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For each prime $p$, let us define $$w_p:=|\{k^{k+1}+(k+1)^k\pmod p:\ k=0,\ldots,p-1\}|,$$ where $a\pmod p$ denotes the residue class $a+p\mathbb Z$.

Based on my computation, I conjecture that $$w_p=(1-e^{-1})p+O(\sqrt{p}).\tag{1}$$ For the $2\times 10^6$-th prime $p=32452843$, we have $$w_p=20519206,\ \ w_p-(1-e^{-1})p\approx 5096.75\ \ \text{and}\ \ \sqrt{p}\approx 5696.74.$$

Question. Does $(1)$ hold? If it is true, how to prove it?

Your comments are welcome!

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    $\begingroup$ In contrast, it seems that $|\{k^k\pmod p:\ k=1,\ldots,p-1\}|/p$ does not have a limit. $\endgroup$ Jan 20, 2020 at 9:12
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    $\begingroup$ let $x_1,\ldots,x_p$ be random independent residues modulo $p$. Then each residue $s$ appears in this sequence with probability $1-(1-1/p)^p\approx 1-e^{-1}$, so the average size of the set $\{x_1,\ldots,x_p\}$ is about $(1-1/e)p$. So your computations confirm that the value of the map $k^{k+1}+(k+1)^k$ do behave as random, and the values of $k^k$ do not. The first is not surprising, but should be difficult to prove, the second comes from restrictions like "each second term is a quadratic residue", and these restrictions come from divisors of $p-1$, thus there is no limit along all primes. $\endgroup$ Jan 20, 2020 at 9:35

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