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A rep-digit number is of the form $$b\cdot \frac{10^a-1}{9}$$ with an integer $b$ with $1\le b\le 9$ and a positive integer $a$. In the decimal expansion , digit $b$ occurs $a$ times and no other digit.

Conjecture : A rep-digit number cannot be a Carmichael-number

I checked upto $a=5\ 000$ ( $5\ 000$ digits ) without finding a Carmichael-number of this form (in fact , not even a Poulet-number). Can we prove that there actually is none ?

We can rule out $b=9$ and $b$ even since a Carmichael number must be odd and squarefree and $b=9$ implies divisor $9$.

Peter
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    My opinion is that this is a case of generalizing too far. Repunits are very sparse and Carmichael numbers are somewhat rare; it's not surprising to see no examples in a finite data set, but I see no reason to conclude that they never occur. – Greg Martin Jul 10 '23 at 15:00

1 Answers1

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Disclaimer: This is not a proof, just a long comment that may result in one.

We know that $b$ cannot be even, as there are no even Carmichael numbers. Also, since Carmichael numbers are squarefree we know that $b$ can never be $9$. Thus, we have $b \in \{ 1,3,5,7\}$.

First, define $$f_b(a) = b \cdot \frac{10^a-1}{9}\,.$$

Now, let us explore $b=5$, we know that the rep-digit $f_5(a)$ is divisible by $5$, for all $a$. Hence, if it is a Carmichael number, $f_5(a)-1$ must be divisible by $4$, therefore we must have $a = 1$ (you can see this holds by using the divisibility trick upon dividing by $4$; you only consider the last two numbers). Since $a=1$ gives $f_5(1)=5$, a non-Carmichael number we know that $b$ cannot equal $5$. Hence $b \in \{ 1,3,7\}$.

If $b=7$, we know that the rep-digit $f_7(a)$ is divisible by $7$, for all $a$. Hence, if it is a Carmichael number, $f_7(a)-1$ must be divisible by $6$, therefore we must have $a \equiv 1 \pmod 3$. Also, $f_7(a)$ must be squarefree in order to be a Carmichael number, thus it can not be divisible by $49$, giving that $a \not\equiv 0 \pmod 6$. If $a \equiv 0 \pmod 2$, we know that $f_7(a)$ is divisible by $11$, thus $f_7(a)-1$ must be divisible by $10$, which is impossible. Hence $a \not\equiv 0 \pmod 2$.

If $b=3$ and $a \equiv 0 \pmod 3$, we have that $f_3(a)$ is divisible by $9$. Thus $a \not \equiv 0 \pmod 3$. As the same argument for $a\equiv 0 \pmod 2$ holds as above, we must have $a \not \equiv 0 \pmod 2$. Thus we must have $(a \bmod 6) \in \{ 1,5 \}$

Now, for $b=1$, things get more difficult. The argument for the divisibility be $11$ does not eliminate cases. However, due to the number needing to be squarefree (note that these relations also hold for $b=3$ and $b=7$), we still know that $a \neq 0 \pmod 9$ because of $3^2$, $a \neq 0 \pmod {22}$ because of $11^2$, and $a \neq 0 \pmod {42}$ because of $7^2$.

For divisibility reasons, we have $f_1(a)$ is divisible by $13$ if $a \equiv 0 \pmod 6$ and $f_1(a)-1$ cannot be divisible by $12$ (unless $a=1$, which is no Carmichael number), thus $a \not\equiv 0 \pmod 6$. Also, because of divisibility by $37$ we have that $a \not \equiv 0 \pmod 3$. Divisibility by $101$ gives $a \not \equiv 0 \pmod 4$. Divisibility by $41$ gives $a \not \equiv 0 \pmod 5$. Divisibility by $4649$ gives $a \not \equiv 0 \pmod 7$. Divisibility by $21649$ gives $a \not \equiv 0 \pmod {11}$. Divisibility by $53$ gives $a \not \equiv 0 \pmod {13}$. I could continue the list, but I think this is enough to encourage a different type of search to be made.

Most of these divisibilities also hold for the cases $b=3$ and $b=7$.

In conclusion, we must have $b \in \{1,3,7\}$.

  • $b=1$, gives that $a$ cannot be divisible by $[2\cdot 2,3,5,7,11,13,17,23\cdot 2,29,37,41,43,47,53,59,61,67,71,73,79,83,89,97,\cdots]$.
  • $b=3$, gives that $a$ cannot be divisible by $[2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,\cdots]$.
  • $b=7$, gives that $a\equiv1\pmod3$ and $a$ cannot be divisible by $[2,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,\cdots]$.
jorisperrenet
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  • Note that if $b=1$, we have that for all $p | f_1(a)$ we must have $p-1 | f_1(a-1)\cdot 10$. – jorisperrenet Jul 11 '23 at 11:15
  • If we let $p=4k+1$ and $p|f_1(a)$ (where $f_1(a)$ is a Carmichael number) we must have $4k|f_1(a-1)\cdot 10$, and as $2 \nmid f_1(a-1)$ we reach a contradiction, a similar argument holds for $p=25k+1$. Therefore primes $p$ of this form may not divide the Carmichael number $f_1(a)$. – jorisperrenet Jul 11 '23 at 11:24
  • Since, we have $f_b(a) | f_b(k\cdot a)$ with $k \in \mathbb N_{\geq 1}$. We get that if $p=4k+1 | f_1(a)$ (or $p=25k+1$ for that matter), then we know that $f_1(k\cdot a)$ cannot be Carmichael for all $k\in \mathbb N_{\geq 1}$. – jorisperrenet Jul 11 '23 at 11:31
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    This is really nice. In line with Greg Martin's comment on the original question I would say that checking up to $n = 2000$ is too little evidence to conclude non-overlap of two sets that are themselves really rare. But now that we have all the conditions from this post the OP might try and check the first 2000 numbers that satisfy all the constraints found by jorisprrenet and I'd say that would give some real information – Vincent Jul 11 '23 at 11:47
  • Similar arguments as above (the $p=4k+1|f_1(a) \implies f_1(k\cdot a)$ cannot be Carmichael for all $k \in \mathbb N_{\geq 1}$) hold for $b=3$ with $p=c\cdot k+1$, where $c\in {3,5,8,11,37,41,53,73,79,\cdots}$. And for $b=7$ with $p = c\cdot k+1$, where $c \in {5,7,11,13,32,37,41,43,53,57,67,71,73,79,89,\cdots}$. The list for $b=1$ with $c\in {4,25}$ is likely to be complete. – jorisperrenet Jul 11 '23 at 12:18
  • @Vincent I would say that the evidence is overwhelming , of course it is no proof , just an evidence (as in the case of Goldbach's conjecture). But considering how rare Carmichael-numbers with more than $2\ 000$ digits are , a hit would be a miracle. So even without all the restrictions testing upto $2\ 000$ makes it very unlikely to miss an example. – Peter Jul 11 '23 at 13:49
  • Wait you mean $n$ the number of digits, not the number itself?!? Yes then it is a completely different story! – Vincent Jul 11 '23 at 15:41
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    @Vincent I see now how the confusion occured , sorry. I wrote I checked upto $n=2\ 000$ , it is however $a=2\ 000$ and this has been extended in the mean time to $a=5\ 000$. I have fixed the error. – Peter Jul 12 '23 at 08:51