Timeline for Solving a recurrence relation for the prime counting function?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 15, 2023 at 14:12 | comment | added | mathoverflowUser | Ok, but numerical computation in GP-Pari and SageMath, suggest that $\gamma$ (oeis.org/A078756) is a zero of $f(z)$, so there must be some convergence issues. | |
| Nov 15, 2023 at 14:07 | comment | added | Peter Taylor | I think what this shows is either that your heuristic about $\exp$ and roots is wrong or that because we're working with formal power series there are convergence issues. From the definition of $b_i$ as $\sum_{i \ge -1} b_i z^i = \frac{\textrm{d}}{\textrm{d}z} \log f(z)$ we easily derive $f(z) = \exp \int \sum_{i \ge -1} b_i z^i \textrm{d}z = \exp \left(b_{-1}\log z + \int \sum_{i \ge 0} b_i z^i \textrm{d}z\right) = z^{b_{-1}} \exp \int \sum_{i \ge 0} b_i z^i \textrm{d}z$ | |
| Nov 15, 2023 at 13:58 | comment | added | mathoverflowUser | Here is a little bit more context: math.stackexchange.com/questions/3164216/… | |
| Nov 15, 2023 at 13:56 | comment | added | mathoverflowUser | By the expression $\Pi(z) = 1/(1-z) \cdot \sum_{p \text{prime}}z^p$ we see that $\gamma$ is a zero of both functions. Does this answer your question above? | |
| Nov 15, 2023 at 13:54 | comment | added | mathoverflowUser | But in the last case, the $\gamma=$ A078756 is a zero of $\Pi(z)$ whereas in the first case of your expression this is not. | |
| Nov 15, 2023 at 13:53 | comment | added | Peter Taylor | The expression $\Pi(z) = z^2 \exp \sum_{i \ge 0} \frac{c_i z^{i+1}}{i+1}$ can also be derived directly from the definition of A307977 and the relationship $\Pi(z) = \frac{1}{1-z} \sum_{p \textrm{ prime}} z^p$. | |
| Nov 15, 2023 at 13:25 | comment | added | Peter Taylor | The definition given for A078756 is that it's the root of the g.f. of the prime indicator function. Is your $H$ the g.f. of the prime indicator function or the prime counting function? | |
| Nov 15, 2023 at 13:15 | comment | added | mathoverflowUser | Thanks for the explanation. In my notation, I wrote $H(t)$ for what you call $\Pi(z)$. Your last equation makes me a bit suspicious, since I do not understand where the zero $\gamma = $ (oeis.org/A078756) of $\Pi(z)$ is gone? It seems like you show, that this function $\Pi(z)=H(z)$ has no zero besides $0$ because of the $\exp$ function? Unfortunately, I can not pin to the place where I think that the confusion arises. | |
| Nov 15, 2023 at 13:04 | comment | added | Peter Taylor | $[z^{n-1}](\Pi(z) C(z)) = [z^{n-3}](z^{-2} \Pi(z) C(z))$. The introduction of the integral might be easier to see backwards: $\int \sum a_i z^i \textrm{d}z = \sum \int a_i z^i \textrm{d}z = \sum \frac{a_i z^{i+1}}{i+1}$ | |
| Nov 15, 2023 at 12:49 | vote | accept | mathoverflowUser | ||
| Nov 15, 2023 at 18:37 | |||||
| Nov 15, 2023 at 12:48 | comment | added | mathoverflowUser | Thanks for your nice answer. I do not understand the step from the choice of the coefficient $[z^{n-1}](\Pi(z) C(z))$ to the integral directly below it and before the words "where the constant of integration must be zero". | |
| Nov 15, 2023 at 8:38 | history | answered | Peter Taylor | CC BY-SA 4.0 |