EDIT: I mean, the bound for second part of the equation is not very clear (randomly generated out of nowhere). Can you please explain how you get the bound for the second part of the equation, especially the part on c^2?

[QUOTE=2M215856352p1;520760]EDIT: I mean, the bound for second part of the equation is not very clear (randomly generated out of nowhere). Can you please explain how you get the bound for the second part of the equation, especially the part on c^2?[/QUOTE]

He tried with a theoretically sound upperbound "8N/(logN)^2" That didn't work out. So he plucked one out of his ***. That is all there is to it.

The thread has moved on from being a failed attempt at GB conjecture proof to masterful trolling.

I suggest we cut our losses.

I meant 10^100 not 10^30.

[QUOTE=axn;520762]He tried with a theoretically sound upperbound "8N/(logN)^2" That didn't work out. So he plucked one out of his ***. That is all there is to it.

The thread has moved on from being a failed attempt at GB conjecture proof to masterful trolling.

I suggest we cut our losses.[/QUOTE]

I strongly recommend that the moderators close this thread. It’s going nowhere now. Should we ban Awojobi for trolling?

[QUOTE=2M215856352p1;520786]Should we ban Awojobi for trolling?[/QUOTE]

Trolling, and being generally obtuse, are not bannable offenses (IMO). [Spam, hate speech, and assorted nastiness are].

Trolling is best dealt with by ...

:dnftt:

OP is welcome to revise his proof to make it more mathy so, I, for one, am not inclined to close this thread.

I'm surprised at your sudden switch. You seemed polite and helpful initially and suddenly you call me a crank and a troll. If a crank is someone who uses secondary school maths to prove an age old conjecture, then I am a crank. Maybe your switch is due to the fact that you are frustrated that your computer simulations backed up my proof within its limits i.e. you couldn't model for sufficiently large N. Your model went for a maximum N of slightly over 1 billion which is minuscule. However, it achieved 100% and number of cycles required was less than the number of primes you used in the 1st quarter. In answer to some of the questions you raised, if only 1 integer was left in the 2nd quarter, the next cycle would have produced it. Your next question concerning how I got the expression for the upper bound can be seen if I use an example. If 10 green bottles are in a row spaced approximately equally apart, is it not clear that an upper bound for the number of a particular spacing, say the spacing between the 2nd and 5th green bottles is 10 green bottles? Now, the total number of repetitions form an arithmetic series added together and this is where the c^2 shows up. By the way, you said you wanted to prove me wrong on something. Still waiting.

[QUOTE=Awojobi;520807]If a crank is someone who uses secondary school maths to prove an age old conjecture, then I am a crank. [/QUOTE]
Go here and rate your score:
[url]https://primes.utm.edu/notes/crackpot.html[/url]

[QUOTE=Uncwilly;520808]Go here and rate your score:
[url]https://primes.utm.edu/notes/crackpot.html[/url][/QUOTE]I count #8, 10, 11, 17, 20, 25 and possibly 9.

Of course, this is not a proof that OP is a crank. It's only a heuristic.

[QUOTE=xilman;520810]I count #8, 10, 11, 17, 20, 25 and possibly 9.

Of course, this is not a proof that OP is a crank. It's only a heuristic.[/QUOTE]
Don't do the OP's homework for them!

:jvang:

[QUOTE=Uncwilly;520811]Don't do the OP's homework for them!

:jvang:[/QUOTE]Good point. :redface:

However, they still need to go to and read the indicated page in order to understand my somewhat cryptic message

I learned something new today. :smile:

In particular, I found out:[quote]
Goldbach arguments based on the density of primes are
known not to work. Here is a simple example explaining
why.

Let T = {4,6,8,.......}

We want to represent all elements of T as the sum of two
elements of a lower density infinite set I will call S.

Goldbach takes S = set of primes.

Consider S1 = {1, 5, 9, 13, 17, 21, ....}

This is the set of odd numbers that are 1 mod 4. Note
that it has POSITIVE density. There are far more
elements of S1 less than a bound B than there are primes
less than B.

However, even though S1 is much denser than the primes,
it is impossible to represent all elements of T as the
sum of two elements of S1. The sum of two elements of
S1 is always 2 mod 4. Half of the elements of T are 0 mod 4.

Density arguments are not sufficient. One must consider
arithmetic properties of the sets as well.[/quote]Perhaps other followers of this thread may also learn.