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Apr
15
comment Sum of products of $(1 - 1/p)$
@user254665 Thanks, I know it diverges; I want to know the rate at which it does so: probably something between $O(\log \log n)$ (which is the asymptotics of the sum of reciprocals of primes) and $O(n)$.
Apr
15
revised Sum of products of $(1 - 1/p)$
now I understand what was messed up -- my $S_n$ is the sum of $\pi(n)$ terms so should probably have been stated $S_{\pi(n)}$
Apr
15
comment Sum of products of $(1 - 1/p)$
@Did Thanks, I have edited the question. I see the ambiguity now! Thanks user1952009 too!
Apr
15
revised Sum of products of $(1 - 1/p)$
clarify
Apr
15
comment Sum of products of $(1 - 1/p)$
@user1952009 I mean $S_N = \sum_{m=1}^{\pi(N)} \prod_{k=1}^m (1-\frac{1}{p_k})$
Apr
15
comment Sum of products of $(1 - 1/p)$
@user1952009 All the terms of the series are different; there is no weighting. I'm trying to understand the reason for the confusion... how could I make it clearer?
Apr
15
comment Sum of products of $(1 - 1/p)$
@Did I'm afraid I don't understand the difference -- I put the final term of $S_n$ only to indicate that we stop at the largest prime $\le n$; otherwise the typical expression is the same, right?
Apr
15
asked Sum of products of $(1 - 1/p)$
Mar
29
awarded  Necromancer
Mar
28
awarded  Revival
Mar
28
answered Intuition behind Erdős proof of the infinitude of prime numbers
Mar
27
comment Time complexity of a modulo operation
There is a bug in the above: The termination condition should not be $s < q$, but running out of digits of $p$.
Mar
27
awarded  Necromancer
Mar
27
awarded  Revival
Mar
18
comment 2015-related question: why are Lucas-Carmichael numbers named after Lucas?
I haven't understood or verified it, but according to mersenneforum.org/showthread.php?t=3359, a Lucas–Carmichael number "is a Lucas pseudoprime to all Lucas sequences". There are also other definitions of Carmichael–Lucas numbers in the literature, so I don't think this terminology is very solid.
Mar
9
awarded  Guru
Feb
21
awarded  Revival
Feb
18
awarded  Benefactor
Feb
18
comment Finding where the tail starts for a probability distribution, from its generating function
Thank you for this excellent and very clear answer and for the link to the Goulden-Jackson cluster method; I will try to learn it.
Feb
16
comment Finding where the tail starts for a probability distribution, from its generating function
@Kirill: Thanks for the explanation; makes sense now. One question (drifting from my original question) is whether we can actually write down bounds for the tail sum (over coefficients $n > l$), beyond just saying that it's asymptotically $C\left(\frac{1}{\alpha}\right)^l$: can we bound the error, for a fixed finite $l$?