mersenneforum.org Incorrect guess based on limited data: Number of Primes 6k-1 > Number of Primes 6k+1
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 2021-08-25, 21:37 #1 Dobri   "Καλός" May 2018 73 Posts Incorrect guess based on limited data: Number of Primes 6k-1 > Number of Primes 6k+1 Guess based on limited data: The number of primes of type 6k-1 is greater than the number of primes of type 6k+1. Range, π(Range), Number of Primes 6k-1, Number of Primes 6k+1, Difference 1,000,000,000, 50,847,534, 25,424,819, 25,422,713, 25,424,819 - 25,422,713 = 2,106 2,000,000,000, 98,222,287, 49,112,582, 49,109,703, 49,112,582 - 49,109,703 = 2,879 3,000,000,000, 144,449,537, 72,226,055, 72,223,480, 72,226,055 - 72,223,480 = 2,575 ... Note: Primes 2 and 3 are counted in π(Range). There is a consistent slight difference of a few thousand primes. Is there a simple way to explain said difference? Also, please share if there are existing references related to this matter. (* Wolfram code *) nrange = 3000000000; nmax = PrimePi[nrange]; tpc = 0; tp5 = 0; tp1 = 0; tpother = 0; n = 1; While[(n <= nmax), p = Prime[n]; tpc++; If[Mod[p, 6] == 5, tp5++;, If[Mod[p, 6] == 1, tp1++;, tpother++;];]; n++]; Print[tpc, ", ", tp5, ", ", tp1, ", ", tpother]; Last fiddled with by Dr Sardonicus on 2021-08-27 at 13:47 Reason: Revise title
 2021-08-25, 22:13 #2 charybdis     Apr 2020 5·163 Posts You've rediscovered a very well-known phenomenon called Chebyshev's bias. Assuming some strong versions of the Riemann hypothesis, it is known that for most N, there are more primes of the form 2 mod 3 than 1 mod 3 up to N. 1 mod 3 does sometimes take the lead; this first happens at N = 608981813029. The effect relates to the fact that numbers of the form 1 mod 3 can be squares while those of the form 2 mod 3 cannot. In some sense, it's actually the numbers of *primes and prime powers* (edit: with a suitable scaling on the prime powers) that we should expect to be balanced between the two classes, but it's not easy to explain why. See here for an overview of the field. The ratio (primes 1 mod 3)/(primes 2 mod 3) tends to 1, by de la Vallee Poussin's theorem that for any k the primes are evenly distributed among the residue classes mod k that are coprime to k. Last fiddled with by charybdis on 2021-08-25 at 22:19
 2021-08-25, 22:20 #3 kriesel     "TF79LL86GIMPS96gpu17" Mar 2017 US midwest 150238 Posts Occasionally, 6k+1 will be exposed to possible factoring by one higher prime than 6k-1 is, as least factor > 1. For example 167 is prime and $sqrt(167)$ = 12.9228...; 169=132. Last fiddled with by kriesel on 2021-08-25 at 22:29
 2021-08-25, 22:25 #4 Dr Sardonicus     Feb 2017 Nowhere 23·739 Posts In the dim past, I posted here, and recommended reading this paper, which is the same as the one linked to in the preceding post to this thread, though at a different URL.
2021-08-26, 06:34   #5
Dobri

"Καλός"
May 2018

73 Posts

Quote:
 Originally Posted by charybdis You've rediscovered a very well-known phenomenon called Chebyshev's bias. Assuming some strong versions of the Riemann hypothesis, it is known that for most N, there are more primes of the form 2 mod 3 than 1 mod 3 up to N. 1 mod 3 does sometimes take the lead; this first happens at N = 608981813029. The effect relates to the fact that numbers of the form 1 mod 3 can be squares while those of the form 2 mod 3 cannot. In some sense, it's actually the numbers of *primes and prime powers* (edit: with a suitable scaling on the prime powers) that we should expect to be balanced between the two classes, but it's not easy to explain why. See here for an overview of the field. The ratio (primes 1 mod 3)/(primes 2 mod 3) tends to 1, by de la Vallee Poussin's theorem that for any k the primes are evenly distributed among the residue classes mod k that are coprime to k.
Thanks, charybdis, this was useful. It would take several days to reach 608981813029 on a Raspberry Pi 4B device with a compiled Wolfram script. Out of curiosity, I may continue further for some time to observe the tendency after the reversal point.

 2021-08-26, 07:20 #6 LaurV Romulan Interpreter     "name field" Jun 2011 Thailand 2×3×1,669 Posts The material is nicely presented, however I tend to be careful when I see Granvilles' name (I don't like him much, after many gaffes he did, and after the story with Beal). Yeah, the two number lines cross infinitely times, however, one "team" is leading "almost" the whole time. Which reminds me of the old joke with two old guys talking on a bench in the park, "how's your sex life these days", "oh, no problem I have sex almost every day", "I don't believe you, I only have it two or three times per year, you are not serious", "of course I am serious, I had sex almost every day", "how come?", "well, Monday I almost had sex, Tuesday I almost had sex, Wednesday I almost had sex, Thursday..." Last fiddled with by LaurV on 2021-08-26 at 07:30
 2021-08-26, 13:25 #7 RomanM   Jun 2021 3×17 Posts 100+ years old talk to Doctor -friend of mine tell me, that he can have sex in row at the one night, and I can not, Help me! -Ok, its very easy to Help! You just can tell him that You can do the same too!
 2021-08-26, 18:44 #8 Dobri   "Καλός" May 2018 5278 Posts The original paper by Bays and Hudson (1978, available in JSTOR, see https://www.jstor.org/stable/2006165...65734d170a779e) has no mention of several cases for small primes when the numbers of primes of the two types π3,2(x) and π3,1(x) are equal. Actually, π3,2(x) = π3,1(x) for x = 2, 3, 7, 13, 19, 37, 43, 79, 163, 223 and 229: π3,2(2) = π3,1(2) = 0 (trivial case) π3,2(3) = π3,1(3) = 0 (trivial case) π3,2(7) = π3,1(7) = 1 π3,2(13) = π3,1(13) = 2 π3,2(19) = π3,1(19) = 3 π3,2(37) = π3,1(37) = 5 π3,2(43) = π3,1(43) = 6 π3,2(79) = π3,1(79) = 10 π3,2(163) = π3,1(163) = 18 π3,2(223) = π3,1(223) = 23 π3,2(229) = π3,1(229) = 24 It seems that the problem is still open for x -> Infinity as there is no strict proof.
2021-08-26, 19:16   #9
Dobri

"Καλός"
May 2018

73 Posts

Quote:
 Originally Posted by LaurV Yeah, the two number lines cross infinitely times, however, one "team" is leading "almost" the whole time.
Indeed, it is unclear (at least to me) what happens at x -> Infinity.
It could be π3,2(x) = π3,1(x) + C for some small non-zero constant C. Or C=0?

2021-08-26, 19:33   #10
Dobri

"Καλός"
May 2018

73 Posts

Quote:
 Originally Posted by Dr Sardonicus In the dim past, I posted here, and recommended reading this paper, which is the same as the one linked to in the preceding post to this thread, though at a different URL.
Quote:
 Originally Posted by Dr Sardonicus According to Prime Races [which I recommend reading], 1 mod 6 doesn't take the lead until p = 608,981,813,029.
The limit of the ratio π3,2(x) / π3,1(x) would tend to 1 as x -> Infinity for the infinite number of reversals.
What I am seeking an answer to is if π3,2(x) = π3,1(x) + C for some small non-zero constant C.

Last fiddled with by Dobri on 2021-08-26 at 19:38

2021-08-26, 19:37   #11
charybdis

Apr 2020

5×163 Posts

Quote:
 Originally Posted by Dobri π3,2(2) = π3,1(2) = 0 (trivial case) π3,2(3) = π3,1(3) = 0 (trivial case) π3,2(7) = π3,1(7) = 1 ...
Don't forget 2 is a prime of the form 2 mod 3.

Quote:
 Originally Posted by Dobri Indeed, it is unclear (at least to me) what happens at x -> Infinity. It could be π3,2(x) = π3,1(x) + C for some small non-zero constant C. Or C=0?
Littlewood's result from 1914, quoted in Granville and Martin's survey, shows that the difference oscillates from positive to negative infinitely many times, and also takes arbitrarily large positive and negative values.

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