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[2014] Representations by 2sinh^[0.5] - Printable Version

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[2014] Representations by 2sinh^[0.5] - tommy1729 - 11/15/2014

As mentioned before Im considering number theory connections to tetration.

One of those ideas is representations.

Every integer M is the sum of 3 triangular number ...

or 4 squares.

Also every integer M is the sum of at most O(ln(M)) powers of 2.

This is all classical and pretty well known.

But there are functions that have growth between polynomials and powers of 2.

So since 2sinh is close to exp and 2sinh^[0.5](0) = 0 , it is natural to ask

2S(n) := floor 2sinh^[0.5](n)

2S-1(n) := floor 2sinh^[-0.5](n)

2S numbers := numbers of type 2S(n)

2S-1 numbers := numbers of type 2S-1(n)

1) Every positive integer M is the sum of at most A(M) 2S numbers.
A(M) = ??

2) Every positive integer M is the sum of at most 2S(M) B numbers.
B numbers := B(n) = ??

Of course we want sharp bounds on A(M) and B(n).

( A(M) = 4 + 2^M works fine but is not intresting for instance )

regards

tommy1729



RE: [2014] Representations by 2sinh^[0.5] - tommy1729 - 11/16/2014

To estimate A(M) I use the following

Tommy's density estimate

***

Let f(n) be a strictly increasing integer function such that f(n)-f(n-1) is also a strictly increasing integer function.

Then to represent a positive density of primes between 2 and M

we need to take T_f(M) elements of f(n).

T_f(M) is about ln(M)/ln(f^[-1](M)).

This is an upper estimate.

***

In this case to represent a positive density of primes between 2 and M we then need about

ln(M)/ln^[3/2](M) 2S numbers.

This is a brute upper estimate.
A(M) is estimated as sqrt( ln(M)/ln^[3/2](M) ).

Improvement should be possible.

regards

tommy1729