Timeline for A seemingly simple combinatorial object that must have an easy generating function
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
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| Mar 8, 2016 at 20:40 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
Added the matrix version
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| Mar 6, 2016 at 14:13 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
added 2 characters in body
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| Mar 6, 2016 at 13:47 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
deleted 9 characters in body
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| Mar 6, 2016 at 13:33 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
edited body
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| Mar 6, 2016 at 13:27 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
deleted 1 character in body
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| Mar 6, 2016 at 13:19 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
Update on the recurrence
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| Mar 1, 2016 at 11:08 | comment | added | მამუკა ჯიბლაძე | @MartinRubey I've added something on that. Can't find any patterns so far | |
| Mar 1, 2016 at 11:06 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
added regrouping
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| Feb 29, 2016 at 20:13 | comment | added | Martin Rubey | Did you try to compute $S_{d_1,\dots,d_k}$ for small $k$? Multiplying by $(1-t^k)$ gives a polynomial, maybe one can do something with these. | |
| Feb 29, 2016 at 11:13 | comment | added | მამუკა ჯიბლაძე | @MartinRubey That seems to be very much to the point, and I am thinking along these lines too. Actually $f_{x,y}$ only depends on $|x-y|$, so it makes sense to redenote $f_{x,y}$ and make it $f_d$ with $d=|x-y|$; then, the sum is that of $f_{d_1}\cdots f_{d_k}S_{d_1,...,d_k}$ over all tuples $d_1,...,d_k$, where$$S_{d_1,...,d_k}(t):=\sum_{|x_{i-1}-x_i|=d_i,\ i=1,...,k}t^{x_0+...+x_k}.$$However I don't see how to approach these $S$es | |
| Feb 29, 2016 at 10:05 | comment | added | Martin Rubey | It's very nice that $\bar f$ is much better to compute! I wonder whether one can exploit the symmetry of $f_{x,y}$ apart from the trivial observation that the summand corresponding to $x_0,\dots,x_k$ is the same as the one corresponding to $x_k,\dots, x_0$. | |
| Feb 29, 2016 at 9:21 | comment | added | მამუკა ჯიბლაძე | Edited that place | |
| Feb 29, 2016 at 9:18 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
made the sequence more clear (hopefully)
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| Feb 29, 2016 at 9:12 | comment | added | მამუკა ჯიბლაძე | @MartinRubey Sorry should be more clear - the sequence starts with $n=1$, and moreover it is halved, so $\bar f=2(t^2+3t^3+10t^4+27t^5+...)$. | |
| Feb 29, 2016 at 9:01 | comment | added | Martin Rubey | I'm confused: if $\bar f=t+3t^2+10t^3+27t^4\dots$, then $1/(1-\bar f) = 1+t+4t^2+17t^3+66t^4+\dots$ which is not the series in the question. | |
| Feb 29, 2016 at 8:33 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
added 20 characters in body
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| Feb 29, 2016 at 8:27 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
Added the sequence for indecomposables
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| Feb 29, 2016 at 0:36 | history | answered | მამუკა ჯიბლაძე | CC BY-SA 3.0 |