mersenneforum.org  

Go Back   mersenneforum.org > Math Stuff > Other Mathematical Topics

Reply
 
Thread Tools
Old 2015-11-27, 09:17   #1
Dubslow
Basketry That Evening!
 
Dubslow's Avatar
 
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

160658 Posts
Default Relatively simple question I've been unable to resolve

I wrote up Clifford Stern's analysis of aliquot sequences in a much more thorough and pedagogical way that explains every step (basically the page I wish I had years ago when I first read his work).

I've found myself unable to prove a small side equality, which while not terribly important for the aliquot analysis, is still of some minor interest to me. It's no doubt fairly easy to prove, so I'd like to bring it to the attention of others on this forum who are far more sophisticated than I.

For now it is available here, though it will be available in more permanent form here when the relevant site is updated with my latest changes.

Please be kind with me
Dubslow is offline   Reply With Quote
Old 2015-11-27, 09:49   #2
Batalov
 
Batalov's Avatar
 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

9,901 Posts
Default

Quote:
Then define the function f as f(n)=[FONT=MathJax_Math][I]x[/I]2[/FONT], and...
...Hmmmm...
Quote:
[During a job interview]
Dr. Lester: Which of these two letters comes first, this one or this one?
Craig Schwartz: The symbol on the left is not a letter, sir?
Dr. Lester: Damn, you're good. I was trying to trick you.
Batalov is offline   Reply With Quote
Old 2015-11-27, 16:27   #3
axn
 
axn's Avatar
 
Jun 2003

5,387 Posts
Default

You want to prove that b(1+p^2+p^4+...p^(2(l-1))) = b(l)

You can proceed as follows:

If l is even, then pair up the terms and pull out (1+p^2)

1+p^2+p^4+...p^(2(l-1)) = (1+p^2)*(1+p^4+p^8+..p^(2(l-2))). Note that p^(2(l-2)) = p^(4(l/2-1))

(1+p^4+p^8+..p^(4(l/2-1))) has l/2 terms. We can proceed this way until we run into an expression with odd number of terms.

The numbers of terms of the form (1+p^2^i) we can pull out depends on the power of two present in l, i.e. b(l).

Next we note that b(1+p^2^i) = 1 (for i>=1). Proof: odd perfect squares are 1 (mod 8). So 1+p^2^i = 2 (mod 8). Hence b()=1
Final factor with odd number of terms will be odd, hence b()=0.

Thus b(1+.p^2+...p^(2(l-1))) = b(l).
axn is offline   Reply With Quote
Old 2015-11-27, 16:53   #4
Dubslow
Basketry That Evening!
 
Dubslow's Avatar
 
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3·29·83 Posts
Default

Brilliant. I had previously noted that if l was even, then we could break it down and I had wondered if there was some recursion -- but this was before I had noted that I could pull out the (1+p) initially (I had done the pairing but not the factoring), so when I had the idea I was looking at (p^(2i)+p^(2i+1))/2^x rather than this recursive (!) pairing and factoring. Wish I had this most important brainwave on my own too -- oh well. (You can see the previous hints in that direction at the second link in the OP, again until such time as it's updated to match the first.

Thanks for the help

Last fiddled with by Dubslow on 2015-11-27 at 16:55
Dubslow is offline   Reply With Quote
Old 2015-11-28, 01:48   #5
LaurV
Romulan Interpreter
 
LaurV's Avatar
 
"name field"
Jun 2011
Thailand

17×19×31 Posts
Default

Quote:
Originally Posted by Dubslow View Post
I had previously noted that if l was even...
no way! you were odd...
LaurV is offline   Reply With Quote
Old 2015-11-28, 06:24   #6
Dubslow
Basketry That Evening!
 
Dubslow's Avatar
 
"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3×29×83 Posts
Default

Quote:
Originally Posted by LaurV View Post
no way! you were odd...
And that's why we use things like TeX to make the typesetting clear...


IlIlIlIlIlIl
Dubslow is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Simple Question Unregistered Information & Answers 3 2012-11-26 02:55
LLRNet linux unable to resolve hostnames nuggetprime No Prime Left Behind 16 2008-12-23 07:46
A simple question Bundu Math 11 2007-09-24 00:02
Simple question... or is it? akruppa Puzzles 28 2006-02-04 03:40
Simple question Suse 9.0 Faraday Linux 13 2005-06-01 02:22

All times are UTC. The time now is 21:09.


Sat Aug 13 21:09:09 UTC 2022 up 37 days, 15:56, 2 users, load averages: 1.06, 1.14, 1.11

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2022, Jelsoft Enterprises Ltd.

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