mersenneforum.org Twin Prime Question
 Register FAQ Search Today's Posts Mark Forums Read

 2010-09-07, 23:51 #1 Unregistered   2×3×13×113 Posts Twin Prime Question I'm interested in twins of the form k*2^n+/-1. One question: What is the probablity that there is a twin of that form with k<1M and n>1M?
2010-09-08, 02:38   #2
TimSorbet
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

11×389 Posts

Quote:
 Originally Posted by Unregistered One question: What is the probablity that there is a twin of that form with k<1M and n>1M?
Probably: 1 (i.e. it's probably certain that at least one exists...unless I'm mistaken)
It's just quite hard to discover with current technology.

2010-09-08, 02:59   #3
CRGreathouse

Aug 2006

135448 Posts

Quote:
 Originally Posted by Mini-Geek Probably: 1 (i.e. it's probably certain that at least one exists...unless I'm mistaken)
I don't think that's known. Heuristically, we expect something like
$\frac{4000}{\log^22}\sum_{n=10^6}^\infty n^{-2}\approx0.008$
examples, right?

Of course the constant factor needs work based on the residue classes 2^n takes on, as well as the factors in k.

Edit:
$\approx21503\sum_{n=10^6}^\infty n^{-2}\approx0.021$
takes the factors of the k-values into account.

Last fiddled with by CRGreathouse on 2010-09-08 at 03:04 Reason: make approximation clearer

2010-09-08, 05:04   #4
Unregistered

253816 Posts

Quote:
 Originally Posted by CRGreathouse I don't think that's known. Heuristically, we expect something like $\frac{4000}{\log^22}\sum_{n=10^6}^\infty n^{-2}\approx0.008$ examples, right? Of course the constant factor needs work based on the residue classes 2^n takes on, as well as the factors in k. Edit: $\approx21503\sum_{n=10^6}^\infty n^{-2}\approx0.021$ takes the factors of the k-values into account.
Where do the 21503 (second figure) and 4000 (first figure) numbers come from?

2010-09-08, 13:00   #5
CRGreathouse

Aug 2006

22×3×499 Posts

Quote:
 Originally Posted by Unregistered Where do the 21503 (second figure) and 4000 (first figure) numbers come from?
4000 is about 1000 candidates * 2^2, where the 2 is because only odd numbers are used (which twice as likely to be prime). Oh! You want k < 1000000, not k < 1000; in that case it's 4000000.

The other one is a calculated number based on the factorizations of k < 1000. Let me try for k < 1000000. OK, I get 14756135.8...
Code:
sum(k=2,10^6-1,ff(k+k)^2,0.)/log(2)^2
So the expectations are
Code:
4e6/log(2)^2*(zeta(2)-sum(n=1,10^6,n^-2,0.))
8.32 (naive)

14.75 (including factorizations)

So with this larger range, the heuristic probability is high: with the Poisson model, it's something like 1 - e-14.75 = 99.99996% likely that such a pair of primes exist.

 Similar Threads Thread Thread Starter Forum Replies Last Post hydeer Lone Mersenne Hunters 9 2018-04-03 22:54 mathPuzzles Math 10 2017-06-24 08:41 cuBerBruce Puzzles 3 2014-12-01 18:15 eepiccolo Math 7 2005-06-04 23:01 1260 Math 13 2003-10-12 09:48

All times are UTC. The time now is 23:01.

Thu Mar 30 23:01:55 UTC 2023 up 224 days, 20:30, 0 users, load averages: 0.98, 0.78, 0.69

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔