sum of all integer digits of all primes between 1 and p = k*p
- This might surprise:
The ratio k_b defined as "the sum of all base-b digits
of all the primes between 1 and n" to "n",
converges to the constant (b-1)/(2 ln(b)) for increasing n.
Sum of all base-b integer digits 1 to b^m-1 = m*b^m*(b-1)/2
Now proportion this value to the number of primes betewwn 1 and n
Sum of all base-b integer digis of all primes between 1 and b^m-1
and pi(b^m-1) ~ (b^m-1)/ln(b^m-1)
and m = log_b(b^m-1) = ln(b^m-1)/ln(b)
~ (b-1)(b^m-1)/(2 ln(b))
hence k_2 = 0.7213476... and k_10 = 1.9543251...
A quick check using a C program the actual value obtained by summing
all the digits of primes between 1 and 99999999977 k_10 = 2.05366 and
similarely in base-2 up to prime 2147483647 k_2 = 0.780761
Is this a known fact or is this a new constant?
- --- In email@example.com, "Anton" wrote:
> Conjecture:It seems to me that this conjecture is true, and your proof is valid,
> The ratio k_b defined as "the sum of all base-b digits
> of all the primes between 1 and n" to "n",
> converges to the constant (b-1)/(2 ln(b)) for increasing n.
> Sum of all base-b integer digis of all primes between 1 and b^m-1
> = (m*b^m*(b-1)/2)*pi(b^m-1)/(b^m-1)
if you assume that primes are equally likely to consist of any digit?
But while we might believe that's true in the long run, it seems like
in the short run it certainly won't be true. (E.g. primes in base 10
won't have so many even digits in them, or 5s, in the short run; but
by the time they have a lot of digits, maybe that last-digit fact
won't matter much any more).
Is there a known proof somewhere that large primes asymptotically have
equal proportion of each digit (in any base)?
It seems very likely to be true, but way beyond what I'm capable of
proving or even intelligently investigating.