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The Landau-Ramanujan Constant is related to the sum of 2 squares. See : http://en.wikipedia.org/wiki/Landau%E2%80%93Ramanujan_constant

Can a similar thing be said for the sum of 4 positive cubes ? Or any other (nontrivial) fixed amount of positive cubes ?

In other words can the Landau-Ramanujan Constant be generalized towards positive cubes ?

In particular Im intrested in the following :

Let $n,m$ be positive integers. Let $f_n(m)$ be the counting function for the sum of $n$ positive cubes.

I assume ( and have been told later by my master ) that

$$f_n(m) = \dfrac {C_n m^{a_n}}{ln(m)^{b_n}ln(ln(m))^{c_n}}+O(1)$$

Where $C_n,a_n,b_n,c_n$ are real numbers depending on $n$ only.

The $O(1)$ implies a Landau-Ramanujan like constant.

I wonder what the values are for $C_4,a_4,b_4,c_4,C_5,a_5,b_5,c_5,C_6,a_6,b_6,c_6$ ?

I read that Davenport proved that

$$ f_3(m) << m^{54/47 + \epsilon}$$ for every $$\epsilon > 0 $$

But I was not able to find more related things.

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  • $\begingroup$ Not sure what you mean by "The $O(1)$ implies a Landau-Ramanujan like constant". That would only require a multiplicative error of $(1 + o(1))$. An additive error of $O(1)$ is far too sharp to expect, and I suspect provably false in the sum-of-squares case. $\endgroup$ – Erick Wong Jan 31 '15 at 16:28
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It is known that every positive integer $n$ not congruent to $4$ or $5$ mod $9$ is the sum of four cubes, allowing negative cubes. It is an open problem if four cubes always suffice (but it is suspected). For $5$ cubes it is known to be true. So the answer depends on one hand how many cubes we ask for. For $5$ cubes the answer is yes, for four cubes we do not know (I think). If we require non-negative cubes, then we need more cubes. It is known that every integer $n≥ exp(524)$ is a sum of seven non negative cubes.

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  • $\begingroup$ That is intresting. But at the same time its a bit like saying not all numbers are the sum of 2 squares. Also trivial to say that it depends on the amount of cubes. It seems that 4,5 or 6 positives cubes are the most intresting cases. Surely you understand this is not a complete answer. But intresting +1. $\endgroup$ – mick Oct 7 '13 at 21:05
  • $\begingroup$ Sure I understand. For a better answer you could make more precise what you mean exactly by "generalized towards cubes". $\endgroup$ – Dietrich Burde Oct 8 '13 at 7:52
  • $\begingroup$ I edited my question. You can extend your answer now if you want. Thanks for your effort. $\endgroup$ – mick Oct 21 '13 at 21:18

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