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The article's opening sentence: "Any positive number can be found at the beginning of a factorial..." is a strong statement but no proof is given. Missing: "..
by notelonmusk 7y ago
The article's opening sentence: "Any positive number can be found at the beginning of a factorial..." is a strong statement but no proof is given.
Missing: "... with high probability, and supported by analysis of the digits of factorials from 1 to 10,000"
The work by Diaconis mentioned but not cited:
[0] https://statweb.stanford.edu/~cgates/PERSI/papers/digits.pdf https://statweb.stanford.edu/~cgates/PERSI/papers/digits.pdf - The distribution of leading digits ad uniform distribution mod 1 - Persi Diaconis
The sequence:
[1] https://oeis.org/A076219 https://oeis.org/A076219 - Smallest positive integer m such that m! begins with n in base 10 ::
1, 2, 9, 8, 7, 3, 6, 14, 96, 27, 22, 5, 15, 42, 25, 89, 69, 76, 63, 16, 87, 113, 54, ...
- SamReidHughes 7y agoI'm not sure what you're getting at, but "any positive number can be found at the beginning of a factorial" is correct. Not with high probability, but with absolute certainty. I don't know which Diaconis proved, but it's pretty easy to construct a non-probabilistic, constructive proof. Edit: The paper you linked does not prove it with high probability, it proves {n!} is a strong Benford sequence. Edit: Which implies absolute certainty.
- notelonmusk 7y ago> I'm not sure what you're getting at I'm sure what I'm not getting at, which is the proof. I'll think about it. Processing ...
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- lurquer 7y agoCould you provide such a proof? (just general outline would suffice...) I can't figure out a "pretty easy" way to do it.
- SamReidHughes 7y agoEvery time you hit a power of ten, fracPart(log10(n!)) gets the unit interval finely iterated by n=10^k, 10^k + 1, 10^k + 2, ..., 10^k + sqrt(2 log(10) 10^k).
- notelonmusk 7y agoSorry, but how is that a proof? It's not clear whether there is a number A with decimal expansion (a_1 a_2 ... a_k ) such that there doesn't exist a positive number X for which X ! = (a_{1} a_{2} ... a_{k} x_{k+1} x_{k+1} ... x_{s} ) for some arbitrary length s in the decimal expansion of X! .
- SamReidHughes 7y agoIt shows {fracPart(log10(n!))} is dense.
- deleted 7y ago[deleted]
- markisus 7y agoI'm also not getting your proof. What does it mean every time you hit a power of 10? The sequence n! does not hit powers of 10 since all terms of the sequence after n=3 are at least divisible by 3. I agree though, density in (0,1) of the fracPart(log10(n!)) is a good thing to try to prove, since 10^fracPart(log10(n!)) should have the same digits as 10^intPart(log10(n!))*10^fracPart(log10(n!))=10^log10(n!)=n!
- SamReidHughes 7y agoEvery time n hits a power of 10, not n!. E.g. https://www.wolframalpha.com/input/?i=Table%5B%5Bfrac%28log10%28%2810%5E6+%2B+n%29%21%29%29%2C+%2810%5E6+%2B+n%29%21%5D%2C%7Bn%2C0.0%2C20.0%2C1.0%7D%5D https://www.wolframalpha.com/input/?i=Table%5B%5Bfrac%28log1... In general for b << 10^k, we have fracPart[log10((10^k + b)!)] ≈ fracPart[log10((10^k)!) + b(b+1)/(2 log(10) 10^k)]. So it takes sqrt(2 log(10) 10^k) steps for these factorial values to work their way around the unit interval. Increase k and you fill it arbitrarily finely. Edit: ^^^ To be clear, the approximation there is formed using the linear approximation log(1 + b/10^k) ≈ b/10^k, and basic log multiplication/addition/division formulas.
- amilein7minutes 7y agoI haven't thought about this carefully but is it correct to say the probability of finding n as the leading digits for a bounded m (but both very large) is a decreasing function of n?
- amilein7minutes 7y agoVery cool! Is there a pattern in the sequence of m's? It feels as though there must be some bounds on the terms of the sequence on m's given that the probability of appearance of n as the first digits decreases with n, in the sequence of factorials.