Breaking the record of Graham's number

Jai Ganesh · 2006-10-04 03:43:37

Let n=6->6->6->6->6
Let n1 equal to the smallest number apart from zero and one which is a perfect square, cube, fourth, fifth, sixth and nth power.
n1+1 is a prime number.
Proof:- n1-1 is divisible to almost any prime number, because of the fact that (a^n-b^n) is divisible by (a-b).
This is a better proof than Graham's Number for Ramsay theory, he he
Such numbers are enormously large.
For example, the smallest number apart from zero and one which is a perfect square, cube, fourth, fifht and sixth power is 1152921504606846976.
(That is how many bytes make an exabyte).
And extrapolating this to tenths power, the result is 3.940842x10^758, a number containing 759 digits.
If this is extrapolated to 20th power, the result is 2.125219 x 10^70077543 approximately, containg 70077544 digits.
And when this extrapolated to 1000th power, the result is 10^10^433, much much larger than a googolplex

Devantè · 2006-10-04 03:56:28

Cool. The modern age just keeps coming up with new things.

Jai Ganesh · 2006-10-04 04:07:47

Prime numbers are never-ending. It is very very difficult to prove that n1+1 is a composite number, well impossible, because of this property i mentioned. However, Graham's Number is a shame to mathematics and to Ramsay theory. The proof isn't sound at all, as said by some other mathematicians. Hence, I was interested in giving a more difficult aspect to prove or disprove by counter-proof.

Ricky · 2006-10-04 04:12:47

Let n=6->6->6->6->6
Let n1 equal to the smallest number apart from zero and one which is a perfect square, cube, fourth, fifth, sixth and nth power.
n1+1 is a prime number.
Proof:- n1-1 is divisible to almost any prime number, because of the fact that (a^n-b^n) is divisible by (a-b).

I don't understand your 6->6->6->6 notation.

I also don't see how stating n1-1 is divible by almost any prime number proves n1+1 is prime.

Jai Ganesh · 2006-10-04 16:43:25

Ricky,
the arrow in the notation denotes John Conway's chained arrow notation which is much much larger than knuth's up-arrow notation.
If n-1 is divisible by almost all prime numbers, n+1 should be prime.
Thats a hypothetic conclusion.

Ricky · 2006-10-04 17:47:52

If n-1 is divisible by almost all prime numbers, n+1 should be prime.
Thats a hypothetic conclusion.

By hypothetic, do you mean one that remains to be unproven?

And what is meant by most?

Jai Ganesh · 2006-10-04 20:37:59

Yes, hypothetic is because, it is next to impossible to prove.
There already exists a proof that (a^n-b^n) is divisible by (a-b).
Therefore, I used the word 'most'.
Like for example, 63 is divisible by 3, 7.
4095 has many prime factors.
These are numbers of the kind I have mentioned in my post #1.

Laterally Speaking · 2007-05-21 05:53:56

I'm pretty sure the reason why people have disclaimed Graham's number is because it is the upper bound to a (rather hairy) problem in Ramsey Theory, the lower bound having been identified as 6 by the same people that "discovered" the upper bound.

As one of these people said, "there's room for improvement".

Laterally Speaking · 2007-05-23 00:33:57

ganesh wrote:

Ricky,
the arrow in the notation denotes John Conway's chained arrow notation which is much much larger than knuth's up-arrow notation.

In fact, The chained-arrow notation here is pretty huge, although I'm not sure I really got the concept of the notation in full.

Math Is Fun Forum

#1 2006-10-04 03:43:37

Breaking the record of Graham's number

#2 2006-10-04 03:56:28

Re: Breaking the record of Graham's number

#3 2006-10-04 04:07:47

Re: Breaking the record of Graham's number

#4 2006-10-04 04:12:47

Re: Breaking the record of Graham's number

#5 2006-10-04 16:43:25

Re: Breaking the record of Graham's number

#6 2006-10-04 17:47:52

Re: Breaking the record of Graham's number

#7 2006-10-04 20:37:59

Re: Breaking the record of Graham's number

#8 2007-05-21 05:53:56

Re: Breaking the record of Graham's number

#9 2007-05-23 00:33:57

Re: Breaking the record of Graham's number

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