Timeline for Counting Bipartitions
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 28, 2017 at 6:19 | vote | accept | Steven Spallone | ||
| Apr 27, 2017 at 10:56 | answer | added | Janne Kokkala | timeline score: 4 | |
| Apr 25, 2017 at 11:07 | comment | added | Amritanshu Prasad | Numerical data also suggests that, for $n>101$, $p_2(n) > n^2 p(n)$, and for $n>321$, $p_2(n)>n^3 p(n)$, and for $n>701$, $p_2(n)>n^4 p(n)$. Could it be that, for every positive integer $k$, $p_2(n) > n^k p(n)$, provided that $n$ is sufficiently large (how large depends on $k$). | |
| Apr 25, 2017 at 9:20 | comment | added | Harry Huang | Numerical data suggests that $p(a)p(b)>p(n)$ when $a+b=n\geq 10$ and $a,b\geq 2$. I haven't found a nice combinatorial proof though. | |
| Apr 25, 2017 at 5:03 | comment | added | Martin Rubey | Just to clarify: I used the definition "square of the generating function for partitions" for the generating function for bipartitions. | |
| Apr 25, 2017 at 4:44 | history | edited | Amritanshu Prasad |
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| Apr 25, 2017 at 4:23 | comment | added | Amritanshu Prasad | If you want more values etc., see oeis.org/A000712 | |
| Apr 25, 2017 at 2:51 | comment | added | T. Amdeberhan | After the OP's edit, the generating function is simply: $\prod_{k=1}^{\infty}\frac1{(1-x^k)^2}$. | |
| Apr 25, 2017 at 2:40 | history | edited | Steven Spallone | CC BY-SA 3.0 |
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| Apr 24, 2017 at 17:42 | comment | added | Martin Rubey | I should add that it is clear that it does satisfy an ADE, what may be surprising is that it would be so nice. | |
| Apr 24, 2017 at 17:39 | comment | added | Martin Rubey | In case it helps, the generating function for the number of bipartitions apparently satisfies a very nice ADE, similar to the one for partitions mathoverflow.net/a/47706/3032: $${{x^2}{{{f(x)}}^2}{{f^{{(iv)}}}(x)}}+{{({8{x^2}{f(x)}{{f'}(x)}}+{5x{{{f(x)}}^2}})}{{f'''}(x)}}-{21{x^2}{f(x)}{{{{f''}(x)}}^2}}+{{({12{x^{2}}{{{{f'}(x)}}^2}}-{15x{f(x)}{{f'}(x)}}+{4{{{f(x)}}^2}})}{{f''}(x)}}+{10x{{{{f'}(x)}}^3}}-{10{f(x)}{{{{f'}(x)}}^2}}=0$$ | |
| Apr 24, 2017 at 16:03 | review | First posts | |||
| Apr 24, 2017 at 16:06 | |||||
| Apr 24, 2017 at 16:00 | history | asked | Steven Spallone | CC BY-SA 3.0 |