# Copeland-Erdős constant: does the cumulative even digit count ever reach or overtake the cumulative odd digit count, and if so, at what prime?

Arthur Herbert Copeland and Paul Erdős proved in 1946 that the Copeland-Erdős constant is a normal number. Since all prime numbers other than 2 are odd, all prime numbers other than 2 end in an odd digit, so one might expect skew of the digit distribution toward odds, since each prime number other than 2 is guaranteed at least 1 odd digit, while there is no such at-least-1-digit guarantee for even digits. So for the constant to be normal, it must be that as the prime numbers go toward infinity, prime numbers become so long digit-wise that the oddness of the last digit becomes negligible.

Now, looking at the first few digits (0.235711131719232931374143...), it is obvious that odd digits far outnumber even digits within the early digits. But since the constant is normal, the evens must "catch up" eventually: either...

• (a) ...the evens asymptomatically approach from below a 50% distribution of all digits, or...

• (b) ...(what seems to me much more likely) which parity of digits is ahead changes infinitely often, though it might take a long time and a very large prime for the evens to first catch up (reminiscent of the very large Skewes's numbers and related numbers wherein π(x) finally catches up to li(x) for the first time), or...

• (c) ...(what seems to me to be unlikely) a combination of the above two cases so that after a finite number of switches of the lead, one parity stays ahead forever while the other stays asymptomatically close.

Does anyone know if there is a proof of which of the three cases is true? If, as I suspect, case (b) is true, what is the smallest prime at which the cumulative even digit count catches up to the odds?

## Numerical Results

Let r(n) be the proportion of even digits after the nth prime. So, since the constant starts out 0.2 3 5 7 11 13 ..., the first few values of r(n) are r(1) = 100%, r(2) = 50%, r(3) = 33.333...%, r(4) = 0.25%, r(5) = 16.666...%, r(6) = 12.5%. Below, when I refer to the "maximum value" of r(n), I am disregarding the trivial r(1) and r(2) values.

I wrote a script to calculate r(n) up to $$n = 7.5 \times 10^7$$ (75 million). For reference as to roughly how large these primes are, the 75,000,000th prime is 1,505,776,939.

For n ≥ 3, r(n) initially falls before starting to rise, before finally tying r(3) = 1/3 at r(380), with r(381) = 444 / (444 + 883) ≈ 33.45% being the first value of r(n) to exceed r(3).

Beyond r(381), r(n) oscillates (obviously), but on average, it rises much more than it falls and initially grows rapidly on average — but as the primes get larger and larger, its average rate of growth falls. r(n) first hits 34% at r(389), hits 35% at r(416), hits 36% at r(654), hits 37% at r(1,106), hits 38% at r(3,097), hits 39% at r(6,861), hits 40% at r(24,613), hits 41% at r(55,426), hits 42% at r(210,117), hits 43% at r(1,790,106), and hits 44% at r(25,609,981).

Anyway, as of the 75th million prime 1,505,776,939, the highest value of r(n) thus far is 44.2537565841856...% at the 46,450,161st prime, 909,090,109. I still do not know if r(n) does ever hit 50%.

• This is neat (+1). You might add some visualization of $r(n)$ if you have one. It sounds very difficult, though. – Jair Taylor May 13 '20 at 0:38
• On second thought: this might not be too difficult, as for very large $n$ most of the digits will be of the $n$th prime number will be the same as the digits of $n \log n$, by the prime number theorem. So you just need to solve this problem for $n \log n$. – Jair Taylor May 13 '20 at 2:48
• A heuristic argument that speaks for (a) [but has to be taken with a generous amount of sodium chloride] is that for $n > 1$ the last digit of $p_n$ is always odd, and there's no obvious reason why the other digits should prefer to be even (or odd). So by this heuristic, the difference between the number of odd and even digits in all primes $\leqslant x$ would be roughly $\pi(x)$. Since the number of digits of the primes is unbounded, that skew wouldn't interfere with the normality of the constant. – Daniel Fischer May 20 '20 at 18:41

Not a proof, but I am fairly certain that $$r(i)$$ will always be strictly smaller than $$0.5$$ for $$i > 2$$. Consider all primes below $$n$$. There are roughly $$\frac{n}{\log(n)}$$ of them, most of them with about $$c \log(n)$$ digits, so there are approximately $$cn$$ digits in total. Of those $$cn$$ digits, $$(1 + o(1))\frac{n}{\log(n)}$$ are guaranteed to be odd, while the other digits can be assumed to be randomly distributed. This means that we can expect $$\frac{cn}{2} - \frac{n}{(2 + o(1))\log(n)} + O(n^{\frac{1}{2} + \epsilon})$$ even digits, and $$\frac{cn}{2} + \frac{n}{(2 + o(1))\log(n)} + O(n^{\frac{1}{2} + \epsilon})$$ odd ones. The latter quantity is bigger for large enough $$n$$ and since you have checked up to $$n = 7.5 \times 10^7$$, I am confident that it holds for all $$n \ge 5$$.