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Let $p_n$ denote the $n$-th odd prime number. Consider the number $(p_n)^n$, the $n$-th power of $p_n$.

The Conjecture

The decimal expansion of $(p_n)^n$ always contains a sequence of consecutive digits whose sum is equal to $p_n$.

Let this sum be $S(p_n)$. Then, the conjecture can be simply stated as:

$S(p_n)=p_n$

Verified Cases

The conjecture has been verified for all odd primes upto $p_{18700}=208739$ with no counter examples found.

(Thanks to vengy and Moxy)

Here are the first ten examples:

$$\begin{array}{|c|c|} \hline n & p_n & (p_n)^{n} & S(p_n)\\ \hline 1 & 3 & 3 &3=3\\\hline 2 & 5 & 25&5=5\\ \hline 3 & 7 &343&3+4=7\\\hline 4 & 11 &14641&1+4+6=11\\ \hline 5 & 13 &371293&3+7+1+2=13\\\hline 6 & 17 &24137569&2+4+1+3+7=17\\ \hline 7 & 19 &893871739&7+3+9=19\\\hline 8 & 23 &78310985281&8+5+2+8=23\\ \hline 9 & 29 &14507145975869&9+7+5+8=29\\ \hline 10 & 31 &819628286980801&6+9+8+0+8=31\\ \hline \end{array}$$

Open Question

While computational evidence supports this conjecture for all tested primes, a rigorous mathematical proof or disproof is currently lacking. The following questions arise:

What underlying mathematical structure ensures that such a sequence always exists in the decimal expansion of $( p_n)^n$ ?

Is there a relationship between the placement of these sequences and properties of $p_n$ or $n$?

This question has been cross-posted on Stack Exchange.

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    $\begingroup$ What is the probability that a random fifteen digit number in base ten has a sequence of digits adding up to 31? $\endgroup$ Commented Jan 11 at 7:28
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    $\begingroup$ @DaveBenson, The probability is 84.83%. $\endgroup$ Commented Jan 11 at 8:36
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    $\begingroup$ It's very easy to come up with unprovable conjectures about prime numbers. The structurelessness of the digit expansions of large primes means that your guess here is (simultaneously!) probably true, impossible to prove, and not particularly interesting as a problem which can be examined with current technology. $\endgroup$ Commented Jan 11 at 9:26
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    $\begingroup$ This question was posted in parallel here and on math.SE. The recent answer was also posted in parallel. This is not good. I suggest to continue the discussion on math.stackexchange.com/questions/5021626 . $\endgroup$ Commented Jan 12 at 12:00

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This is not an answer but long comment on some observation motivated by this question which shows that this conjecture might be a special case of a more generic properties of numbers. Looks like numbers are up to something mischivious again.

Let $n \ge 2$ and $a_n$ be the smallest exponent such that the decimal representation of $n^{a_n}$ contains a at least string of consecutive digits which sums up to $n$. I plot the graph of $n$ vs $a_n$ for $n \le 20000$.

enter image description here

The scatterplot clearly shows four distinct lines categories depending on the last digit of $n^{a_n}$. Moreover the growth rate of each line seems to be proportional to $\dfrac{n}{\log n}$ and only differ by the proportionality constant. More specifically, for example for lowest line i.e. numbers not ending in $0$, we have

$$ a_n \approx \dfrac{n}{2\log n} $$

It follows from OP's question that if $a_n$ is the smallest exponent for a prime $p_n$, then there is at least one more exponent close to $2a_n$ which is the one OP has observed.

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