I watched this video recently about the Copeland-Erdos constant.
There following text was included below the video:
Write the primes in a row — 2, 3, 5, 7, 11, 13, and on forever — put a decimal point in front, and you get a single number: 0.235711131719232931… the Copeland–Erdős constant. In 1946, Copeland and Erdős proved it is normal in base ten: every finite string of digits appears in its expansion, and appears exactly as often as pure chance would predict — each digit one tenth of the time, each pair one hundredth, every block its fair share. So the opening of pi (314159) is in there, at digit 6,903. So is 42, at digit 214. So is every birthday, phone number, and message ever written.
What makes this remarkable: normality is the rule, not the exception — a randomly chosen real number is normal with probability one (Borel, 1909) — yet we still cannot prove that pi, e, or the square root of two are normal. The first explicit example was Champernowne's constant (0.123456789101112…), normal because it was built by pasting in every integer on purpose. The primes do no such favor: they thin out, follow no pattern, and were never designed to be tidy — and their concatenation is normal anyway. This video builds the number, defines normality, shows why the famous constants resist proof, and walks the intuition (via the prime number theorem) for why the patternless primes still even out in the limit.
The position in the Copeland-Erdos constant
Target '28239' first appears at position: 61014
What's of interest is whether a number can be equal to its index. I got Gemini to investigate and here is what it came up with:
While no fixed point where the 1-based index $P(N)$ equals $N$ itself has been found among smaller integers in the Copeland–Erdős constant, number-theoretic density principles dictate that such numbers must exist among larger composite numbers, even though they are mathematically impossible among prime numbers.
To see why this dichotomy exists, we have to look at the growth rate of the string indices for primes versus composite numbers.
Why Primes Never Equal Their Index
For any prime number $p$, its "natural" appearance occurs when $p$ itself is concatenated into the decimal stream. The 1-based index where $p$ begins is equal to $1$ plus the total number of digits of all primes strictly smaller than $p$:
$$P_{\text{natural}}(p) = 1 + \sum_{q < p} \lfloor \log_{10}(q) + 1 \rfloor$$Using Chebyshev's function $\theta(p) = \sum_{q \le p} \ln(q)$, which by the Prime Number Theorem is asymptotically equivalent to $p$, we can evaluate the sum of the base-10 logarithms of all primes up to $p$:
$$\sum_{q < p} \log_{10}(q) = \frac{1}{\ln 10} \sum_{q < p} \ln(q) \sim \frac{p}{\ln 10} \approx 0.4343 \times p$$Because $\frac{1}{\ln 10} \approx 0.4343 < 1$, the cumulative number of digits written into the stream grows at less than half the rate of the prime numbers themselves. By the time we concatenate a large prime $p$, its starting position is only around the $0.4343p$ mark. If $p$ happens to appear even earlier as an accidental substring across prime boundaries, its index is smaller still.
Therefore, for all primes, $P(p)$ is strictly less than $p$, making a fixed point $P(p) = p$ impossible.
Why Composites Can Form Fixed Points
Composite numbers behave completely differently because they never get a "natural" turn in the prime concatenation. They can only appear in two ways:
- As an accidental boundary string formed across the junction of two or more adjacent primes (for example,
4first appears at index 21, spanning across the end of $3\mathbf{7}$ and the start of $\mathbf{4}1$). - As a substring entirely within a much larger prime $q > N$.
Because the Copeland–Erdős constant is normal in base 10 (proven by Copeland and Erdős in 1946), every finite decimal string is guaranteed to appear eventually. However, the expected waiting time to see a specific $k$-digit composite string by chance in a normal sequence is roughly $10^k \ln 10 \approx 2.3 \times 10^k$.
Because $N$ itself sits between $10^{k-1}$ and $10^k$, a typical composite number that does not get an early boundary match will appear at an index much larger than $N$.
The Oscillation of $P(N) - N$
We can see this sharp split by examining the first few integers in the constant (OEIS A229190):
| Integer $N$ | Index $P(N)$ | Type | Behavior |
|---|---|---|---|
| 2 | 1 | Prime | $P(N) < N$ |
| 3 | 2 | Prime | $P(N) < N$ |
| 4 | 21 | Composite | $P(N) > N$ |
| 5 | 3 | Prime | $P(N) < N$ |
| 6 | 31 | Composite | $P(N) > N$ |
| 7 | 4 | Prime | $P(N) < N$ |
| 8 | 41 | Composite | $P(N) > N$ |
| 9 | 12 | Composite | $P(N) > N$ |
| 10 | 47 | Composite | $P(N) > N$ |
Notice how consistently the primes lag behind their own values ($P(N) < N$), while the composite numbers overshoot ($P(N) > N$).
As we examine larger numbers, "lucky" composite numbers will occasionally appear unusually early due to boundary concatenations, while "unlucky" ones will appear late. Because the difference function $f(N) = P(N) - N$ oscillates wildly between negative values (for primes and early-matching composites) and large positive values (for late-matching composites), the graph of $P(N)$ repeatedly jumps across the line $Y = N$.
While finding a specific integer where $P(N) = N$ requires extensive string-searching through millions of digits, this continuous crossing confirms that fixed points exist among composite numbers.
