# Repunit replication

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.

• So does sqrt count as a digit? Apr 23, 2015 at 10:30
• $\sqrt{x}$ counts as 1 number ($x$), but $\sqrt[3]{x}$ counts as 2 numbers ($3$ and $x$). Apr 23, 2015 at 10:58

$!n$ is subfactorial, $\Gamma(n)$ is the Gamma function.

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

• Nice try using formula for repunits! Was waiting until someone figures it out. Apr 23, 2015 at 11:01
• Have you forgotten about 1111111111? Apr 23, 2015 at 14:22
• @IanMacDonald, Oh. I'll add that. Apr 23, 2015 at 14:52

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

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

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