Is there a pattern/formula to this?

Question

I came across a nice "sequence" on a mathematics Facebook page and it intrigued me somewhat. This is not homework or anything of the sort, just something fun to start the year off I guess.

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Observe the following:

\begin{align} (1)^1 & = 1 \\[5 mm] (8 + 1)^2 & = 81 \\[5 mm] (5 + 1 + 2)^3 & = 512 \\[5 mm] (2 + 4 + 0 + 1)^4 & = 2401 \end{align}

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Some questions came to my mind almost immediately.

1. In general, is there a pattern/formula to this? For example, can we find the next (5-digit) number easily? Does it even exist? In particular, can we find a formula for the n-digit number?

2. For some of the numbers, the solutions are non-unique. For example, for the first power, obviously any digit would fit. Can we generalise this observation? In particular, do we know if the solution for every n-digit number, if it exists, is always non-unique?

P.S. The many brilliant minds on Math SE have never failed to amaze me, so I thought I should put this one up here to see what you guys can come up with :)

Answer 1

Notice that we always have:

$$10^k = 1\underbrace{00...0}_{k \text { zeroes}}$$

which has $k+1$ digits. Hence for $n^k$ to have exactly $n$ digits, we must have $n < 10$.

For fifth powers, the first $9$ fifth powers are (http://oeis.org/A000584):

$$1, 32, 243, 1024, 3125, 7776, 16807, 32768, 59049$$

Now we look at the digit sums of $16807, 32768, 59049$, which are $22, 26, 27$ respectively, and are not equal to $7,8,9$. Hence we cannot have $(1+6+8+0+7)^5 = 16807$, etc.

A further question is, are there any more numbers in this sequence? If $9^k$ has less than $k$ digits, there are no $k$-digit $k$-th power. The number of digits (decimal) of a number $N$ is given by:

$$\lfloor\log_{10}N\rfloor+1$$

so if $9^k$ has less than $k$ digits, we have:

$$\lfloor\log_{10}9^k\rfloor+1 < k$$ $$k\log_{10}9 < k-1$$ $$k > \frac 1{1- \log_{10}9}\approx 21.854$$

hence we just need to verify the result up to $22$nd powers. (Observe that $9^{22}$ has only $21$ digits).

EDIT: the rest of the candidates are:

\begin{align} 7^6 &= 117649\\ 8^6 &= 262144\\ 8^7 &= 2097152\\ 8^8 &= 16777216\\ 8^9 &= 134217728\\ 8^{10} &= 1073741824\\ 9^6 &= 531441\\ 9^7 &= 4782969\\ 9^8 &= 43046721\\ 9^9 &= 387420489\\ 9^{10} &= 3486784401\\ 9^{11} &= 31381059609\\ 9^{12} &= 282429536481\\ 9^{13} &= 2541865828329\\ 9^{14} &= 22876792454961\\ 9^{15} &= 205891132094649\\ 9^{16} &= 1853020188851841\\ 9^{17} &= 16677181699666569\\ 9^{18} &= 150094635296999121\\ 9^{19} &= 1350851717672992089\\ 9^{20} &= 12157665459056928801\\ 9^{21} &= 109418989131512359209 \end{align}

and none of these powers have digit sums equal to their bases.