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This Carmichaelnumber with $39$ digits $$145410193191244273054310497291961592961$$ is not pandigital , the digit $8$ is missing.

What is the largest known Carmichael-number not being pandigital (with at least one digit missing in the decimal expansion) ?

Peter
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    $$502968961661095567282810561851297609678229070560076298749007885246826871726668089189601$$ is an example with $87$ digits. Digit $3$ is missing. – Peter Jun 29 '23 at 22:28
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    factordb – Peter Jun 29 '23 at 22:31
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    An example with $97$ digits , $5$ missing , is $$1200678011889802716621401397429179829919803917070231914778418048382642472819240078429963681472321$$ – Peter Jun 29 '23 at 22:42
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    factordb – Peter Jun 29 '23 at 22:45
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    The number $n=(6k+1)(12k+1)(18k+1)$ with k equal to 294555313376703770324072440550486 has $101$ digits and is a Carmichael number (it misses the digit $7$). Another $101$-digit example for n can be constructed in the same way with k equal to 202826390084431567861373760569275, missing the digit $6$. – jorisperrenet Jun 30 '23 at 07:48
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    Using the same format k=901966164698621693452028481838495 gives an $102$-digit $n$ with the digit $6$ missing. – jorisperrenet Jun 30 '23 at 07:49
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    @jorisperrenet I used the same idea because it is the best way to quickly construct many Carmichael numbers. I wonder whether there are infinite many such Carmichael numbers. – Peter Jun 30 '23 at 07:51
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    Again, using the same format k=77827388506854892505748442714276512918005 gives a $126$-digit Carmichael number with missing digit $2$. – jorisperrenet Jun 30 '23 at 13:11

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