The reciprocal sum of the prime-prefix-free numbers (https://oeis.org/A287117) converges to a number less than 5*10^14, conditional on the Riemann Hypothesis.
An equivalent version: if we start with 1 and then output a stream of random bits, reading the number as a big-endian binary number at each step (so each time a bit arrives, the number is multiplied by 2 and 1 is either added or not), the expected time until the number is an odd prime is finite.
Anyway, I think it's weird it depends on the Riemann Hypothesis.
Do you have some numerical test for intervals like sum up to 1000, up to 10000, up to 100000, up to 1000000, ... ?
jdb1729 4 days ago [-]
Yes, see the table in Remark 7.3 on page 5, it exceeds 3.5, with growth slowing to a crawl. But the calculations mean little, sum(1/p) grows as divergent log(log(n)), so it also has the appearance of convergence on that basis. Many on math.SE argued for divergence (answers since deleted)! Although the proved upper bound is 5e14, heuristically it should be less than 4. I doubt RH is truly necessary. But even relying on RH, the exact value of the sum is elusive.
gus_massa 3 days ago [-]
Sorry for the delay. Now I had some time to skim the proof, but obviously not time to verify all the results. Some assorted remarks:
* I totally forgot the second log in sum(1/primes) ~= log(log(N)). It's nice to see numerical experiments, but now I realize I had to agree that it's difficult to get a huge number even in the well known case that is infinite.
* The article says that the result of the version with the binary prefix is finite but version with the ternary prefix is infinite. Do you have some numerical experiments? I'd love to see a graphic with the correct amount of logs in both axes to show the difference of behaviour.
* IIUC, the result of the version with quaternary prefix is infinite too, but the result should be comparable to the result of the binary prefix. At least quaternary(N)>binary(N). [I'm not sure if ¿ternary(N)>binary(N)?. Looks difficult.] So it's totally posible (and perhaps obvious) that quaternary(N) is unbounded in spite binary(N) is bounded. It's not very intuitive, but I think I saw something very slightly similar in the past and I got surprised too.
* I'm still not sure why it uses the RH, but it looks like you really thought about it (importing lemma 2.1 and remark 7.1), so I guess I will not be able to remove the RH skimming the paper.
jdb1729 3 days ago [-]
To be forthright it wasn't me doing most of the thinking! I'm taking my time to understand it. The issue with the bases as I understand it is that we need log(b) < 1 for convergence which only works for b = 2 < e. Check out the other math.SE answer which explains the heuristic but reaches the wrong conclusion by being off by a factor of 2.
39 minutes ago [-]
laichzeit0 5 minutes ago [-]
Ok? Why is this significant?
nextaccountic 53 minutes ago [-]
that's a result that says more about the Riemann hypothesis than this specific problem right?
jdb1729 33 minutes ago [-]
Well, not really, the contrapositive is that if the series diverges the Riemann Hypothesis would be refuted. But few doubt that the Riemann Hypothesis is true. So using it as an assumption merely makes the convergence slightly iffy. For another example of its use, see the deterministic Miller primality test: https://en.wikipedia.org/wiki/Miller–Rabin_primality_test
yzydserd 2 hours ago [-]
“Author of The Da Vinci Code”
?
sorokod 10 minutes ago [-]
That would be Leonardo di ser Piero da Vinci
jdb1729 1 hours ago [-]
A nom de OOM.
0976jzhs 1 hours ago [-]
out of memory?
dash2 1 hours ago [-]
Why does it matter to hn? Is it because Dan Brown wrote it?
0976jzhs 1 hours ago [-]
Because the proof and Lean formalization have been produced by a clanker.
Rendered at 21:40:32 GMT+0000 (Coordinated Universal Time) with Vercel.
This Lean-verified proof answers a question I posed 10 years ago: https://math.stackexchange.com/questions/2288648/does-the-su...
An equivalent version: if we start with 1 and then output a stream of random bits, reading the number as a big-endian binary number at each step (so each time a bit arrives, the number is multiplied by 2 and 1 is either added or not), the expected time until the number is an odd prime is finite.
Anyway, I think it's weird it depends on the Riemann Hypothesis.
Do you have some numerical test for intervals like sum up to 1000, up to 10000, up to 100000, up to 1000000, ... ?
* I totally forgot the second log in sum(1/primes) ~= log(log(N)). It's nice to see numerical experiments, but now I realize I had to agree that it's difficult to get a huge number even in the well known case that is infinite.
* The article says that the result of the version with the binary prefix is finite but version with the ternary prefix is infinite. Do you have some numerical experiments? I'd love to see a graphic with the correct amount of logs in both axes to show the difference of behaviour.
* IIUC, the result of the version with quaternary prefix is infinite too, but the result should be comparable to the result of the binary prefix. At least quaternary(N)>binary(N). [I'm not sure if ¿ternary(N)>binary(N)?. Looks difficult.] So it's totally posible (and perhaps obvious) that quaternary(N) is unbounded in spite binary(N) is bounded. It's not very intuitive, but I think I saw something very slightly similar in the past and I got surprised too.
* I'm still not sure why it uses the RH, but it looks like you really thought about it (importing lemma 2.1 and remark 7.1), so I guess I will not be able to remove the RH skimming the paper.
?