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Oct 24, 2023 at 14:50 answer added G. Melfi timeline score: 2
Sep 18, 2023 at 2:11 vote accept Adithya Chakravarthy
Apr 9, 2023 at 13:59 comment added colt_browning The relevant OEIS entry is oeis.org/A055507 , but the references there are apparently mostly about asymptotic properties.
Apr 9, 2023 at 13:58 comment added Adithya Chakravarthy @colt_browning Thanks for catching that - I fixed it so that $k$ runs from $k=1, \dots, n-1$.
Apr 9, 2023 at 13:57 history edited Adithya Chakravarthy CC BY-SA 4.0
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Apr 9, 2023 at 13:57 comment added colt_browning @AdithyaChakravarthy Yes, and the 2nd factor is $\sigma_0(n-k)$, which becomes $\sigma_0(0)$ for $k=n$.
Apr 9, 2023 at 13:53 comment added Adithya Chakravarthy @colt_browning Nope, I don't think it involves $\sigma_0(0)$. The summation over $k$ in the definition of $S(n)$ is for $k=1,\dots,n$.
Apr 9, 2023 at 13:51 comment added colt_browning This definition apparently involves $\sigma_0(0)$. Is that right?
Apr 9, 2023 at 9:17 history became hot network question
Apr 9, 2023 at 0:01 answer added kodlu timeline score: 6
Apr 8, 2023 at 22:44 comment added Adithya Chakravarthy Sure, here's my motivation: I'm studying a product of two Eisenstein series. The Fourier coefficients of Eisenstein series are divisor functions, and it turns out the Fourier coefficients of the product of Eisenstein series is a convolution of divisor functions. I'm interested in how the Fourier coefficients of this product of Eisenstein series are distributed mod $p$, which leads me to ask how this convolution of divisor sums is distributed mod $p$.
Apr 8, 2023 at 22:37 comment added Anurag Sahay You probably want to use additive characters mod 𝑝 here instead of multiplicative characters. Do you mind saying a bit about your motivation for asking this question? Is it just curiosity, or did it arise from some other considerations?
Apr 8, 2023 at 22:23 history edited Adithya Chakravarthy CC BY-SA 4.0
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Apr 8, 2023 at 22:10 history asked Adithya Chakravarthy CC BY-SA 4.0