For a positive integer $k$ let $\gamma\_k(n)$ be the number of representations of $n$ as a sum of strictly increasing perfect $k^{\text{th}}$ powers. For example $\gamma\_k(2)=0$ for any $k$. Now is the following true?



> 
> For any $x\in \mathbb{R}^\ast\_+$ there's an integer $N$ such that for all $n\ge N$
>  $$\dfrac{\max\_{1\le i\le n}\;\gamma\_k(i)}{n}>x$$
> 
> 
> 


Motivation :
This is motivated from an easier problem :
Let $\gamma^{\ast}\_k(n)$ be number of representations of $n$ as sum of distinct $k^{\text{th}}$ powers, where order doesn't matter. Then $\exists \; n$ such that 
$$\gamma\_k^{\ast}(n)>nx$$ for any positive real $x$. Can someone shed some light on this conjecture of mine? I have tried to check as much as possible,and I have found no way out of it. Thanks for all help. 



Arbitrarily large number of representations
Yes, the statement is true.


For any positive integer $m$ there are $2^m$ subsets of $\{1^k,2^k,\ldots,m^k\}$. Each subset has sum bounded by $m^{k+1}$, so by the pigeonhole principle
$$
\max\_{1\leq i\leq m^{k+1}} \gamma\_k(i)\geq \frac{2^m}{m^{k+1}}.
$$
This shows that
$$
\frac{\max\_{1\leq i\leq n}\gamma\_k(i)}{n}\geq \frac{2^{\lfloor\sqrt[k+1]{n}\rfloor}}{\lceil\sqrt[k+1]{n}\rceil^{k+1} n},
$$
which goes to $\infty$ as $n\to\infty$.


