Last time I asked, was told that normal numbers and Kolmogorov complexity are not necessarily related, and basically are independent definitions of randomness. Certainly Pi is a good candidate for and example of it (if only we could prove it’s normal)

from mpmath import mp

mp.dps = 100000
s = str(mp.pi)

digits = 4

counts = []
for i in range(0, 10**digits):
    counts.append(0)

new_found = 0
all_found = False
for i in range(2, len(s) - digits + 1):
    d = int(s[i:i + digits])
    counts[d] += 1
    if counts[d] == 1:
        new_found += 1
        if new_found == 10**digits:
            all_found = True
            print("Last %d digit number in decimal expansion of pi is %s, at position %d" % (digits, d, i - 1))
            break
if not all_found:
    raise "oops, increase precision"

It does “feel” like there should be a connection, but I don’t see what form it could take.

For a number to be normal, it is not enough for every finite string to appear in its expansion, it must appear with the correct frequency! It is quite possible (in the logical sense) that every finite string of digits appears in the expansion of pi, but that pi is not normal. Nobody considers that a realistic possibility, but we are still unable to rule it out. In fact, it is still possible that from some point on the only digits that appear in pi are 0 and 3, say. IIRC, in Carl Sagan’s book Contact, some apparent non-normality of pi is discovered and this has implications for the plot.

There are many numbers known to be irrational but whose expansions are not normal and do not contain every finite string. For example, let a_i be 1 if there is a multiple of sqrt{3} between i and i+1, and let a_i be 0 otherwise. Then the number 0.a_1a_2a_3… is irrational (since it is not repeating and not finite), but you will never see two consecutive 0′s, nor three consecutive 1′s.

Using the definition of “normal number” at the link John provided — i.e. a number where the limiting frequency of any sequence of k digits is 1/b^k in base b — this will also be true of any normal number. If the complete works of Mark Twain requires k digits to encode, then once in every b^k subsequences of length k (on average) you will get the complete works of Mark Twain.

So, if pi is normal, then the answer is yes. I’m not clever enough to know whether, if pi is not normal, the answer might still be yes.

so 6275th string position should be the 6274th decimal, not 6276th?
Or did I misunderstand something?

There must be no upper limit to the number of digits in x; the only stipulation is that it is finite.

Let f(n) be the integer truncation of pi times 10 to the n. So f(3) is 3,141 and f(5) is 314,159.

For any n, f(n) must appear somewhere in pi. Given the above, this can be trivially proven through induction.

Does this imply that the decimal digits of pi must include pi?

That seems wrong, because it violates the stipulation that the length of x’s digit string must be finite. but if it’s wrong, that means there exists some n where f(n) is not included in the decimal digits of pi.

There’s probably some property of infinite sets that I’m forgetting from college that makes this question silly, and I’ve never been much for higher maths. But I’m having a bit of trouble turning it over in my head.

The Smartest Man in Babylon

(note: official abstrusegoose is currently in downtime)