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For the formula below, is there a way to estimate the average minimum value of $n$?

$$x = \sum_{i=1}^n p_i^y$$

Where:

  • $x, y, n \in \Bbb N$
  • $p$ is a prime number and $p \le 1619$

Examples:

  1. For $x = 34$, $n = 2$: $$34 = 31^1 + 3^1$$
  2. For $x = 112$, $n = 2$: $$112 = 109^1 + 3^1$$
  3. For $x = 1850$, $n = 2$: $$1850 = 43^2 + 2^0$$ Incorrect: $$1850 = 1619^1 + 229^1 + 2^1$$

Rephrasing the question, what value does "minimum" $n$ approach as an average for all $x$?

From the examples it seems to be $2$ but that's obviously not sound reasoning.

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    Why don't you have any odd values of $x$? Why do none of your examples specify $y$? Are you only asking about $y=1$? – 2017-02-25
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    In any case, these type of problems are studied extensively under the name Waring-Goldbach problem. The case $y=1$, Goldbach's conjecture has received a vast amount of attention. You should read the survey paper by Kumchev and Tolev: https://arxiv.org/abs/math/0412220 – 2017-02-25

1 Answers 1

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I've not succeeded in getting exact average, but my thoughts may be helpful for someone on the road.

Denote $g(x)$ as the function giving this minimal $n$. As we can have $y = 1$, we can note that $$g(x) \le \min_{k}(g(k) + g(x-k))$$ as we can sum expansions from $k$ and $x-k$ and get an expansion of $x$.

Denote also $\frac1{x}\sum_{k=1}^{x}g(k) = A(x)$ as its average over $m$ successive integers. From the inequality above we have

$g(x) \le \min_{k}(g(k) + g(x-k)) = \frac1{x-1}\sum_{k=1}^{x-1}(g(k) + g(x-k)) = \frac2{x-1}\sum_{k=1}^{x-1}g(k) = 2A(x-1)$

Then lets look at $A(x)$

$$A(x) = \frac1m\sum_{x=1}^mg(x) \le \frac1{m}\sum_{k=1}^{m-1}g(k) + \frac1{m}g(m) \le \frac{m-1}mA(m-1)+\frac2{m}A(m-1) = \frac{m+1}mA(m-1).$$

So

$A(m) \le \frac{m+1}mA(m-1) \le \dots \le (m+1)A(1)$

Not so good, one could think of finer upper bound on $$g(x) \le \min_{k}(g(k) + g(x-k)) \le 1 + \min_{i=2\dots5}g(x-i)$$