- a := sum ({n,1,k}, prime_n^(1/n) – prime_(n+1)^(1/(n+1))) , {n,1,k} means n=1 to k.

For k <= 1,000,000, observe a < 1, and a-->1.

What is the largest k tested? Does a >1 ever happen? What does this imply if it does or does not happen? Can this be proved?

John W. Nicholson

[Non-text portions of this message have been removed] - I do not know why the extra characters were place in, but here it is again.

a := sum ({n,1,k}, prime_n^(1/n) prime_(n+1)^(1/(n+1))) , {n,1,k} means n=1 to k.

For k <= 1,000,000, observe a < 1, and a-->1.

What is the largest k tested? Does a >1 ever happen? What does this imply if it does or does not happen? Can this be proved?

John W. Nicholson

--- In primenumbers@yahoogroups.com, "John W. Nicholson" <reddwarf2956@...> wrote:

>

> a := sum ({n,1,k},Â prime_n^(1/n) â" prime_(n+1)^(1/(n+1)))Â ,Â Â {n,1,k} means n=1 to k.

> For k <= 1,000,000, observe a < 1, and a-->1.

>

> Â

> What is the largest k tested? Does a >1 ever happen? What does this imply if it does or does not happen? Can this be proved?

> Â

> John W. Nicholson

>

> [Non-text portions of this message have been removed]

> - --- In primenumbers@yahoogroups.com,

"John" <reddwarf2956@...> wrote:

> a := sum ({n,1,k}, prime_n^(1/n) prime_(n+1)^(1/(n+1)))

which was an unhelpful way of defining

a(k) = 2 - prime(k+1)^(1/(k+1))

At large k, the PNT gives

(1 - a(k))*log(k)/k = 1 + O(log(log(k))/log(k))

David - --- In primenumbers@yahoogroups.com,

"djbroadhurst" <d.broadhurst@...> wrote:

> a(k) = 2 - prime(k+1)^(1/(k+1))

but meant to write

> (1 - a(k))*log(k)/k = 1 + O(log(log(k))/log(k))

(1 - a(k)) = (log(k)/k)*(1 + O(log(log(k))/log(k)))