I was messing around with numbers, trying to find patterns, and came across this one:
91 isn't prime, 991 is, 9991 isn't, 99991 is...
If you keep adding 9's, does this pattern hold infinitely?
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4$999999991$ is not prime, it is divisible by $67$. – projectilemotion Jun 28 '17 at 22:36
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5The terms with an odd number of 9's will always be composite, because they're equal to $10^{2n} - 9 = (10^n-3)(10^n+3)$. – Daniel Schepler Jun 28 '17 at 22:39
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Wolfram Alpha says that from $10^{11} - 9$ onwards until $10^{31}-9$ (when the exponent is odd), the numbers are all composite. However, $10^{33}-9$ (https://www.wolframalpha.com/input/?i=is+10%5E33++-9++prime) is a prime, where the pattern breaks again. – Toby Mak Jun 28 '17 at 22:46
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Since the infinitude of Mersenne (or Fermat) primes has not been proved, I guess there is no easy proof about the existence of infinite primes of the form $10^k-9$. – Jack D'Aurizio Jun 28 '17 at 22:48
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No, the pattern breaks as $999999991=67\cdot 14925373.$
In fact, your numbers $f(n)=10^n-9$ are prime exactly for $n=3, 5, 7, 33, 45, 105, 197, 199, 281, 301, 317, 1107, 1657$ when $n\le 2000.$
Reiner Martin
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2Moreover, $10^n-9$ is divisible by $67$ whenever $n \equiv 9 \mod 33$. – Robert Israel Jun 28 '17 at 22:57
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Similar question: does function (10^n+1)/11 except for n=3 always produce primes when n is odd? – sirous Jun 29 '17 at 07:58