mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Miscellaneous Math

Reply
 
Thread Tools
Old 2011-06-18, 18:25   #1
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts
Default question on a few things

most of you probably know of A002812 (a.k.a a(n) = 2a(n-1)^2 - 1, or a(n) = A003010(n)/2 ), and I want to ask a few questions about it:

1) has a(n) ever been expressed as a algebraic equation, especially using the fact that a(1) ( or a(2) depending on how you count it) is M3

and

2) if not, could this ever be used to find a pattern in the Mersenne numbers with prime exponents, that work out to be Mersenne primes. ( I mention this because they intersect and so we have a basis to form a relationship off which to base the relations between the two).

Last fiddled with by science_man_88 on 2011-06-18 at 18:29 Reason: added in and then bolded the bad word everyone hates
science_man_88 is offline   Reply With Quote
Old 2011-06-18, 20:56   #2
CRGreathouse
 
CRGreathouse's Avatar
 
Aug 2006

172116 Posts
Default

Quote:
Originally Posted by science_man_88 View Post
has a(n) ever been expressed as a algebraic equation
No doubt. I imagine it's not too dissimilar to that of A003010.
CRGreathouse is offline   Reply With Quote
Old 2011-06-18, 21:07   #3
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts
Default

Quote:
Originally Posted by CRGreathouse View Post
No doubt. I imagine it's not too dissimilar to that of A003010.
it's based on it.
science_man_88 is offline   Reply With Quote
Old 2011-06-18, 21:47   #4
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts
Default

Quote:
Originally Posted by science_man_88 View Post
it's based on it.
a(n) = 2a(n-1)^2 - 1 comes about by x^2-2=4y^2 -2 when x is even, and so \frac{4y^2-2}{2} = 2y^2-1 but I was more leaning towards in relation to p=3 as a(1)=2^3-1 =7.

Last fiddled with by science_man_88 on 2011-06-18 at 21:50
science_man_88 is offline   Reply With Quote
Old 2011-06-18, 22:21   #5
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts
Default

Quote:
Originally Posted by science_man_88 View Post
a(n) = 2a(n-1)^2 - 1 comes about by x^2-2=4y^2 -2 when x is even, and so \frac{4y^2-2}{2} = 2y^2-1 but I was more leaning towards in relation to p=3 as a(1)=2^3-1 =7.
A000225 and A002812 have 7 in common from that common value on using algebra can we figure out what elements will divide each other ?


2*(2^p-1)^2-1 = 2(2^{2p}-2^{p+1}+1)-1 = 2^{2p+1}-2^{p+2}+2-1 = 2^{2p+1}-2^{p+2}+1 I went one step further into A002818 than this but I'll currently save you on that part.

so I'm trying to relate 2^{p+k}-1 to this sequence is their a general formula outside of just doing (a+b+c)^2 = a^2+ab+ac+ba+b^2+bc+ca+cb+c^2 =a^2+b^2+c^2+2(ab)+2(ac)+2(bc) to relate it back to the known p=3 they rely on ?
science_man_88 is offline   Reply With Quote
Old 2011-06-18, 23:18   #6
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

838410 Posts
Default

anybody confused about what I mean ?

Last fiddled with by science_man_88 on 2011-06-18 at 23:39
science_man_88 is offline   Reply With Quote
Old 2011-06-19, 00:31   #7
R.D. Silverman
 
R.D. Silverman's Avatar
 
Nov 2003

164448 Posts
Default

Quote:
Originally Posted by science_man_88 View Post
anybody confused about what I mean ?
You certainly are.
R.D. Silverman is offline   Reply With Quote
Old 2011-06-19, 00:32   #8
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts
Default

Quote:
Originally Posted by science_man_88 View Post
anybody confused about what I mean ?
Quote:
0, 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023,
Quote:
2, 7, 97, 18817, 708158977, 1002978273411373057,
the bolder term is where they meet any term in either after this should be able to be related back to that term and hence to terms of the other (especially since these sequences can be used with each other). 2^k(2^p-1)+(2^k-1) = Mp+k so is there a simple relation for the other that we can then make a relation between the relations from ?

Last fiddled with by science_man_88 on 2011-06-19 at 00:37
science_man_88 is offline   Reply With Quote
Old 2011-06-19, 00:34   #9
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts
Default

Quote:
Originally Posted by R.D. Silverman View Post
You certainly are.
Doh! , I knew that was coming. I'm more confusing , mostly to others but, sometimes I confuse myself.
science_man_88 is offline   Reply With Quote
Old 2011-06-19, 12:33   #10
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts
Default

however I know I'm not confused just having trouble getting my point across, no surprises there.
science_man_88 is offline   Reply With Quote
Old 2011-06-19, 16:31   #11
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

20C016 Posts
Default

Quote:
Originally Posted by R.D. Silverman View Post
You certainly are.
What's your reasoning for that comment? I'd love to know!
science_man_88 is offline   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
Weird E-shaped things(serious question) jasong Homework Help 36 2020-05-25 10:58
Some things Une Personne Information & Answers 8 2012-06-12 18:10
Some important things that I should know about Raman Lounge 8 2010-10-20 16:35
All Things Skype moo Lounge 34 2006-01-12 10:20
How things are set up... Xyzzy Lounge 22 2003-08-10 14:36

All times are UTC. The time now is 00:21.

Sat Sep 19 00:21:35 UTC 2020 up 8 days, 21:32, 0 users, load averages: 1.34, 1.50, 1.52

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, 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.