Exploring the Limiting Behavior of $(n + 1)^n$ in Base $ n $ as $ n $ Approaches Infinity

Question

Hello Math Stack Exchange Community,

I've been investigating an interesting pattern in the polynomial expansion of $(n + 1)^n$ when expressed in base $ n $, specifically the emergence of a consistent pattern where the leftmost digit is '2', and the rightmost two digits are '01', with the middle coefficients varying. This is observed for finite values of $ n $. My question revolves around understanding the behavior of this expression as $ n $ approaches infinity.

We know the following:

1. $ n^n (1 + \frac{1}{n})^n = (n + 1)^n $
2. $n^n$ shifts the decimal point of $ (1 + \frac{1}{n})^n $ $n$-many places to the right in base $n$, leaving the digits in tact
3. $(n + 1)^n$ in base $n$ is ($n + 1$)-many digits (i.e. coefficients) long for all $n > 2$
4. As $ n $ grows larger, $ (1 + \frac{1}{n})^n $ approaches Euler's number $ e $
5. The constant $2 \ldots 01_n$ pattern reflects the symmetric $1_n$ and $10_n$ that bookend the expansion (as a consequence of the binomial theorem), and the carries of the "$1$" of $10_n$

In the limit as $ n $ approaches infinity, how can we interpret this pattern of $2 \ldots 01_n$ in base $ n $ in relation to $ e $? Is there a meaningful way to discuss the concept of $ e $ in an "infinite base" system, or does the notion become abstract or theoretical?

There is also an apparent pattern in the three terminal digits, of "$101$" for odd $ n $ and $ m + $ "$01$", where $ m = \frac{n}{2} + 1 $ for even $ n $. Can we prove this holds?

I am also curious about any relevant mathematical literature or research that delves into this area. So far, my searches have not yielded much, and I would greatly appreciate any insights or references that could guide my understanding of this phenomenon.

Here is the Mathematica code for any one interested:

Table[{n, IntegerDigits[(n + 1)^n, n]}, {n, 2, 100}]

Thank you for your assistance!

Answer 1

In response to the question "Can we prove [the '101'] pattern holds?", here is a visual proof in Wolfram Language. The triangular numbers can be defined as the sequence of (duplicate) successive odd numbers, where each odd number is scaled by $n$ or $n + 1$ depending on its parity in the sequence. These are listed in the last two columns. The "$1$" from the $n$'s place of $10_{n}$ at $n \choose n - 1$ gets carried to the ones place of the triangular number, $n \choose n - 2$, which either holds a "$0$" (yielding the leading "$1$" of $101$ for odd $n$) or the $\frac{n}{2}$th natural number (which gets incremented to yield the pattern $m+$"$01$", where $m = \frac{n}{2} + 1$) for even $n$. Note that $n \choose n - 2$ is the $(n - 1)$th triangular number, not the $n$th. To simplify discussion, I'm considering the reflection of the triangular number in the binomial expansion, as $\binom{n}{k}$ is symmetric and equal to $ \binom{n}{n-k} $. Many thanks to the commenters for your guidance and insights:

(*Define the table*)
table = Table[{n,(*Column 1:n*)2 n + 1,(*Column 2:
    2n+1*)(2 n + 1)*n,(*Column 3:(2n+1)*
    n*)(2 n + 1)*(n + 1),(*Column 4:(2n+1)*(n+1)*)
    BaseForm[(2 n + 1)*n, 2 n],(*Column 5:(2n+1)*n in base (2n+1)*)
    BaseForm[(2 n + 1)*(n + 1), 
     2 n + 1] (*Column 6:(2n+1)*(n+1) in base 2n*)}, {n, 0, 
    36} (*Adjust the range as needed*)];

(*Label and display the table*)
labels = {"n", "2n + 1", "(2n+ 1) * n", "(2n+ 1) * (n+1)", 
   "(2n+ 1) * n in base 2n", "(2n+ 1) * (n+1) in base (2n+1)"};
labeledTable = Prepend[table, labels];

(*Display the table,for example,using Grid for formatting*)
Grid[labeledTable, Frame -> All]

Answer 2

The $2$ is explained as follows: As $n$ gets large, the first digits of $(n+1)^n\sim en^n$ will approach those corresponding to $2.71828...$ base ten. In the varying base $n$ the $.71728...$ part has varying digits, but the integer $2$ ahead of the decimal point remains just $2$ in any base $\ge 3$.

You should also see some regularity in the second digit base $n$; plot it as a function of $n$ for let's say $n$ equal to powers of $2$ up to $2^{16}$ and note what happens (choosing these bases allows you to work with just the binary representation). Can you explain the apparent asymptotic behavior?

Answer 3

The final $0$ and $1$ is because the binomial theorem gives you

$$ (n+1)^n = \sum_{k=0}^n {n \choose k} n^k $$ The terms for $k \ge 2$ are all divisible by $n^2$, the $k=1$ term is $n^2$ because ${n \choose 1} = n$, and the $k=0$ term is $1$.

BTW, since the $k=2$ term is $n^3 (n-1)/2$, when $n$ is odd the third-last digit is $1$, while if $n$ is even it is $n/2 + 1$.