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domotorp
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Helmut
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The following are the 4 matrices for n=5:

[[ 0. 0. 0. 0. 1.]
[ 1. 0. 0. 1. 0.]
[ 2. 0. 1. 0. 0.]
[ 3. 1. 0. 0. 0.]
[ 5. 0. 0. 0. 0.]

[[ 0. 0. 0. 0. 1.]
[ 1. 0. 0. 1. 0.]
[ 1. 2. 0. 0. 0.]
[ 3. 1. 0. 0. 0.]
[ 5. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 1.]
[ 0. 1. 1. 0. 0.]
[ 2. 0. 1. 0. 0.]
[ 3. 1. 0. 0. 0.]
[ 5. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 1.]
[ 0. 1. 1. 0. 0.]
[ 1. 2. 0. 0. 0.]
[ 3. 1. 0. 0. 0.]
[ 5. 0. 0. 0. 0.]]

The following are the 11 matrices for n=6:

[[ 0. 0. 0. 0. 0. 1.]
[ 1. 0. 0. 0. 1. 0.]
[ 2. 0. 0. 1. 0. 0.]
[ 3. 0. 1. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 1. 0. 0. 0. 1. 0.]
[ 2. 0. 0. 1. 0. 0.]
[ 2. 2. 0. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 1. 0. 0. 0. 1. 0.]
[ 1. 1. 1. 0. 0. 0.]
[ 3. 0. 1. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 1. 0. 0. 0. 1. 0.]
[ 1. 1. 1. 0. 0. 0.]
[ 2. 2. 0. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 0. 1. 0. 1. 0. 0.]
[ 2. 0. 0. 1. 0. 0.]
[ 3. 0. 1. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 0. 1. 0. 1. 0. 0.]
[ 2. 0. 0. 1. 0. 0.]
[ 2. 2. 0. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 0. 1. 0. 1. 0. 0.]
[ 1. 1. 1. 0. 0. 0.]
[ 3. 0. 1. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 0. 1. 0. 1. 0. 0.]
[ 1. 1. 1. 0. 0. 0.]
[ 2. 2. 0. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 0. 1. 0. 1. 0. 0.]
[ 0. 3. 0. 0. 0. 0.]
[ 2. 2. 0. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 0. 0. 2. 0. 0. 0.]
[ 1. 1. 1. 0. 0. 0.]
[ 2. 2. 0. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 0. 0. 2. 0. 0. 0.]
[ 1. 1. 1. 0. 0. 0.]
[ 3. 0. 1. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

The following are the 4 matrices for n=5:

[[ 0. 0. 0. 0. 1.]
[ 1. 0. 0. 1. 0.]
[ 2. 0. 1. 0. 0.]
[ 3. 1. 0. 0. 0.]
[ 5. 0. 0. 0. 0.]

[[ 0. 0. 0. 0. 1.]
[ 1. 0. 0. 1. 0.]
[ 1. 2. 0. 0. 0.]
[ 3. 1. 0. 0. 0.]
[ 5. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 1.]
[ 0. 1. 1. 0. 0.]
[ 2. 0. 1. 0. 0.]
[ 3. 1. 0. 0. 0.]
[ 5. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 1.]
[ 0. 1. 1. 0. 0.]
[ 1. 2. 0. 0. 0.]
[ 3. 1. 0. 0. 0.]
[ 5. 0. 0. 0. 0.]]

The following are the 11 matrices for n=6:

[[ 0. 0. 0. 0. 0. 1.]
[ 1. 0. 0. 0. 1. 0.]
[ 2. 0. 0. 1. 0. 0.]
[ 3. 0. 1. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 1. 0. 0. 0. 1. 0.]
[ 2. 0. 0. 1. 0. 0.]
[ 2. 2. 0. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 1. 0. 0. 0. 1. 0.]
[ 1. 1. 1. 0. 0. 0.]
[ 3. 0. 1. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 1. 0. 0. 0. 1. 0.]
[ 1. 1. 1. 0. 0. 0.]
[ 2. 2. 0. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 0. 1. 0. 1. 0. 0.]
[ 2. 0. 0. 1. 0. 0.]
[ 3. 0. 1. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 0. 1. 0. 1. 0. 0.]
[ 2. 0. 0. 1. 0. 0.]
[ 2. 2. 0. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 0. 1. 0. 1. 0. 0.]
[ 1. 1. 1. 0. 0. 0.]
[ 3. 0. 1. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 0. 1. 0. 1. 0. 0.]
[ 1. 1. 1. 0. 0. 0.]
[ 2. 2. 0. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 0. 1. 0. 1. 0. 0.]
[ 0. 3. 0. 0. 0. 0.]
[ 2. 2. 0. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 0. 0. 2. 0. 0. 0.]
[ 1. 1. 1. 0. 0. 0.]
[ 2. 2. 0. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

[[ 0. 0. 0. 0. 0. 1.]
[ 0. 0. 2. 0. 0. 0.]
[ 1. 1. 1. 0. 0. 0.]
[ 3. 0. 1. 0. 0. 0.]
[ 4. 1. 0. 0. 0. 0.]
[ 6. 0. 0. 0. 0. 0.]]

deleted 119 characters in body
Source Link
Helmut
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Sequence A002846 in https://oeis.org/A002846 (OEIS) gives, for each positive integer $n$, the number $a(n)$ of ways of transforming a set of $n$ indistinguishable objects into $n$ singletons via a sequence of $n-1$ refinements, i.e. the number of ways of transforming a set of $n$ unlabelled objects into $n$ singletons via a sequence of $n-1$ binary partitions.

For a given $n$, each sequence of refinements can be represented by an $n\times{}n$ triangular matrix. The $(i,j)^{th}$ element of the matrix represents the number of sets after $i-1$ partitions that contain $j$ elements. The sum of the $i^{th}$ row is simply the number of sets after $i-1$ partitions, so is clearly $=i$. I am interested in linear combinations of these $a(n)$ matrices. For convenience, suppose that the coefficients total to 1. Then obviously if a matrix $M$ is formed by such a linear combination, the sum of the $i^{th}$ row is still $i$. However, a number of trials have convinced me that if $M$ is non-singular, the sum of the $i^{th}$ row of $M^{-1}$ is $\frac{i}{n}.$

To allow you to easily experiment, a list of matrices for n=4 to 9 can be obtained at github.com/helmutsimon/coalescent_tree_data . The file matrix_list_9 etc can be downloaded and then opened with pickle and gzip. The file then contains a python list of matrices represented as numpy arrays. At present, I have no idea how to go about proving this conjecture.

I have attached a document showing the matrices for n=5 and n=6.

Sequence A002846 in https://oeis.org/A002846 (OEIS) gives, for each positive integer $n$, the number $a(n)$ of ways of transforming a set of $n$ indistinguishable objects into $n$ singletons via a sequence of $n-1$ refinements, i.e. the number of ways of transforming a set of $n$ unlabelled objects into $n$ singletons via a sequence of $n-1$ binary partitions.

For a given $n$, each sequence of refinements can be represented by an $n\times{}n$ triangular matrix. The $(i,j)^{th}$ element of the matrix represents the number of sets after $i-1$ partitions that contain $j$ elements. The sum of the $i^{th}$ row is simply the number of sets after $i-1$ partitions, so is clearly $=i$. I am interested in linear combinations of these $a(n)$ matrices. For convenience, suppose that the coefficients total to 1. Then obviously if a matrix $M$ is formed by such a linear combination, the sum of the $i^{th}$ row is still $i$. However, a number of trials have convinced me that if $M$ is non-singular, the sum of the $i^{th}$ row of $M^{-1}$ is $\frac{i}{n}.$

To allow you to easily experiment, a list of matrices for n=4 to 9 can be obtained at github.com/helmutsimon/coalescent_tree_data . The file matrix_list_9 etc can be downloaded and then opened with pickle and gzip. The file then contains a python list of matrices represented as numpy arrays. At present, I have no idea how to go about proving this conjecture.

I have attached a document showing the matrices for n=5 and n=6.

Sequence A002846 in https://oeis.org/A002846 (OEIS) gives, for each positive integer $n$, the number $a(n)$ of ways of transforming a set of $n$ indistinguishable objects into $n$ singletons via a sequence of $n-1$ refinements, i.e. the number of ways of transforming a set of $n$ unlabelled objects into $n$ singletons via a sequence of $n-1$ binary partitions.

For a given $n$, each sequence of refinements can be represented by an $n\times{}n$ triangular matrix. The $(i,j)^{th}$ element of the matrix represents the number of sets after $i-1$ partitions that contain $j$ elements. The sum of the $i^{th}$ row is simply the number of sets after $i-1$ partitions, so is clearly $=i$. I am interested in linear combinations of these $a(n)$ matrices. For convenience, suppose that the coefficients total to 1. Then obviously if a matrix $M$ is formed by such a linear combination, the sum of the $i^{th}$ row is still $i$. However, a number of trials have convinced me that if $M$ is non-singular, the sum of the $i^{th}$ row of $M^{-1}$ is $\frac{i}{n}.$

To allow you to easily experiment, a list of matrices for n=4 to 9 can be obtained at github.com/helmutsimon/coalescent_tree_data . The file matrix_list_9 etc can be downloaded and then opened with pickle and gzip. The file then contains a python list of matrices represented as numpy arrays. At present, I have no idea how to go about proving this conjecture.

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Helmut
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