Timeline for Number of representations of an integer as an (arbitrary) sum of products
Current License: CC BY-SA 3.0
19 events
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| Mar 27, 2014 at 17:03 | comment | added | David Ellis | Yes, thanks very much for this. Not being too familiar with the method, it took a while for me to go through the details. Thanks again. | |
| Mar 27, 2014 at 16:56 | vote | accept | David Ellis | ||
| Mar 25, 2014 at 18:13 | comment | added | Lucia | I think my answer below provides a complete solution to your problem. Were you looking for something more? Or perhaps there was something unclear with what I wrote? | |
| Mar 18, 2014 at 15:41 | answer | added | Lucia | timeline score: 8 | |
| Mar 16, 2014 at 14:25 | comment | added | Lucia | One can get an asymptotic formula for problems like this by using the saddle point method. This is classical, but one reference may be these course notes: math.berkeley.edu/~moorxu/oldsite/notes/155/155main.pdf , see page 44 and following. That deals with the partition problem with generating function $\Gamma(s)\zeta(s)\zeta(s+1)$ (see page 47); you need to make similar calculations with $\Gamma(s)\zeta(s)^2\zeta(s+1)$. There won't be the modular forms miracle as with partitions, but one doesn't need that for an asymptotic. | |
| Mar 14, 2014 at 21:22 | answer | added | Jan-Christoph Schlage-Puchta | timeline score: 3 | |
| Mar 12, 2014 at 12:21 | history | edited | David Ellis | CC BY-SA 3.0 |
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| Mar 11, 2014 at 4:00 | comment | added | Gerry Myerson | I think this is oeis.org/A006171 and a reference and some formulas are given there, but I see no asymptotics. | |
| Mar 10, 2014 at 19:36 | history | edited | David Ellis | CC BY-SA 3.0 |
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| Mar 10, 2014 at 19:30 | history | edited | David Ellis | CC BY-SA 3.0 |
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| Mar 10, 2014 at 19:13 | history | edited | David Ellis | CC BY-SA 3.0 |
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| Mar 10, 2014 at 19:05 | history | edited | David Ellis | CC BY-SA 3.0 |
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| Mar 10, 2014 at 18:24 | comment | added | David Ellis | My apologies, I meant products of pairs. (I have edited my original post accordingly.) | |
| Mar 10, 2014 at 18:23 | history | edited | David Ellis | CC BY-SA 3.0 |
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| Mar 10, 2014 at 17:07 | comment | added | The Masked Avenger | I just realized that I assumed a product always has the form ab for a and b some numbers. If ab*c is allowed, then I don't know how to approach r(n) at all. | |
| Mar 10, 2014 at 17:00 | comment | added | The Masked Avenger | As another twist, consider the related function s(n) which counts those representation that do not have 1*1 as a summand, so s(3)=2. Then r(n) = 1 + sum s(i) for i at most n. Perhaps s(n) will be tractable. | |
| Mar 10, 2014 at 16:53 | comment | added | The Masked Avenger | It seems r(n) is the sum over all partitions p of n of the product d(a) for a ranging over the members of p, and d being the divisor function. As d(1)=1, perhaps there is a simplification of the sum or the product terms that may give you some good estimates? | |
| Mar 10, 2014 at 16:39 | review | First posts | |||
| Mar 10, 2014 at 16:51 | |||||
| Mar 10, 2014 at 16:22 | history | asked | David Ellis | CC BY-SA 3.0 |