User talk:Alexander Adam

A bound on sigma

You write:

$\sigma _{1}\left(n\right)=\sum _{d|n}d=\sum _{d|n}{\frac {n}{d}}\leq 1+n+\sum _{i=2}^{\sigma _{0}\left(n\right)-2}\left\lfloor {\frac {n}{i}}\right\rfloor$

I think you intended $\sigma _{0}(n)-1$, yes? That bound is tight for primes (of course) and 4, 6, 8, 12. (I think the list is complete but didn't prove it.)

Charles R Greathouse IV 05:12, 9 January 2013 (UTC)

• Yes, that is true - thank you. I shifted the summation index but I forgot the upper bound. Regarding the list, number 12 is already special and for a proof one maybe has to check for such corner cases - could be a good excercise :). Alexander Adam 01:29, 10 January 2013 (UTC)

Prime prediction conjecture

I don't understand this conjecture. First of all, what is an "exponential domain" in this context? (Perhaps I am misparsing the sentence.)

Second, it is surely not true that $p_{2a_{2}-a_{1}}\approx p_{a_{2}}^{2}p_{a_{1}}^{-1}$ on all of $2<p_{a_{1}}<p_{a_{2}},$ (unless you have an exceptionally generous interpretation of $\approx$.) For example, $p_{2n-2}$ is far smaller than $p_{n}^{2}/3$ for any reasonably large value of n. Your examples all have $a_{2}-a_{1}$ small but I don't see the precise condition. (You say that you can "probably" choose O of the square root, but that's not part of the statement itself.) Of course if the numbers are close $p_{a_{1}}\approx p_{a_{2}}$ and so $p_{2a_{2}-a_{1}}\approx p_{a_{2}}\approx p_{a_{2}}^{2}/p_{a_{1}}.$

The usual approximation would be

$p_{2a_{2}-a_{1}}\approx (2a_{2}-a_{1})\log(2a_{2}-a_{1})$

which is

$p_{2a_{2}-a_{1}}\approx (2a_{2}-a_{1})\log a_{2}$

when $a_{1}\approx a_{2}.$

Charles R Greathouse IV 19:02, 15 January 2013 (UTC)

• Nice that you are interested. I meant that the prime number function behaves there like an exponential function: $\prod _{i=1}^{n}\exp \left(a_{i}\right)=\exp \left(\left\langle a\right\rangle \right)^{n}$. This is a functional equation of the exponential function and i want to pick the domain in a way that the prime number function fulfills it too. I'm not sure about the exact definition of the domain but I figured that the elements $a$ should not be too far apart. You can prove the asymptotic correctness of this formula with the approximation formula $p_{a_{i}}\approx {a_{i}}\ln {a_{i}}$ (just checked it for the case where n=2). This is what I meant with exponential domain. Does it help?
• I choose $n=2$ and then $\epsilon =1$ (the conjecture only states that there is an $\epsilon >0$, the asymptotic analysis should give more inside) which makes it an instance of the conjecture. The value of $a_{1}$ should be in general big because for your corner case $p_{2n-2}$ we have $a_{1}=2$ so by conjecture you are only allowed to choose $n\approx 2+{\sqrt {2}}$.
• Your reasoning for $p_{a_{1}}\approx p_{a_{2}}$ is of course true and holds for every continuous function (if I am not mistaken), that is why I came up with this formula in the first place and I asked myself about the conditions on $a$. How far may I extend the indices $a$? It seems that the result predicts the primes better than the usual approximation if $a_{1}$ is big but it might be not too big, but right now I can prove the correctness only in the limit $a_{1}\rightarrow \infty$ and I cannot say if it will always be better than the usual approximation (I guess not). Alexander Adam 22:56, 15 January 2013 (UTC)

I think I'm starting to see how the parts are fitting together. Part of the answer to "why does this seem to be better?" is that the log approximation isn't that close to the truth, and although li would do better using the prime itself is obviously superior. You might compare the result to $p_{a_{2}}+{\frac {a_{2}-a_{1}}{2}}\log a_{1}a_{2}$ which removes this difference.

Charles R Greathouse IV 00:43, 16 January 2013 (UTC)

I also thought by using the primes itself one is able to decode their relative relations with more precision. By the way, your formula works a lot better but how did you come up with it? I thought about it and I could not find a good answer. Alexander Adam 18:30, 16 January 2013 (UTC)

I started with the lower prime and took the number of primes to skip. I then took the average (geometric mean, in this case) of the two numbers to get an idea of the order of magnitude of the average gap between primes and multiplied it by the number of primes to move forward. Charles R Greathouse IV 18:59, 16 January 2013 (UTC)

Yes, I can also see it now. You took the logarithmic of the geometric mean of the two indices because this is equal to the mean of the local densities $\ln a_{1}$ and $\ln a_{2}$ of the two prime numbers $p_{a_{1}}$ and $p_{a_{2}}$. Then you assumed that this average density holds also for the range $a_{2}-a_{1}$ of primes we want to skip - this is nice. Alexander Adam 19:20, 16 January 2013 (UTC)