Timeline for Number of integer partitions modulo 3
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 18, 2019 at 9:58 | vote | accept | Martin Rubey | ||
| Feb 16, 2019 at 14:41 | answer | added | Somos | timeline score: 2 | |
| Feb 16, 2019 at 10:10 | answer | added | Fedor Petrov | timeline score: 3 | |
| Feb 16, 2019 at 8:05 | comment | added | Martin Rubey | @Somos, I did not verify by hand, but your computation looks OK, since $6 + O(x) = O(x)\mod 3$. | |
| Feb 16, 2019 at 2:51 | comment | added | Somos | I am sorry, but I can't verify your ADE. If $\,f(x) = a_0 + a_1\,x + O(x^2)\,$ then your differential operator produces $\,(2a_0^2+2a_1)+O(x).\,$ Your power series has $\,a_0=1, a_1=2\,$ which gives $\,6+O(x)\,$ by my calculations instead of $0$. | |
| Feb 15, 2019 at 22:56 | comment | added | Martin Rubey | @FedorPetrov, indeed, this is quite remarkable. The equation for $1/f$ is simply $f(x)^2 f^{''}(x) + xf'(x)^3 + 2f(x)f'(x)^2=0$. | |
| Feb 15, 2019 at 22:39 | comment | added | Fedor Petrov | Would you please rewrite the same equation for $1/f=\sum_{k=-\infty}^{\infty} (-1)^k x^{k(3k+1)/2}$? It looks on the first glance that the key property of 3 is that the derivative of $1/f$ has simplified (compared to other localisations) formula: $(2/f)'=\sum_{k=-\infty}^{\infty} (-1)^k k x^{k(3k+1)/2}$. | |
| Feb 15, 2019 at 22:20 | comment | added | Martin Rubey | I should add: for the parity of the partition function, no such differential equation was found - at least not using the first few hundred terms. | |
| Feb 15, 2019 at 21:52 | history | asked | Martin Rubey | CC BY-SA 4.0 |