wolfson

OK by means of Mersenne Primes: Mersenne Primes relate to exponent (e) or time ten to the power of x. The equation, is in relation to 2^n-1 is divisible by 2^n-1, when finding out whether a number (e) is a Mersenne prime we then have to use equation(s): Since I have not fathomed out how to get my equation editor to work on scienceforums (due to my incompetence), you will have to bear with me: So where was I oh yes equations to finalize correspondence between a number (e) and a Mersenne Prime: Ok x and y are primes If y is divisible by Mx = 2x-1, then: X = [+]/[-] –1 (mod8) and y = 2kp +1. K being integer. E.g. x =3 (mod4) be prime, then 2x +1 is also prime if it divides by Mx. M(p) = 2p-1 and P(p) = 2^p-1(2^p-1) p is know to be a “perfect number” it has to be, to be a Mersenne. So divisible e numbers are: 2, 3, 5, 7, ….. until m12 (m= Mersenne), where perfect numbers are higher and much harder to solve, think of calculating all number 0 to 1.0x10^1666 (exclusion of 6), remember it must be divisible by Mx^ms, so well you would be there for ever, I know that they have computer programs for Mersenne Primes, but even that take forever, that is the reason for the ‘lack’ of Mersenne Primes. Although they are starting to be produced in a more increased dynamic, it will be a while for the 41st Mp I think around next year (maybe no hopes). I hope this helped explain the Mersenne number to you, if you have any more questions by all means ask away. And I haven’t forgotten about the equation for the normal primes hopefully tomorrow ill manage to crack that. And I’m also looking for triangle inside triangle pictures if anyone can get me them I would be very grateful (see post named FORMULA for more details).

Could someone please help me find a picture of a triangle's with a triangle, a sequence of top line 1 triangle next two and so on.... This would be much appreciated

You cant have a reactant and a product the same they wouldnt react. And yes bond energies are never the same relating to principal quantum number(s).

Ok say^3 why dont you ask Maths Professor, and a physics Professor and then identify the answers, there is not ONE correct answer!!!!!!!

Try incorporating oxides of Strontium Aluminate (SrAl) as the phosphor, that should work better, they used to use Zinc Sulphide with copper sufactants for making photoluminescent paint. GL.

Yes as i already stated there are two possible answers the Physics one and the mathematical one, I chose to represent the mathematical version you chose the physics!!!!!!!!

lol the number are in multiple's of three

Try http://images.google.co.uk/images?svnum=10&hl=en&ie=UTF-8&oe=UTF-8&q=Bose-Einstein+Condensate+&spell=1 i hope it helps.

That blike number is not a normal prime number it's the 40th Mersenne prime, Mersenne primes are realted by 2^p-1, and since there are only 40 Mersenne Primes then u won't be testing any of them at the moment, as for normal prime number, there is an equation it is quite long and highly defined, transposing the Nth term of the prime squenece would be eaier. I'll have to work hard and see if i can solve the Nth term.

Ok in your calculation’s you get n16 (middle number) being $9xx.xx, if you add them all up they won’t add up, the only way you could do that would be to have a higher first term, which is not possible as n16 is suppose to be highest. Take a look at my calculations and try and pinpoint your error. As for temperature calculation the “psychics-wise” answer is yours, “mathematically-wise” YT and I have the correct answer.

i got 5 lol

Yes just EXPAND, 1,2,3, is on the right step. Listen Sn. is right you will not be able to lower the number, unless you lower n and k.

You could still tell if there had been a reaction even if energies were the same, think bonds, think product, think property change. I think Pantano's Rule relate's to thermodynamics of chemical reactants, but take Ab [ab] / [bA]x[Ab] following equilibrium constant(s) in a reversible reaction "theoretically" the same energy would be re-created on its 2nd channel. (aB)--(Ba)---(cD)---(Dc)---(aB). or xy realting to yx.

The equation is just a reagular Sn. for a squenece of muliple progressions the variable's are constant n always being the squence value, and k being the squence constant!

#ABC = (xS^t+yS^p+rS-(S mod 2)) / f The letters relate to colums if you send me the picture or attach the picture of the traingle(s), then ill be happy to transpose the equation for you..

The closed theory is all about the conservation of entropy, i can't see the universe coming to a rest, well at least for another 1.0e100^m. Try looking at Geometry of the Universe that usually gives a better explenation than i can.gl

Stoichiometry and the mole concept , that would be best as soon you will be using it alot

The reduction of acetaldehyde requires a low activation energy level it's coupled with oxidation so its a exergonic reaction, again although low it's still there.

0 C is 32 F, and half of this is 16 F, which is -8.9 C i did the same as you.

I dont mean to be rude, but your answer is incorrect, my equation which is widley used for sequences gives the correct answer as $3000, not $939.39. That answer confused me!

Try looking at http://www.csse.monash.edu.au/~lloyd/tildeAlgDS/Recn/ and http://www.csse.monash.edu.au/~lloyd/tildeAlgDS/Recn/Perm/ and maybe www.cs.princeton.edu/courses/archive/spr01s126/demo/powerpoint/demo-enumerate.ppt

You need a recursion to generate all the permutions for your lotto lines, off the top of my head I can’t think of the recursion but if you give me time I will try and locate one for you.

Why would you want to test the number, would you like an equation?

n_C_k = n!/ k!(n - k)! !=Fractionating This is a simpliar way of writing it out for you, i used the from the final transposed sections of n^2

Your odds are 1 in 133.5 chance is 7.49E-3 (0.00749), the equation I used is M1(x) = Si(xi-E(x))nP(xi) =SixiP(xi)-Sie(x)P(xi)= E(X)-E(X)SiP(Xi)=0. Though this would not be worth doing as spending £134 on lotto tickets to win £10 is just not ethical. The equation goes on a bit more till you transpose to reach N^2 which ends as E(X^2)-EP(X^2).