User joriki - MathOverflow

Answer by joriki for Asymptotics for primality of sum of three consecutive primes

2011-08-10T06:17:52Z

I verified Noam's calculation of the factor $\lambda=2.30096\ldots$ and Álvaro's computations, extended the latter and calculated the corresponding ratios:

$$ \begin{array}{|c|c|c|c|c|} n & R(n) & 2n/\log n & \lambda n / \log n & R(n)\log n/n\\ \hline 10 & 7 & 9 & 10 & 1.61181\\ 100 & 44 & 43 & 50 & 2.02627\\ 1000 & 339 & 290 & 333 & 2.34173\\ 10000 & 2437 & 2171 & 2498 & 2.24456\\ 100000 & 18892 & 17372 & 19986 & 2.17502\\ 1000000 & 157183 & 144765 & 166549 & 2.17156\\ 10000000 & 1346797 & 1240841 & 1427564 & 2.17078\\ 30000000 & 3784831 & 3484987 & 4009410 & 2.17208\\ \end{array} $$

(The values in the third and fourth columns are rounded to the nearest integer, the values in the last column are rounded to 5 digits after the decimal point.)

I don't think we can deduce anything from the ratio in this form, however, since it shows convergence in the "random" fluctuations but not with respect to the asymptotic approximations made, e.g. dropping a term $\log\log n$, which at this stage is still comparable to $\log n$; a more detailed analysis will be required to test the independence hypothesis in this case.

[Update:] With reference to Noam's comments below, here are some data for the relative frequencies of the sum of three consecutive primes being divisible by the first four odd primes. These are averaged over samples of $400,000$ primes beginning at powers of ten, which are given in the first column; note that these refer to the numbers $x$ themselves, not the indices $n$ of the primes.

$$ \begin{array}{|c|c|c|c|c|} \log_{10}x&3&5&7&11\\ \hline 8 &0.183&0.165&0.130&0.087\\ 9 &0.189&0.169&0.131&0.087\\ 10&0.195&0.170&0.133&0.087\\ 11&0.198&0.172&0.133&0.088\\ 12&0.203&0.173&0.133&0.088\\ 13&0.208&0.175&0.134&0.087\\ \hline \text{limit?}&0.250&0.188&0.139&0.090 \end{array} $$

I also looked at the joint distribution of the residues modulo $3$ for the three primes. There's a significant preference for avoiding repeated residues; for instance, at $x=10^9$, the repeating patterns $1,1,1$ and $2,2,2$ have relative frequencies around $0.095$, the alternating patterns $1,2,1$ and $2,1,2$ have relative frequencies around $0.150$, and the remaining mixed patterns have relative frequencies around $0.128$, which is almost completely explained by $1,1$ and $2,2$ having relative frequencies $0.445$ and $1,2$ and $2,1$ having relative frequencies $0.555$. I'm trying to work out a probabilistic model for these effects.

Answer by joriki for Polylogarithm inequality

2011-02-07T14:17:11Z

The inequality seems to be true of the partial sums as well (though I haven't checked that thoroughly), so you might be able to prove it by induction, but I don't quite see how to do that.

Here's an idea for a different proof:

Convert to a common denominator (dropping a factor of 3):

\[{\def\Li{\text{Li}}}\Re\left[\frac{3\Li_1(z)\Li_3(z)-2\Li_2(z)\Li_2(z)}{\Li_2(z)\Li_3(z)}\right]\]

Substitute the definition:

\[\def\sumty{\sum_{i=1}^\infty}\Re\left[\left(3\sumty\frac{z^n}{n}\sumty\frac{z^n}{n^3}-2\sumty\frac{z^n}{n^2}\sumty\frac{z^n}{n^2}\right)/\left( \sumty\frac{z^n}{n^2}\sumty\frac{z^n}{n^3}\right)\right]\]

Gather powers of $z$:

\[\def\sumk{\sum_{k=2}^\infty}\def\suml{\sum_{l=1}^{k-1}}\Re\left[\left( 3\sumk\left(\suml\frac{1}{(k-l)l^3}\right)z^k- 2\sumk\left(\suml\frac{1}{(k-l)^2l^2}\right)z^k \right)/\left( \sumk\left(\suml\frac{1}{(k-l)^2l^3}\right)z^k \right)\right]\]

Combine and convert to a common demoninator in the numerator:

\[\def\sumk{\sum_{k=2}^\infty}\def\suml{\sum_{l=1}^{k-1}}\Re\left[\left( \sumk\left(\suml\frac{3(k-l)-2l}{(k-l)^2l^3}\right)z^k \right)/\left( \sumk\left(\suml\frac{1}{(k-l)^2l^3}\right)z^k \right)\right]\]

Divide out the leading term $z^2$, and take a $1$ out of the sum (thereby removing the constant term in the numerator):

\[\def\sumkk#1{\sum_{k=#1}^\infty}\def\suml{\sum_{l=1}^{k-1}} 1 + \Re\left[\left( \sumkk3\left(\suml\frac{3(k-l)-2l-1}{(k-l)^2l^3}\right)z^{k-2} \right)/\left( \sumkk2\left(\suml\frac{1}{(k-l)^2l^3}\right)z^{k-2} \right)\right]\]

The coefficients in the numerator decay with $1/k$, the ones in the denominator with $1/k^2$. The leading terms are (courtesy of WolframAlpha):

\[ 1 + \Re\left[\frac{\frac{1}{2}z+\frac{475}{864}z^2+\frac{445}{864}z^3+...}{1+\frac{3}{8}z+\frac{155}{864}z^2+\frac{175}{1728}z^3+...}\right]\]

Note that the leading coefficients in the numerator are all close to $1/2$. Thus, if we multiply the numerator by $1-z$, the leading terms will nearly cancel, and the coefficients will decay with $1/k^2$, as in the denominator. Doing that (and also pulling out a factor of $z/2$ from the numerator) yields:

\[\def\sumkk#1{\sum_{k=#1}^\infty}\def\suml{\sum_{l=1}^{k-1}} 1 + \Re\left[\frac{z}{2(1-z)}\left( \sumkk3\left(2\Delta_k\suml\frac{3(k-l)-2l-1}{(k-l)^2l^3}\right)z^{k-3} \right)/\left( \sumkk2\left(\suml\frac{1}{(k-l)^2l^3}\right)z^{k-2} \right)\right]\]

where $\Delta_k f(k) := f(k) - f (k-1)$. The leading terms are now

\[ 1 + \Re\left[\frac{z}{2(1-z)} \frac{1+\frac{43}{432}z- \frac{5}{72}z^2+...}{1+\frac{3}{8}z+\frac{155}{864}z^2+\frac{175}{1728}z^3+...}\right]\]

I think from there you may be able to show by distinguishing cases and making use of the $1/k^2$ decay of the coefficients that the real part in the second term can never be less than or equal to $-1$. If the term $3/8z$ in the denominator is too big and causes trouble, it might help to multiply through by $1-3/8z$, as in the numerator.

I'd appreciate if you let me know what you're using this for :-)

Comment by joriki

2013-04-20T12:38:06Z

I proved your assertion about the 5-vertex case in this math.SE post: math.stackexchange.com/questions/367287.

Comment by joriki

2013-03-09T01:54:50Z

@Qiaochu: The distances correspond to the barycentric coordinates of the point. More generally, choosing a point uniformly in a $k$-simplex corresponds to independently uniformly choosing $k$ points in the unit interval and using the resulting $k+1$ interval lengths as barycentric coordinates.

Comment by joriki

2012-12-12T16:25:02Z

cross-posted to math.stackexchange.com/questions/256598

Comment by joriki

2012-08-07T23:08:33Z

@mjqxxxx: I've now updated my answer at math.SE to give a more rigorous proof of the asymptotic $2R^2\log R$ behaviour (which shows that the lower bound is tight).

Comment by joriki

2012-08-07T14:21:53Z

That first link is now dead; the article is at www1.math.american.edu/People/kalman/pdffiles/….

Comment by joriki

2012-08-07T10:46:02Z

@fedja: See [my answer at math.SE](math.stackexchange.com/a/179859/6622), where I derive an expansion up to $O(R^2)$ and compare with simulation results.

Comment by joriki

2012-02-29T12:15:45Z

Answered at math.SE.

Comment by joriki

2011-10-11T16:29:49Z

Sorry for taking too long to write up the answer at math.SE -- David beat me to it by 20 minutes :-). Here it is: math.stackexchange.com/questions/71235/…

Comment by joriki

2011-09-13T21:17:55Z

@ARupinski: No need for the curly brackets; \mathbb R also produces $\mathbb R$.

Comment by joriki

2011-08-24T15:21:47Z

It's good style to tell people that you're cross-posting (math.stackexchange.com/questions/59470/…); else efforts will be unnecessarily duplicated.

Comment by joriki

2011-08-10T17:45:47Z

@Noam: Yes, sorry, $1/4$. Regarding you other questions, see above. Yes, seeing the ratio converge to $2.30$ isn't feasible, I think; what might be feasible, though, is to model these effects well enough to support the conclusion that they eventually disappear.

Comment by joriki

2011-08-10T15:22:09Z

@Noam: Yes, I made the same observations and was just doing some experiments on that -- the approach is quite slow; even at the end of the table the probability for $l=3$ is only about $0.2$ instead of the asymptotic $3/4$; this is due to repeated residues being significantly less likely than alternating residues, and this effect only decays slowly as the gaps between the primes become wider.

Comment by joriki

2011-08-10T06:43:28Z

@Will: That seems like an overvaluation of proofs to me. It can't be proved, but it seems highly likely that it does, and something interesting might be learned from what the limit seems to be quite independent of whether its existence can be proved.