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Repunit is a number consisting only of $1$s (here in base $10$).

Examples - $1$, $11$, $11111$.

Goal

Create 10 first repunits $n$-digits long from least possible number instances of numbers $n$.
For example (these are not minimal),
$11 = 2\times2\times2\times2-2-2-2/2$
$111 = 3\times(3\times3\times3 + 3 + 3 + 3) + 3$.

Allowed operators

  • Arithmetic $+$, $-$, $\times$, $/$
  • Exponentiation of positive rational numbers
  • Square root, $n$-th roots of positive rational numbers
  • Factorial, Gamma function, subfactorial of integers

Number concatenation is not allowed!

Note: actually, you need only 9 repunits (first ($1$) is trivial, so no need to write it down) - starting with 11 and ending with 1111111111.

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  • $\begingroup$ So does sqrt count as a digit? $\endgroup$ – Ben Frankel Apr 23 '15 at 10:30
  • $\begingroup$ $\sqrt{x}$ counts as 1 number ($x$), but $\sqrt[3]{x}$ counts as 2 numbers ($3$ and $x$). $\endgroup$ – Somnium Apr 23 '15 at 10:58
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$!n$ is subfactorial, $\Gamma(n)$ is the Gamma function.


$11 =\;!(2+2)+2$
$111 = \Gamma(3!)-3\cdot3$
$1111 = 4\cdot(4^4+4!)-!4$
$11111 = \frac{(5+5)^5-5/5}{!(5-5/5)}$
$111111 = \frac{(\Gamma(6)/(6+6))^6-6/6}{6+(6\cdot6)/(6+6)}$
$1111111 = \frac{(7+(7+7+7)/7)^7-7/7}{7+(7+7)/7}$
$11111111 = \frac{(8+\sqrt{\sqrt{8+8}})^8-8/8}{!\sqrt{8+8}}$
$111111111 = \frac{(9+!!\sqrt{9})^9-!!\sqrt{9}}{9}$
$1111111111 = \frac{10^{10}-10/10}{10-10/10}$

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  • $\begingroup$ Nice try using formula for repunits! Was waiting until someone figures it out. $\endgroup$ – Somnium Apr 23 '15 at 11:01
  • $\begingroup$ Have you forgotten about 1111111111? $\endgroup$ – Ian MacDonald Apr 23 '15 at 14:22
  • $\begingroup$ @IanMacDonald, Oh. I'll add that. $\endgroup$ – Ben Frankel Apr 23 '15 at 14:52
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This doesn't really address the challenge, but I would like to share an expression that allows to write any $n$-th repunit using (at most) 15 copies of $n$.

$$\frac{\left(!\left(\frac{n+n+n+n}n\right)+\frac n n\right)^n-\frac n n}{!\left(\frac{n+n+n+n}n\right)}$$

Or, in Wolfram Alpha code

((subfactorial((n+n+n+n)/n)+n/n)^n-n/n)/subfactorial((n+n+n+n)/n)

Which simplifies to the general formula

$$ \frac{10^n-1}9 $$

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I'll add to this as time goes by, if that's ok

$11: \frac{(2^2)!}{2} - \Gamma(2)$. Four twos, giving $\frac{24}{2} - 1$ for $11$

$111: 3^3 \times 3 + 3^3 + 3$. Six threes. Pretty sure this can be shortened

$1111: 4^4 \times 4 + (4! \times 4) -\space!4$. Six fours. Again, can probably be shortened.

Still working on the rest, but this gives people a starting point

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