Revisions to A binomial determinant fomula

Corrected top line of the second factor

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edited Jan 12, 2014 at 20:17

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I don't have a proof yet, but it's likely that this matrix has a simple LU decomposition which makes the determinant obviously $1$.

\begin{multline} \scriptsize\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\\ 4 & 5 & 7 & 9 & 11 & 13 \\\ 9 & 15 & 28 & 45 & 66 & 91 \\\ 16 & 35 & 84 & 165 & 286 & 455 \\\ 25 & 75 & 210 & 495 & 1001 & 1820 \\\ 36 & 141 & 463 & 1287 & 3003 & 6188 \end{pmatrix} \\ \scriptsize\hskip1cm= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\\ 1 & 1 & 0 & 0 & 0 & 0 \\\ 1 & 3 & 1 & 0 & 0 & 0 \\\ 1 & 5 & 6 & 1 & 0 & 0 \\\ 1 & 7 & 15 & 10 & 1 & 0 \\\ 1 & 9 & 28 & 35 & 15 & 1 \end{pmatrix} \begin{pmatrix}1 & 1 & 1 & 1 & 1 & 1 \\\ 0 & 1 & 6 & 20 & 50 & 105 \\\ 0 & 0 & 1 & 8 & 35 & 112 \\\ 0 & 0 & 0 & 1 & 10 & 54 \\\ 0 & 0 & 0 & 0 & 1 & 12 \\\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{multline}\begin{multline} \scriptsize\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\\ 4 & 5 & 7 & 9 & 11 & 13 \\\ 9 & 15 & 28 & 45 & 66 & 91 \\\ 16 & 35 & 84 & 165 & 286 & 455 \\\ 25 & 75 & 210 & 495 & 1001 & 1820 \\\ 36 & 141 & 463 & 1287 & 3003 & 6188 \end{pmatrix} \\ \scriptsize\hskip1cm= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\\ 1 & 1 & 0 & 0 & 0 & 0 \\\ 1 & 3 & 1 & 0 & 0 & 0 \\\ 1 & 5 & 6 & 1 & 0 & 0 \\\ 1 & 7 & 15 & 10 & 1 & 0 \\\ 1 & 9 & 28 & 35 & 15 & 1 \end{pmatrix} \begin{pmatrix}1 & 4 & 9 & 16 & 25 & 36 \\\ 0 & 1 & 6 & 20 & 50 & 105 \\\ 0 & 0 & 1 & 8 & 35 & 112 \\\ 0 & 0 & 0 & 1 & 10 & 54 \\\ 0 & 0 & 0 & 0 & 1 & 12 \\\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{multline}

The first factor appears to be A054142 as a lower diagonal matrix. The second factor appears to be A156308 as an upper diagonal matrix.

I don't have a proof yet, but it's likely that this matrix has a simple LU decomposition which makes the determinant obviously $1$.

\begin{multline} \scriptsize\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\\ 4 & 5 & 7 & 9 & 11 & 13 \\\ 9 & 15 & 28 & 45 & 66 & 91 \\\ 16 & 35 & 84 & 165 & 286 & 455 \\\ 25 & 75 & 210 & 495 & 1001 & 1820 \\\ 36 & 141 & 463 & 1287 & 3003 & 6188 \end{pmatrix} \\ \scriptsize\hskip1cm= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\\ 1 & 1 & 0 & 0 & 0 & 0 \\\ 1 & 3 & 1 & 0 & 0 & 0 \\\ 1 & 5 & 6 & 1 & 0 & 0 \\\ 1 & 7 & 15 & 10 & 1 & 0 \\\ 1 & 9 & 28 & 35 & 15 & 1 \end{pmatrix} \begin{pmatrix}1 & 1 & 1 & 1 & 1 & 1 \\\ 0 & 1 & 6 & 20 & 50 & 105 \\\ 0 & 0 & 1 & 8 & 35 & 112 \\\ 0 & 0 & 0 & 1 & 10 & 54 \\\ 0 & 0 & 0 & 0 & 1 & 12 \\\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{multline}

The first factor appears to be A054142 as a lower diagonal matrix. The second factor appears to be A156308 as an upper diagonal matrix.

I don't have a proof yet, but it's likely that this matrix has a simple LU decomposition which makes the determinant obviously $1$.

\begin{multline} \scriptsize\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\\ 4 & 5 & 7 & 9 & 11 & 13 \\\ 9 & 15 & 28 & 45 & 66 & 91 \\\ 16 & 35 & 84 & 165 & 286 & 455 \\\ 25 & 75 & 210 & 495 & 1001 & 1820 \\\ 36 & 141 & 463 & 1287 & 3003 & 6188 \end{pmatrix} \\ \scriptsize\hskip1cm= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\\ 1 & 1 & 0 & 0 & 0 & 0 \\\ 1 & 3 & 1 & 0 & 0 & 0 \\\ 1 & 5 & 6 & 1 & 0 & 0 \\\ 1 & 7 & 15 & 10 & 1 & 0 \\\ 1 & 9 & 28 & 35 & 15 & 1 \end{pmatrix} \begin{pmatrix}1 & 4 & 9 & 16 & 25 & 36 \\\ 0 & 1 & 6 & 20 & 50 & 105 \\\ 0 & 0 & 1 & 8 & 35 & 112 \\\ 0 & 0 & 0 & 1 & 10 & 54 \\\ 0 & 0 & 0 & 0 & 1 & 12 \\\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{multline}

The first factor appears to be A054142 as a lower diagonal matrix. The second factor appears to be A156308 as an upper diagonal matrix.

added 10 characters in body

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edited Jan 12, 2014 at 19:53

Mariano Suárez-Álvarez

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I don't have a proof yet, but it's likely that this matrix has a simple LU decomposition which makes the determinant obviously $1$.

$$\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\\ 4 & 5 & 7 & 9 & 11 & 13 \\\ 9 & 15 & 28 & 45 & 66 & 91 \\\ 16 & 35 & 84 & 165 & 286 & 455 \\\ 25 & 75 & 210 & 495 & 1001 & 1820 \\\ 36 & 141 & 463 & 1287 & 3003 & 6188 \end{pmatrix} \\ = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\\ 1 & 1 & 0 & 0 & 0 & 0 \\\ 1 & 3 & 1 & 0 & 0 & 0 \\\ 1 & 5 & 6 & 1 & 0 & 0 \\\ 1 & 7 & 15 & 10 & 1 & 0 \\\ 1 & 9 & 28 & 35 & 15 & 1 \end{pmatrix} \begin{pmatrix}1 & 1 & 1 & 1 & 1 & 1 \\\ 0 & 1 & 6 & 20 & 50 & 105 \\\ 0 & 0 & 1 & 8 & 35 & 112 \\\ 0 & 0 & 0 & 1 & 10 & 54 \\\ 0 & 0 & 0 & 0 & 1 & 12 \\\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}$$\begin{multline} \scriptsize\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\\ 4 & 5 & 7 & 9 & 11 & 13 \\\ 9 & 15 & 28 & 45 & 66 & 91 \\\ 16 & 35 & 84 & 165 & 286 & 455 \\\ 25 & 75 & 210 & 495 & 1001 & 1820 \\\ 36 & 141 & 463 & 1287 & 3003 & 6188 \end{pmatrix} \\ \scriptsize\hskip1cm= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\\ 1 & 1 & 0 & 0 & 0 & 0 \\\ 1 & 3 & 1 & 0 & 0 & 0 \\\ 1 & 5 & 6 & 1 & 0 & 0 \\\ 1 & 7 & 15 & 10 & 1 & 0 \\\ 1 & 9 & 28 & 35 & 15 & 1 \end{pmatrix} \begin{pmatrix}1 & 1 & 1 & 1 & 1 & 1 \\\ 0 & 1 & 6 & 20 & 50 & 105 \\\ 0 & 0 & 1 & 8 & 35 & 112 \\\ 0 & 0 & 0 & 1 & 10 & 54 \\\ 0 & 0 & 0 & 0 & 1 & 12 \\\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{multline}

The first factor appears to be A054142 as a lower diagonal matrix. The second factor appears to be A156308 as an upper diagonal matrix.

I don't have a proof yet, but it's likely that this matrix has a simple LU decomposition which makes the determinant obviously $1$.

$$\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\\ 4 & 5 & 7 & 9 & 11 & 13 \\\ 9 & 15 & 28 & 45 & 66 & 91 \\\ 16 & 35 & 84 & 165 & 286 & 455 \\\ 25 & 75 & 210 & 495 & 1001 & 1820 \\\ 36 & 141 & 463 & 1287 & 3003 & 6188 \end{pmatrix} \\ = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\\ 1 & 1 & 0 & 0 & 0 & 0 \\\ 1 & 3 & 1 & 0 & 0 & 0 \\\ 1 & 5 & 6 & 1 & 0 & 0 \\\ 1 & 7 & 15 & 10 & 1 & 0 \\\ 1 & 9 & 28 & 35 & 15 & 1 \end{pmatrix} \begin{pmatrix}1 & 1 & 1 & 1 & 1 & 1 \\\ 0 & 1 & 6 & 20 & 50 & 105 \\\ 0 & 0 & 1 & 8 & 35 & 112 \\\ 0 & 0 & 0 & 1 & 10 & 54 \\\ 0 & 0 & 0 & 0 & 1 & 12 \\\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}$$

The first factor appears to be A054142 as a lower diagonal matrix. The second factor appears to be A156308 as an upper diagonal matrix.

I don't have a proof yet, but it's likely that this matrix has a simple LU decomposition which makes the determinant obviously $1$.

\begin{multline} \scriptsize\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\\ 4 & 5 & 7 & 9 & 11 & 13 \\\ 9 & 15 & 28 & 45 & 66 & 91 \\\ 16 & 35 & 84 & 165 & 286 & 455 \\\ 25 & 75 & 210 & 495 & 1001 & 1820 \\\ 36 & 141 & 463 & 1287 & 3003 & 6188 \end{pmatrix} \\ \scriptsize\hskip1cm= \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\\ 1 & 1 & 0 & 0 & 0 & 0 \\\ 1 & 3 & 1 & 0 & 0 & 0 \\\ 1 & 5 & 6 & 1 & 0 & 0 \\\ 1 & 7 & 15 & 10 & 1 & 0 \\\ 1 & 9 & 28 & 35 & 15 & 1 \end{pmatrix} \begin{pmatrix}1 & 1 & 1 & 1 & 1 & 1 \\\ 0 & 1 & 6 & 20 & 50 & 105 \\\ 0 & 0 & 1 & 8 & 35 & 112 \\\ 0 & 0 & 0 & 1 & 10 & 54 \\\ 0 & 0 & 0 & 0 & 1 & 12 \\\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} \end{multline}

The first factor appears to be A054142 as a lower diagonal matrix. The second factor appears to be A156308 as an upper diagonal matrix.

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answered Jan 12, 2014 at 19:51

Douglas Zare

28.2k
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132

I don't have a proof yet, but it's likely that this matrix has a simple LU decomposition which makes the determinant obviously $1$.

$$\begin{pmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\\ 4 & 5 & 7 & 9 & 11 & 13 \\\ 9 & 15 & 28 & 45 & 66 & 91 \\\ 16 & 35 & 84 & 165 & 286 & 455 \\\ 25 & 75 & 210 & 495 & 1001 & 1820 \\\ 36 & 141 & 463 & 1287 & 3003 & 6188 \end{pmatrix} \\ = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 \\\ 1 & 1 & 0 & 0 & 0 & 0 \\\ 1 & 3 & 1 & 0 & 0 & 0 \\\ 1 & 5 & 6 & 1 & 0 & 0 \\\ 1 & 7 & 15 & 10 & 1 & 0 \\\ 1 & 9 & 28 & 35 & 15 & 1 \end{pmatrix} \begin{pmatrix}1 & 1 & 1 & 1 & 1 & 1 \\\ 0 & 1 & 6 & 20 & 50 & 105 \\\ 0 & 0 & 1 & 8 & 35 & 112 \\\ 0 & 0 & 0 & 1 & 10 & 54 \\\ 0 & 0 & 0 & 0 & 1 & 12 \\\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}$$

The first factor appears to be A054142 as a lower diagonal matrix. The second factor appears to be A156308 as an upper diagonal matrix.

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