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Indeed,

Conjecture   Let $\ p\ $ be a prime, $\ n>1\, $ -- a natural number, and let the largest prime divisor $\ q\ $ of $\ s(p^n) :=\sum_{k=0}^n\,p^k\ $ satisfy $ q<p.\ $ Then $\ p^n\ =\ 7^3.$

The unique(?) exception would be

$$ s(7^3)\ =\ 400\ = 2^4\cdot5^2 $$

EDIT A micro-observation: when primes $\ p\ q\ r\ $ satisfy $\ q|s(p^{r-1})\ $ then $\ p>r\ $ and $\ q\ge r$.

More generally, $\ p>\rho $ and $\ q\ge \rho\ $ when $\ \rho\ $ is the smallest prime divisor of $\ r\,\ $ where this time $\ r\ $ is not assumed to be a prime.

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    $\begingroup$ A numerical search for $2\leq p\leq 1000$ and $2\leq n\leq 20$ turned up 79 counterexamples. $\endgroup$
    – Julian Rosen
    Commented Jul 13, 2021 at 5:07
  • $\begingroup$ Thank you. I am spectacularly wrong. $\endgroup$
    – Wlod AA
    Commented Jul 13, 2021 at 5:20
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    $\begingroup$ Would someone kindly add these counterexamples to OEIS? I feel that these examples are essential to the theory of perfect and baroque (multi-perfect) numbers. $\endgroup$
    – Wlod AA
    Commented Jul 13, 2021 at 5:23
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    $\begingroup$ The counterexample values $p^n$ up to one million are $343, 4489, 6241, 18769, 22201, 26569, 32761, 36481, 44521, 52441, 68921, 69169, 72361, 76729, 79507, 97969, 103823, 139129, 185761, 192721, 249001, 271441, 326041, 389017, 398161, 426409, 571787, 619369, 654481, 657721, 674041, 677329, 844561, 896809, 942841, 982081, 994009$. I'll leave it to someone else to post to OEIS if they are inclined. $\endgroup$
    – Julian Rosen
    Commented Jul 13, 2021 at 5:39

1 Answer 1

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There's also

$$s(67^2) = 1 + 67 + 67^2 = 4557 = 3 \cdot 7^2 \cdot 31 \tag{1}\label{eq1A}$$

In this case, $q = 31 \lt 67$ meets your criteria. I haven't checked further, but I'm fairly certain there are additional exceptions.

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    $\begingroup$ Thank you. This is a surprise(!) due to the even exponent, $\ 67^2.$ $\endgroup$
    – Wlod AA
    Commented Jul 13, 2021 at 5:04
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    $\begingroup$ @WlodAA For small $n$, there are lots and lots of examples. -- But there is also e.g. $s(493397^6) = 29^2 \cdot 127 \cdot 1163 \cdot 2129 \cdot 4229 \cdot 26041 \cdot 50177 \cdot 71359 \cdot 138349$. $\endgroup$
    – Stefan Kohl ♦
    Commented Jul 14, 2021 at 18:39
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    $\begingroup$ @StefanKohl, impressive! $\endgroup$
    – Wlod AA
    Commented Jul 15, 2021 at 5:35

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