Distribution of 0 digits in factorial series

Large Factorials and the Distribution of '0' in base 10 digits.

About the task

We can see that some features of factorial numbers (the series of numbers 1!, 2!, 3!, ...) come about because such numbers are the product of a series of counting numbers, and so those products have predictable factors. For example, all factorials above 1! are even numbers, since they have 2 as a factor. Similarly, all factorials from 5! up end in a 0, because they have 5 and 2 as factors, and thus have 10 as a factor. In fact, the factorial integers add another 0 at the end of the factorial for every step of 5 upward: 5! = 120, 10! = 3628800, 15! = 1307674368000, 16! = 20922789888000 and so on.

Because factorial numbers, which quickly become quite large, continue to have another terminal 0 on the right hand side of the number for every 5 increase in the factorial base number, one might think that the proportion of zeros in a base 10 factorial number might be close to 1/5. However, though the factorial products add another terminating 0 every 5, as the numbers become quite large, the number of digits in the factorial product expands exponentially, and the number above the terminating zeros tends toward 10% of each digit from 0 to 1, as the factorial becomes larger. Thus, as the factorials become larger, the proportion of 0 digits in the factorial products shifts slowly from around 1/5 toward 1/10, since the number of terminating zeros in n! increases only in proportion to n, whereas the number of digits of n in base 10 increases exponentially.

The task

Create a function to calculate the mean of the proportions of 0 digits out of the total digits found in each factorial product from 1! to N!. This proportion of 0 digits in base 10 should be calculated using the number as printed as a base 10 integer.

Example: for 1 to 6 we have 1!, 2!, 3!, 4!, 5!, 6!, or (1, 2, 6, 24, 120, 720), so we need the mean of (0/1, 0/1, 0/1, 0/2, 1/3, 1/3) = (2/3) (totals of each proportion) / 6 (= N), or 0.1111111...

Example: for 1 to 25 the mean of the proportions of 0 digits in the factorial products series of N! with N from 1 to 25 is 0.26787.

Do this task for 1 to N where N is in (100, 1000, and 10000), so, compute the mean of the proportion of 0 digits for each product in the series of each of the factorials from 1 to 100, 1 to 1000, and 1 to 10000.

Stretch task

Find the N in 10000 < N < 50000 where the mean of the proportions of 0 digits in the factorial products from 1 to N permanently falls below 0.16.

Python

<lang python>def facpropzeros(N, verbose = True):

   proportions = [0.0] * N
   fac, psum = 1, 0.0
   for i in range(N):
       fac *= i + 1
       d = list(str(fac))
       psum += sum(map(lambda x: x == '0', d)) / len(d)
       proportions[i] = psum / (i + 1)

   if verbose:
       print("The mean proportion of 0 in factorials from 1 to {} is {}.".format(N, psum / N))

   return proportions

for n in [100, 1000, 10000]:

   facpropzeros(n)

props = facpropzeros(47500, False) n = (next(i for i in reversed(range(len(props))) if props[i] > 0.16))

print("The mean proportion dips permanently below 0.16 at {}.".format(n + 2))

</lang>

Output:

The mean proportion of 0 in factorials from 1 to 100 is 0.24675318616743216.
The mean proportion of 0 in factorials from 1 to 1000 is 0.20354455110316458.
The mean proportion of 0 in factorials from 1 to 10000 is 0.17300384824186707.
The mean proportion dips permanently below 0.16 at 47332.

Wren

Library: Wren-big

Library: Wren-fmt

Very slow indeed, 10.75 minutes to reach N = 10,000. <lang ecmascript>import "/big" for BigInt import "/fmt" for Fmt

var fact = BigInt.one var sum = 0 System.print("The mean proportion of zero digits in factorials up to the following are:") for (n in 1..10000) {

   fact = fact * n
   var bytes = fact.toString.bytes
   var digits = bytes.count
   var zeros  = bytes.count { |b| b == 48 }
   sum = sum + zeros / digits
   if (n == 100 || n == 1000 || n == 10000) {
       Fmt.print("$,6d = $12.10f", n, sum / n)
   }

}</lang>

Output:

The mean proportion of zero digits in factorials up to the following are:
   100 = 0.2467531862
 1,000 = 0.2035445511
10,000 = 0.1730038482