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Conjecture:

All prime numbers greater than $109$ is of the form $\displaystyle\sum_{k=1}^5q_k^k$, where all $q_k$ are primes.

The conjecture is extracted from the question and answer here: Annoying primes

Tested for all primes $<70,000$.

The exceptional primes seems to be
$\{2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,109\}$

Are there heuristic arguments for the conjecture?

Also:
All primes except $\{2,3,5,7,11,13,17,19,23,29,37,43,53,61,67\}$ seems to be of the form $\displaystyle\sum_{k=1}^4q_k^k$.
All primes except $\{2,3,5,7,11,13,37,61,127\}$ seems to be of the form $\displaystyle\sum_{k=1}^3q_k^k$.

Both cases tested for primes less than $10,000$.

Lehs
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  • I neither understand the downvote nor the closevote. – Peter Oct 26 '17 at 11:05
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    @Peter: Maybe some kind of misunderstanding about a duplicate? – Lehs Oct 26 '17 at 11:15
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    If every odd number $n>351$ can be expressed by $n=a^3+b^2+c$, where $a,b,c$ are primes, the claim is true. And chances are not bad that this is the case. According to my claulations, for $351<n\le 10^6$, such a representation exists. – Peter Oct 26 '17 at 11:29
  • @Peter: maybe you should post your interesting conjecture as a separate question? – Lehs Oct 26 '17 at 11:34
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    It was down voted 3 hour ago and it was edited only 1 hour ago. So I will remove down vote now. – nonuser Oct 26 '17 at 12:23
  • @Peter I was wondering the same thing. $p^3+q^2+r$ doesn’t seem to be enough for the evens though (it would need to happen via $q=2$) so I think you need at least 4 to get all numbers. – Charles Oct 26 '17 at 12:48
  • @Charles We do not need the even numbers, we can set $q_4=q_5=2$ – Peter Oct 26 '17 at 18:18
  • @Peter It’s just a comment about minimality — if you wanted to get all (large enough) numbers, not just primes, that speaks to the exponent needed. – Charles Oct 26 '17 at 19:30
  • @Charles The conjecture is that all PRIMES have the desired form. – Peter Oct 26 '17 at 19:33
  • @Peter Yes — I was making a broader statement about a more general question, as indeed you were in suggesting looking at representations of the form $p^3+q^2+r$. (Other broader questions are relevant, such as representations by $p+q^3.$) – Charles Oct 26 '17 at 19:37

1 Answers1

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Here’s one small heuristic argument: there are no local obstructions, so lacking a good reason to the contrary, all large enough numbers should be of this form.

Another: There are $\gg x^{137/60-\varepsilon}$ representations up to $x$, and so each number has on average $x^{77/60-\varepsilon}$ representations. Since $$ \int \exp-x^{-76/60} $$ converges, the number of exceptions should be finite.

Charles
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    Please explain the second argument. How did you estimate the number of representations ? – Peter Oct 26 '17 at 18:24