|Jun19-05, 09:48 AM||#1|
Alsings hypothesis of integers bigger than 2
I have a found a hypothesis which I would like you to look at, and perhabs (dis)prove..
All integers (n) bigger than 2 (3, 4, 5, 6, ...) be descriped as:
n = (p_1 * p_2 * ...) + k
where all p and k are primes, but also include 1. Notice that k < (p_1 * p_2 * ...), and that no prime may occure more than once.
Here is what I found out untill now:
1) All values of the equation can be decriped as 2x added an optional value of 1:
All of the primes (p_i and k) bigger than 2, can be descriped with the equation aswell. I you repeat doing this you will end up with an equation consisted of 2's and 1's added and multiplied together. Check the example:
10 = 7+3
3 = 2+1
7 = 2*3+1 = 2*(2+1)+1
10 = (2*(2+1)+1)+(2+1) = (2*2+2*1)+1+2+1 = (2*3)+2+(1+1) = 2*4+2 = 2*5
This can be done with all the equations of n, but sometimes 2x is added the value of 1..
2) None of n's prime factors occure in the equation for n:
Let p_n be a prime factor to n. If p_n occures in (p_1 * p_2 * ...) the result would also be divisible by p_n. This means that no k-value exists (obeying the requirements), which would make n divisible by p_n (k cannot be p_n or divisible to p_n). Since this is a contradiction in terms, p_n cannot occure in (p_1 * p_2 * ...).
Now try to apply the value of p_n to k. Then (p_1 * p_2 * ...) cannot be p_n or divisible to p_n (since the prime cannot occure twice). This gives a value for n, which is not divisible by p_n and results in a contradiction in terms, onces again.
I have showed that all values of the equation can be broken down to 2x added an optional value of 1. But to prove this hypothesis you have to show that all values of 2x added an optional value of 1, also can be descriped with my equation!
|Jun19-05, 02:21 PM||#2|
By the way.. I am now finished running a homemade program made to test if this hypothesis is true for all numbers up to 500; and indeed it is!
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