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Sturgeon
Sturgeon
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I have checked the following combinatorial identity for several cases and it seems to work. I would like to know if this is known or if there is a counter-example. Note, i is a given constant.

$$(i-1+d)\sum_{k\geq d} a_k\binom{k-d}{i-1} = \sum_{l=1}^{d} n_l \sum_{k\geq d} b_{k,l}\binom{k-d}{i-1}$$$$(i+d)\sum_{k\geq d} a_k\binom{k-d}{i} = \sum_{l=1}^{d} n_l \sum_{k\geq d} b_{k,l}\binom{k-d}{i}$$

$$a_k=| \{ (i_1,..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k \}|$$

$$b_{k,l}=|\{(i_1,..i_{l-1},n_l,i_{i+1},..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k\}|$$

I have been thinking about the left as some weighting on the compositions that come from cutting the space by a hyperplane. The right looks like some kind of weighted projection summation.

I have checked the following combinatorial identity for several cases and it seems to work. I would like to know if this is known or if there is a counter-example. Note, i is a given constant.

$$(i-1+d)\sum_{k\geq d} a_k\binom{k-d}{i-1} = \sum_{l=1}^{d} n_l \sum_{k\geq d} b_{k,l}\binom{k-d}{i-1}$$

$$a_k=| \{ (i_1,..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k \}|$$

$$b_{k,l}=|\{(i_1,..i_{l-1},n_l,i_{i+1},..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k\}|$$

I have been thinking about the left as some weighting on the compositions that come from cutting the space by a hyperplane. The right looks like some kind of weighted projection summation.

I have checked the following combinatorial identity for several cases and it seems to work. I would like to know if this is known or if there is a counter-example. Note, i is a given constant.

$$(i+d)\sum_{k\geq d} a_k\binom{k-d}{i} = \sum_{l=1}^{d} n_l \sum_{k\geq d} b_{k,l}\binom{k-d}{i}$$

$$a_k=| \{ (i_1,..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k \}|$$

$$b_{k,l}=|\{(i_1,..i_{l-1},n_l,i_{i+1},..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k\}|$$

I have been thinking about the left as some weighting on the compositions that come from cutting the space by a hyperplane. The right looks like some kind of weighted projection summation.

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Sturgeon
Sturgeon
  • 11
  • 2

I have checked the following combinatorial identity for several cases and it seems to work. I would like to know if this is known or if there is a counter-example. Note, i is a given constant.

$$(i-1+d)\sum_{k\geq0} a_k\binom{k-1}{i-1} = \sum_{l=1}^{d} n_l \sum_{k\geq0} b_{k-n_l,l}\binom{k-1}{i-1}$$$$(i-1+d)\sum_{k\geq d} a_k\binom{k-d}{i-1} = \sum_{l=1}^{d} n_l \sum_{k\geq d} b_{k,l}\binom{k-d}{i-1}$$

$$a_k=| \{ (i_1,..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k \}|$$

$$b_{k-n_l,l}=|\{(i_1,..i_{l-1},n_l,i_{i+1},..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k\}|$$$$b_{k,l}=|\{(i_1,..i_{l-1},n_l,i_{i+1},..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k\}|$$

I have been thinking about the left as some weighting on the compositions that come from cutting the space by a hyperplane. The right looks like some kind of weighted projection summation.

I have checked the following combinatorial identity for several cases and it seems to work. I would like to know if this is known or if there is a counter-example. Note, i is a given constant.

$$(i-1+d)\sum_{k\geq0} a_k\binom{k-1}{i-1} = \sum_{l=1}^{d} n_l \sum_{k\geq0} b_{k-n_l,l}\binom{k-1}{i-1}$$

$$a_k=| \{ (i_1,..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k \}|$$

$$b_{k-n_l,l}=|\{(i_1,..i_{l-1},n_l,i_{i+1},..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k\}|$$

I have been thinking about the left as some weighting on the compositions that come from cutting the space by a hyperplane. The right looks like some kind of weighted projection summation.

I have checked the following combinatorial identity for several cases and it seems to work. I would like to know if this is known or if there is a counter-example. Note, i is a given constant.

$$(i-1+d)\sum_{k\geq d} a_k\binom{k-d}{i-1} = \sum_{l=1}^{d} n_l \sum_{k\geq d} b_{k,l}\binom{k-d}{i-1}$$

$$a_k=| \{ (i_1,..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k \}|$$

$$b_{k,l}=|\{(i_1,..i_{l-1},n_l,i_{i+1},..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k\}|$$

I have been thinking about the left as some weighting on the compositions that come from cutting the space by a hyperplane. The right looks like some kind of weighted projection summation.

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Sturgeon
Sturgeon
  • 11
  • 2

Counting hyperplane cuts vs. projections. Combinatorial identity

I have checked the following combinatorial identity for several cases and it seems to work. I would like to know if this is known or if there is a counter-example. Note, i is a given constant.

$$(i-1+d)\sum_{k\geq0} a_k\binom{k-1}{i-1} = \sum_{l=1}^{d} n_l \sum_{k\geq0} b_{k-n_l,l}\binom{k-1}{i-1}$$

$$a_k=| \{ (i_1,..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k \}|$$

$$b_{k-n_l,l}=|\{(i_1,..i_{l-1},n_l,i_{i+1},..,i_d) \in [n_1]\times...\times[n_d]:\sum_{j=1}^d i_j=k\}|$$

I have been thinking about the left as some weighting on the compositions that come from cutting the space by a hyperplane. The right looks like some kind of weighted projection summation.