems:
Equation system 1: ${A*x_{\mathrm{1}}=i_{\mathrm{1}}}$ :
$\begin{array}{ccccccccccccc}a_{\mathrm{11}}*x_{\mathrm{11}}&+&0&+&0&+&0&+&0&+&0&=&1\\ a_{\mathrm{21}}*x_{\mathrm{11}}&+&a_{\mathrm{22}}*x_{\mathrm{21}}&+&0&+&0&+&0&+&0&=&0\\ a_{\mathrm{31}}*x_{\mathrm{11}}&+&a_{\mathrm{32}}*x_{\mathrm{21}}&+&a_{\mathrm{33}}*x_{\mathrm{31}}&+&0&+&0&+&0&=&0\\ a_{\mathrm{41}}*x_{\mathrm{11}}&+&a_{\mathrm{42}}*x_{\mathrm{21}}&+&a_{\mathrm{43}}*x_{\mathrm{31}}&+&a_{\mathrm{44}}*x_{\mathrm{41}}&+&0&+&0&=&0\\ a_{\mathrm{51}}*x_{\mathrm{11}}&+&a_{\mathrm{52}}*x_{\mathrm{21}}&+&a_{\mathrm{53}}*x_{\mathrm{31}}&+&a_{\mathrm{54}}*x_{\mathrm{41}}&+&a_{\mathrm{55}}*x_{\mathrm{51}}&+&0&=&0\\ a_{\mathrm{61}}*x_{\mathrm{11}}&+&a_{\mathrm{62}}*x_{\mathrm{21}}&+&a_{\mathrm{63}}*x_{\mathrm{31}}&+&a_{\mathrm{64}}*x_{\mathrm{41}}&+&a_{\mathrm{65}}*x_{\mathrm{51}}&+&a_{\mathrm{66}}*x_{\mathrm{61}}&=&0\end{array}$

Equation system 2: ${A*x_{\mathrm{2}}=i_{\mathrm{2}}}$ :
$\begin{array}{ccccccccccccc}0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&a_{\mathrm{22}}*x_{\mathrm{22}}&+&0&+&0&+&0&+&0&=&1\\ 0&+&a_{\mathrm{32}}*x_{\mathrm{22}}&+&a_{\mathrm{33}}*x_{\mathrm{32}}&+&0&+&0&+&0&=&0\\ 0&+&a_{\mathrm{42}}*x_{\mathrm{22}}&+&a_{\mathrm{43}}*x_{\mathrm{32}}&+&a_{\mathrm{44}}*x_{\mathrm{42}}&+&0&+&0&=&0\\ 0&+&a_{\mathrm{52}}*x_{\mathrm{22}}&+&a_{\mathrm{53}}*x_{\mathrm{32}}&+&a_{\mathrm{54}}*x_{\mathrm{42}}&+&a_{\mathrm{55}}*x_{\mathrm{52}}&+&0&=&0\\ 0&+&a_{\mathrm{62}}*x_{\mathrm{22}}&+&a_{\mathrm{63}}*x_{\mathrm{32}}&+&a_{\mathrm{64}}*x_{\mathrm{42}}&+&a_{\mathrm{65}}*x_{\mathrm{52}}&+&a_{\mathrm{66}}*x_{\mathrm{62}}&=&0\end{array}$

Equation system 3: ${A*x_{\mathrm{3}}=i_{\mathrm{3}}}$ :
$\begin{array}{ccccccccccccc}0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&0&+&a_{\mathrm{33}}*x_{\mathrm{33}}&+&0&+&0&+&0&=&1\\ 0&+&0&+&a_{\mathrm{43}}*x_{\mathrm{33}}&+&a_{\mathrm{44}}*x_{\mathrm{43}}&+&0&+&0&=&0\\ 0&+&0&+&a_{\mathrm{53}}*x_{\mathrm{33}}&+&a_{\mathrm{54}}*x_{\mathrm{43}}&+&a_{\mathrm{55}}*x_{\mathrm{53}}&+&0&=&0\\ 0&+&0&+&a_{\mathrm{63}}*x_{\mathrm{33}}&+&a_{\mathrm{64}}*x_{\mathrm{43}}&+&a_{\mathrm{65}}*x_{\mathrm{53}}&+&a_{\mathrm{66}}*x_{\mathrm{63}}&=&0\end{array}$

Equation system 4: ${A*x_{\mathrm{4}}=i_{\mathrm{4}}}$ :
$\begin{array}{ccccccccccccc}0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&0&+&0&+&a_{\mathrm{44}}*x_{\mathrm{44}}&+&0&+&0&=&1\\ 0&+&0&+&0&+&a_{\mathrm{54}}*x_{\mathrm{44}}&+&a_{\mathrm{55}}*x_{\mathrm{54}}&+&0&=&0\\ 0&+&0&+&0&+&a_{\mathrm{64}}*x_{\mathrm{44}}&+&a_{\mathrm{65}}*x_{\mathrm{54}}&+&a_{\mathrm{66}}*x_{\mathrm{64}}&=&0\end{array}$

Equation system 5: ${A*x_{\mathrm{5}}=i_{\mathrm{5}}}$ :
$\begin{array}{ccccccccccccc}0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&0&+&0&+&0&+&a_{\mathrm{55}}*x_{\mathrm{55}}&+&0&=&1\\ 0&+&0&+&0&+&0&+&a_{\mathrm{65}}*x_{\mathrm{55}}&+&a_{\mathrm{66}}*x_{\mathrm{65}}&=&0\end{array}$

Equation system 6: ${A*x_{\mathrm{6}}=i_{\mathrm{6}}}$ :
$\begin{array}{ccccccccccccc}0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&0&+&0&+&0&+&0&+&0&=&0\\ 0&+&0&+&0&+&0&+&0&+&a_{\mathrm{66}}*x_{\mathrm{66}}&=&1\end{array}$

These equation systems are then solved by forward substitution. See wikipedia Triangular matrix, forward substitution . If the matrix to be inverted would have been upper triangular then we would have arrived at similar equations to be solved by back substitution.

Mats Granvik mats.granvik(AT)abo.fi

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Mobius function “recursion”

Posted on December 12, 2010 by Mats Granvik

Update 18.2.2017:

The recurrence for the Möbius function is:
$\mu(1)=1$
$n>1:$
$\mu(n) = -\underset{\underset{d<n}{d|n}}{\sum} \Lambda(d)$

The recurrence for the von Mangoldt function is:
$\Lambda(n) = \log(n)-\underset{\underset{d<n}{d|n}}{\sum} \Lambda(d)$

The recurrence for the Euler totient function is:
$\phi(n) = n-\underset{\underset{d<n}{d|n}}{\sum}\phi(d)$

The recurrence for the Dirichlet inverse of the totient function $a(n)$ is:
$\frac{a(n)}{n} = \frac{1}{n}-\underset{\underset{d<n}{d|n}}{\sum} a(d)$

The recurrence for the Liouville lambda function is:
$\lambda(n) = \text{sgn}\left(a-\underset{\underset{d<n}{d|n}}{\sum} \lambda(d)\right)$

Where $\text{sgn}$ is the sign function and $0<a<1$

Link to math se answer:
http://math.stackexchange.com/a/853300/8530

—————————–
Someone searched the Internet for "mobius function recursion" and found this blog so this post will be about how to view the Mobius function in terms of recursion, ~~although the mobius function “mu” doesn’t have a recurrence/recursion.~~ (I am not sure any more. I read in Wikipedia that it has a recursion.) Anyhow here follows examples of what the “recursion” looks like:

mu(1) = 1

mu(2)=-mu(1)=-1

mu(3)=-mu(1)=-1

mu(4)=-mu(1)-mu(2)=-1-(-1)=-1+1=0

mu(5)=-mu(1)=-1

mu(6)=-mu(1)-mu(2)-mu(3)=-1-(-1)-(-1)=-1+1+1=1

mu(7)=-mu(1)=-1

mu(8)=-mu(1)-mu(2)-mu(4)=-1-(-1)-0=-1+1-0=0

mu(9)=-mu(1)-mu(3)=-1-(-1)=-1+1=0

mu(10)=-mu(1)-mu(2)-mu(5)=-1-(-1)-(-1)=-1+1+1=1

mu(11)=-mu(1)=-1

mu(12)=-mu(1)-mu(2)-mu(3)-mu(4)-mu(6)=-1-(-1)-(-1)-0-(1)=-1+1+1-0-1=0

mu(13)=-mu(1)=-1

mu(14)=-mu(1)-mu(2)-mu(7)=-1-(-1)-(-1)=-1+1+1=1

mu(15)=-mu(1)-mu(3)-mu(5)=-1-(-1)-(-1)=-1+1+1=1

mu(16)=-mu(1)-mu(2)-mu(4)-mu(8)=-1-(-1)-0-0=-1+1-0-0=0

mu(17)=-mu(1)=-1

mu(18)=-mu(1)-mu(2)-mu(3)-mu(6)-mu(9)=-1-(-1)-(-1)-(1)-0=-1+1+1-1-0=0

mu(19)=-mu(1)=-1

mu(20)=-mu(1)-mu(2)-mu(5)-mu(10)=-1-(-1)-(-1)-(1)=-1+1+1-1=0

mu(21)=-mu(1)-mu(3)-mu(7)=-1-(-1)-(-1)=-1+1+1=1

mu(22)=-mu(1)-mu(2)-mu(11)=-1-(-1)-(-1)=-1+1+1=1

mu(23)=-mu(1)=-1

mu(24)=-mu(1)-mu(2)-mu(3)-mu(4)-mu(6)-mu(8)-mu(12)=-1-(-1)-(-1)-0-(1)-0-0=-1+1+1-0-1-0-0=0

mu(25)=-mu(1)-mu(5)=-1-(-1)=-1+1=0

mu(26)=-mu(1)-mu(2)-mu(13)=-1-(-1)-(-1)=-1+1+1=1

mu(27)=-mu(1)-mu(3)-mu(9)=-1-(-1)-0=-1+1-0=0

mu(28)=-mu(1)-mu(2)-mu(4)-mu(7)-mu(14)=-1-(-1)-0-(-1)-(1)=-1+1-0+1-1=0

mu(29)=-mu(1)=-1

mu(30)=-mu(1)-mu(2)-mu(3)-mu(5)-mu(6)-mu(10)-mu(15)=-1-(-1)-(-1)-(-1)-(1)-(1)-(1)=-1+1+1+1-1-1-1=-1

A European dot comma style Excel formula for the above:

=IF(COLUMN()=1;1;IF(ROW()=COLUMN();-SUM(INDIRECT(ADDRESS(ROW();1)&":"&ADDRESS(ROW();COLUMN()-1)));IF(ROW()>COLUMN();INDIRECT(ADDRESS(ROW()-COLUMN();COLUMN()));0)))

which is a recurrence for the Mobius function.

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The inverse of a triangular matrix as a binomial series

Posted on December 8, 2010 by Mats Granvik

Consider the lower triangular matrix A:

${A=}\left(\begin{array}{cccccc}a_{\mathrm{11}}&0&0&0&0&0\\ a_{\mathrm{21}}&a_{\mathrm{22}}&0&0&0&0\\ a_{\mathrm{31}}&a_{\mathrm{32}}&a_{\mathrm{33}}&0&0&0\\ a_{\mathrm{41}}&a_{\mathrm{42}}&a_{\mathrm{43}}&a_{\mathrm{44}}&0&0\\ a_{\mathrm{51}}&a_{\mathrm{52}}&a_{\mathrm{53}}&a_{\mathrm{54}}&a_{\mathrm{55}}&0\\ a_{\mathrm{61}}&a_{\mathrm{62}}&a_{\mathrm{63}}&a_{\mathrm{64}}&a_{\mathrm{65}}&a_{\mathrm{66}}\end{array}\right)$

Divide the COLUMNS with the diagonal elements in matrix A:

$\left(\begin{array}{cccccc}a_{\mathrm{11}}/a_{\mathrm{11}}&0&0&0&0&0\\ a_{\mathrm{21}}/a_{\mathrm{11}}&a_{\mathrm{22}}/a_{\mathrm{22}}&0&0&0&0\\ a_{\mathrm{31}}/a_{\mathrm{11}}&a_{\mathrm{32}}/a_{\mathrm{22}}&a_{\mathrm{33}}/a_{\mathrm{33}}&0&0&0\\ a_{\mathrm{41}}/a_{\mathrm{11}}&a_{\mathrm{42}}/a_{\mathrm{22}}&a_{\mathrm{43}}/a_{\mathrm{33}}&a_{\mathrm{44}}/a_{\mathrm{44}}&0&0\\ a_{\mathrm{51}}/a_{\mathrm{11}}&a_{\mathrm{52}}/a_{\mathrm{22}}&a_{\mathrm{53}}/a_{\mathrm{33}}&a_{\mathrm{54}}/a_{\mathrm{44}}&a_{\mathrm{55}}/a_{\mathrm{55}}&0\\ a_{\mathrm{61}}/a_{\mathrm{11}}&a_{\mathrm{62}}/a_{\mathrm{22}}&a_{\mathrm{63}}/a_{\mathrm{33}}&a_{\mathrm{64}}/a_{\mathrm{44}}&a_{\mathrm{65}}/a_{\mathrm{55}}&a_{\mathrm{66}}/a_{\mathrm{66}}\end{array}\right)$

Which gives us matrix B:

${B=}\left(\begin{array}{cccccc}1&0&0&0&0&0\\ a_{\mathrm{21}}/a_{\mathrm{11}}&1&0&0&0&0\\ a_{\mathrm{31}}/a_{\mathrm{11}}&a_{\mathrm{32}}/a_{\mathrm{22}}&1&0&0&0\\ a_{\mathrm{41}}/a_{\mathrm{11}}&a_{\mathrm{42}}/a_{\mathrm{22}}&a_{\mathrm{43}}/a_{\mathrm{33}}&1&0&0\\ a_{\mathrm{51}}/a_{\mathrm{11}}&a_{\mathrm{52}}/a_{\mathrm{22}}&a_{\mathrm{53}}/a_{\mathrm{33}}&a_{\mathrm{54}}/a_{\mathrm{44}}&1&0\\ a_{\mathrm{61}}/a_{\mathrm{11}}&a_{\mathrm{62}}/a_{\mathrm{22}}&a_{\mathrm{63}}/a_{\mathrm{33}}&a_{\mathrm{64}}/a_{\mathrm{44}}&a_{\mathrm{65}}/a_{\mathrm{55}}&1\end{array}\right)$

Now replace the all the ones on the main diagonal with zeros:

${C=}\left(\begin{array}{cccccc}0&0&0&0&0&0\\ a_{\mathrm{21}}/a_{\mathrm{11}}&0&0&0&0&0\\ a_{\mathrm{31}}/a_{\mathrm{11}}&a_{\mathrm{32}}/a_{\mathrm{22}}&0&0&0&0\\ a_{\mathrm{41}}/a_{\mathrm{11}}&a_{\mathrm{42}}/a_{\mathrm{22}}&a_{\mathrm{43}}/a_{\mathrm{33}}&0&0&0\\ a_{\mathrm{51}}/a_{\mathrm{11}}&a_{\mathrm{52}}/a_{\mathrm{22}}&a_{\mathrm{53}}/a_{\mathrm{33}}&a_{\mathrm{54}}/a_{\mathrm{44}}&0&0\\ a_{\mathrm{61}}/a_{\mathrm{11}}&a_{\mathrm{62}}/a_{\mathrm{22}}&a_{\mathrm{63}}/a_{\mathrm{33}}&a_{\mathrm{64}}/a_{\mathrm{44}}&a_{\mathrm{65}}/a_{\mathrm{55}}&0\end{array}\right)$

Then calculate matrix powers as follows: ${D=C^0-C^1+C^2-C^3+C^4-C^5+C^6-C^7+C^8-...}$ Which is exactly as the binomial series: ${(1+x)^{-1}=1-x+x^2-x^3+x^4-x^5+x^6-x^7+x^8-...}$ except that here it is applied to a triangular matrix and the result is a new triangular matrix D. Notice that ${C^0}$ is the identity matrix.

And then finally divide the ROWS in matrix D with the diagonal elements in A:

${A^{-1}=}\left(\begin{array}{cccccc}d_{\mathrm{11}}/a_{\mathrm{11}}&0&0&0&0&0\\ d_{\mathrm{21}}/a_{\mathrm{22}}&d_{\mathrm{22}}/a_{\mathrm{22}}&0&0&0&0\\ d_{\mathrm{31}}/a_{\mathrm{33}}&d_{\mathrm{32}}/a_{\mathrm{33}}&d_{\mathrm{33}}/a_{\mathrm{33}}&0&0&0\\ d_{\mathrm{41}}/a_{\mathrm{44}}&d_{\mathrm{42}}/a_{\mathrm{44}}&d_{\mathrm{43}}/a_{\mathrm{44}}&d_{\mathrm{44}}/a_{\mathrm{44}}&0&0\\ d_{\mathrm{51}}/a_{\mathrm{55}}&d_{\mathrm{52}}/a_{\mathrm{55}}&d_{\mathrm{53}}/a_{\mathrm{55}}&d_{\mathrm{54}}/a_{\mathrm{55}}&d_{\mathrm{55}}/a_{\mathrm{55}}&0\\ d_{\mathrm{61}}/a_{\mathrm{66}}&d_{\mathrm{62}}/a_{\mathrm{66}}&d_{\mathrm{63}}/a_{\mathrm{66}}&d_{\mathrm{64}}/a_{\mathrm{66}}&d_{\mathrm{65}}/a_{\mathrm{66}}&d_{\mathrm{66}}/a_{\mathrm{66}}\end{array}\right)$

mats.granvik(AT)abo.fi

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Relation to the silver ratios

Posted on November 25, 2010 by Mats Granvik

Mathworld has a page about the silver ratios. It can be found here: http://mathworld.wolfram.com/SilverRatio.html

Some time ago I noticed that the Mobius function is somewhat related to the silver ratios. I published my results in the OEIS: http://oeis.org/A178536, http://oeis.org/A181434, http://oeis.org/A181435

Wolfram|Alpha disproof

Posted on November 6, 2010 by Mats Granvik

In my previous post I wrote that I was not sure if the expression:

a*sin(b*x)-c*tan(pi/2*x)+d*(cos(e*x)-1)+f*(1/cos(pi/2*x)-1) (1)

still could be used in some form as an ordinary generating function for the Möbius function..

I have now proven using Wolfram|Alpha that it is wrong, it can’t be used as an o.g.f, and here is the proof starting by entering this line into Wolfram|Alpha:

expand a*sin(b*x)-c*tan(pi/2*x)+d*(cos(e*x)-1)+f*(1/cos(pi/2*x)-1)

Then from the block: Series expansion at x=0:

form the six equations: (a*b-(pi*c)/2)=1, (pi^2*f-4*d*e^2)/8=-1, 1/24*(-4*a*b^3-pi^3*c)=-1, 1/384*(16*d*e^4+5*pi^4*f)=0, 1/240*(2*a*b^5-pi^5*c)=-1, (61*pi^6*f-64*d*e^6))/46080=1

I.e the first six values of the Möbius function with six unknowns. Separate them into two groups to be entered separately into Wolfram|Alpha:

(a*b-(pi*c)/2)=1, 1/24*(-4*a*b^3-pi^3*c)=-1, 1/240*(2*a*b^5-pi^5*c)=-1

and

(pi^2*f-4*d*e^2)/8=-1, 1/384*(16*d*e^4+5*pi^4*f)=0, (61*pi^6*f-64*d*e^6))/46080=1

In the first case more than one solution is possible. But in the second case no solutions exist and thereby expression (1) can’t be expanded into the Möbius function, which was to be proven.

Mats Granvik, mats.granvik(AT)abo.fi

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About my generating function attempt

Posted on November 6, 2010 by Mats Granvik

A few days ago I made a suggestion for an ordinary generating function (o.g.f.) for the Möbius function. Perhaps a silly suggestion. Might not even be possible. Anyways you can find it here: http://list.seqfan.eu/pipermail/seqfan/2010-November/006354.html

The o.g.f. I suggested was: a*sin(b*x)-c*tan(pi/2*x)+d*(cos(e*x)-1)+f*(1/cos(pi/2*x)-1).

Here b and e should perhaps be pi times some number. I tried b=e=pi and b=e=pi/2 but the fit from the the regression got worse. Yes I used regression to try to find the generating function for the power series. After a email response to my comment on the seqfan forum I am now more aware than before that the idea of an o.g.f. is to be precise. And I know my suggestion is incorrect.

Still I can’t completely give up thinking about whether it is possible to express the o.g.f for the Möbius function with trigonometric expressions.

Mats Granvik mats.granvik(AT)abo.fi

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