20210825, 21:37  #1 
"Καλός"
May 2018
101010110_{2} Posts 
Incorrect guess based on limited data: Number of Primes 6k1 > Number of Primes 6k+1
Guess based on limited data: The number of primes of type 6k1 is greater than the number of primes of type 6k+1.
Range, π(Range), Number of Primes 6k1, 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 20210827 at 13:47 Reason: Revise title 
20210825, 22:13  #2 
Apr 2020
802_{10} Posts 
You've rediscovered a very wellknown 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 20210825 at 22:19 
20210825, 22:20  #3 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
1101000001000_{2} Posts 
Occasionally, 6k+1 will be exposed to possible factoring by one higher prime than 6k1 is, as least factor > 1. For example 167 is prime and = 12.9228...; 169=13^{2}.
Last fiddled with by kriesel on 20210825 at 22:29 
20210825, 22:25  #4 
Feb 2017
Nowhere
170A_{16} 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.

20210826, 06:34  #5  
"Καλός"
May 2018
2·3^{2}·19 Posts 
Quote:


20210826, 07:20  #6 
Romulan Interpreter
"name field"
Jun 2011
Thailand
17·19·31 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 20210826 at 07:30 
20210826, 13:25  #7 
Jun 2021
33_{16} 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! 
20210826, 18:44  #8 
"Καλός"
May 2018
101010110_{2} 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 π_{3,2}(x) and π_{3,1}(x) are equal.
Actually, π_{3,2}(x) = π_{3,1}(x) for x = 2, 3, 7, 13, 19, 37, 43, 79, 163, 223 and 229: π_{3,2}(2) = π_{3,1}(2) = 0 (trivial case) π_{3,2}(3) = π_{3,1}(3) = 0 (trivial case) π_{3,2}(7) = π_{3,1}(7) = 1 π_{3,2}(13) = π_{3,1}(13) = 2 π_{3,2}(19) = π_{3,1}(19) = 3 π_{3,2}(37) = π_{3,1}(37) = 5 π_{3,2}(43) = π_{3,1}(43) = 6 π_{3,2}(79) = π_{3,1}(79) = 10 π_{3,2}(163) = π_{3,1}(163) = 18 π_{3,2}(223) = π_{3,1}(223) = 23 π_{3,2}(229) = π_{3,1}(229) = 24 It seems that the problem is still open for x > Infinity as there is no strict proof. 
20210826, 19:16  #9 
"Καλός"
May 2018
2×3^{2}×19 Posts 

20210826, 19:33  #10  
"Καλός"
May 2018
2·3^{2}·19 Posts 
Quote:
Quote:
What I am seeking an answer to is if π_{3,2}(x) = π_{3,1}(x) + C for some small nonzero constant C. Last fiddled with by Dobri on 20210826 at 19:38 

20210826, 19:37  #11  
Apr 2020
2·401 Posts 
Quote:
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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