mersenneforum.org question on a few things
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 2011-06-18, 18:25 #1 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26×131 Posts 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
2011-06-18, 20:56   #2
CRGreathouse

Aug 2006

175B16 Posts

Quote:
 Originally Posted by science_man_88 has a(n) ever been expressed as a algebraic equation
No doubt. I imagine it's not too dissimilar to that of A003010.

2011-06-18, 21:07   #3
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by CRGreathouse No doubt. I imagine it's not too dissimilar to that of A003010.
it's based on it.

2011-06-18, 21:47   #4
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

203008 Posts

Quote:
 Originally Posted by science_man_88 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

2011-06-18, 22:21   #5
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by science_man_88 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 ?

 2011-06-18, 23:18 #6 science_man_88     "Forget I exist" Jul 2009 Dumbassville 20C016 Posts anybody confused about what I mean ? Last fiddled with by science_man_88 on 2011-06-18 at 23:39
2011-06-19, 00:31   #7
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by science_man_88 anybody confused about what I mean ?
You certainly are.

2011-06-19, 00:32   #8
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

20C016 Posts

Quote:
 Originally Posted by science_man_88 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

2011-06-19, 00:34   #9
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

20C016 Posts

Quote:
 Originally Posted by R.D. Silverman You certainly are.
Doh! , I knew that was coming. I'm more confusing , mostly to others but, sometimes I confuse myself.

 2011-06-19, 12:33 #10 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26·131 Posts however I know I'm not confused just having trouble getting my point across, no surprises there.
2011-06-19, 16:31   #11
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by R.D. Silverman You certainly are.
What's your reasoning for that comment? I'd love to know!

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