# $\left(n^n\right)_b = \left(n\right)_b\left(n\right)_b\ldots\left(n\right)_b$

A friend of mine asked me this today and I was not able to give him an answer. Given a base $b$, find (or show the absence of) an integer $n>1$ s.t. $$\left(n^n\right)_b=\left(n\right)_b\left(n\right)_b\ldots \left(n\right)_b.$$

The notation $\left(n\right)_b \left(m\right)_b$ means concatenating the representation of the number $n$ in base $b$ with the representation of the number $m$ in base $b$. The $\ldots$ in the above should be taken to mean "any number of" consecutive $\left(n\right)_b$.

Edit:

@alex.jordan points out the interesting relaxation (see comments):

$$\left(n^n\right)_b=\left(\underbrace{0...0}_k\,n\right)_b\left(\underbrace{0...0}_k\,n\right)_b\ldots \left(\underbrace{0...0}_k\,n\right)_b.$$

where $\underbrace{0...0}_k$ should be understood to mean $k\geq0$ leading zeros.

An answer to either is acceptable. The construction of such an $n$ giving an affirmative answer for the first question will (naturally) give an affirmative one to the second with $k=0$.

Edit 2:

The general problem is prohibitively difficult. I will accept an answer for $b=10$.

• Should the left side also be in base b? – Tobias Kildetoft Jan 4 '15 at 8:24
• Are leading zeros permitted? For instance, $(3^3)_2$ is $11011$, which we could see as $011011$, parse as $(011)(011)$, and have $(3)_2(3)_2$. – alex.jordan Jan 4 '15 at 8:33
• @par What's the significance of writing the left-hand side in base $b$? – Tunococ Jan 4 '15 at 8:55
• @Tunococ: the significance of writing the left-hand side in base $b$ is to have the equation makes sense as an equation of strings in base $b$. Now it is an equation between a number and a concatenated sequence of strings, which makes no sense. Unless one adds an operator to the right hand side that reinterprets the concatenated string as a number in base $b$; it is certainly not obvious that such an interpretation is meant. In fact with the left hand side being a number, it is natural to interpret juxtaposition on the right as multiplication (but then expression in base $b$ is meaningless). – Marc van Leeuwen Jan 4 '15 at 10:54
• The construction similar to the one suggested by @alex.jordan can be used to find (trivial) solutions for some other bases: If $b=n^{n-1}-1$ with $n\geq 3$, we have $(n^n)_b = (n)_b(n)_b$ since $(n)_b$ consists of just one $b$-ary digit. – Peter Košinár Jan 5 '15 at 20:05

If $$(n)_b = (d_1d_2\ldots d_t)_b \\ (n^n)_b = (n)_b(n)_b\cdots(n)_b$$ for some digits $d_1,d_2,\ldots,d_t$ in base $b$, then $(n)_B$ is a single digit and $$(n^n)_B=(n)_B(n)_B\cdots(n)_B$$ is a repeated single digit base $B=b^t$. This is also true if some $d_i=0$, allowing for leading zeros.
So for the example given by @alex.jordan in the comments $3^3=011011_2$ is equivalent to $3^3=33_8$.
Thus it suffices to look for solutions where $n<B$ is represented by a single base $B$ digit. In that case we have $$n^n = nB^{l-1} + nB^{l-2} + \cdots + nB + n \\ n^{n-1} = \frac{B^l-1}{B-1}$$
For $l=1$ there are no solutions with $n>1$. For $l=2$ we get the pattern noted by @PeterKošinár in the comments, solutions with $B = n^{n-1}-1$.
For $l\ge3$ we have $(B-1)n^{n-1} + 1 = B^l$. Let $a=1,~b=(B-1)n^{n-1},~c=B^l$ then $$a + b = c \\ \operatorname{rad}(abc) \le nB(B-1) < l B^2 \log B < \left(B^l\right)^{1/(1+\epsilon)} \text{ for some }\epsilon>0,n>3$$ where $\operatorname{rad}(abc)$ is the product of distinct primes dividing $abc$. Assuming the abc conjecture there can only be finitely many pairs $n,B$ like this, so there generally won't be a solution for a given $B$. In fact I'd conjecture there are no solutions like this, but am not able to rule out exceptions.
Now if there is a solution with $n^n = n(B+1)$ and $B=b^t$ with $n>2,t>1$ then it can also be a solution in base $b$ only if $n^{n-1}=b^t+1$. By Catalan's conjecture (Mihailescu's theorem) the only powers differing by $1$ are $8,9$, so conditioned on my conjecture of the previous section $3^3=011011_2$ would be the only multi-digit solution.