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Powers of two satisfy Benford's law. In order for powers of an integer k to not satisfy Benford's law, k would have to be a rational power of 10; in fact it wo

Powers of two satisfy Benford's law. In order for powers of an integer k to not satisfy Benford's law, k would have to be a rational power of 10; in fact it would have to be an integer power of 10 because if 10^(m/n) is rational then it's an integer. But a quick calculation makes it look like powers of two don't satisfy Benford's law, For example the first power of 2 which begins with 7 is 2^46, and the first power of 2 which begins with 9 is 2^53. This takes so long, roughly speaking, because 2^10 = 1024 is so close to a power of ten.

tgb 9y ago
That surprised me since I saw the "generalization to digits beyond the first" in the article and it's clear that powers of 2 or 5 clearly don't have a Benford-style distribution of their last digits. However, that was a misunderstanding on my part, since that generalization says something about the first n digits, and any fixed n will eventually not include the last digit of the number.
- madcaptenor 9y ago
  In fact powers of any integer don't have a Benford-style distribution of their last digits, since those digits will repeat periodicially. For example powers of 2 end in 2, 4, 8, 6, 2, 4, 8, 6, ...
mark-r 9y ago
Yes they do. Python is a good resource for quick back-of-the-envelope calculations like this, I just tested the first digit of the powers of 2 from 0 to 9999. The percentages were 30.10, 17.61, 12.49, 9.70, 7.91, 6.70, 5.79, 5.12, and 4.58. These agree almost precisely with the Wikipedia numbers. Ditto for powers of 5.
- madcaptenor 9y ago
  I agree. But for powers from 1 to 50 those same percentages are 30, 20, 10, 10, 10, 8, 2, 10, 0. In particular there is only one power of 2 in this range starting in 7 and none that ends in 9. This might make someone doing hand computation suspicious. (But really you can't tell much from such a small sample.)

tgb 9y ago

That surprised me since I saw the "generalization to digits beyond the first" in the article and it's clear that powers of 2 or 5 clearly don't have a Benford-style distribution of their last digits. However, that was a misunderstanding on my part, since that generalization says something about the first n digits, and any fixed n will eventually not include the last digit of the number.

madcaptenor 9y ago
In fact powers of any integer don't have a Benford-style distribution of their last digits, since those digits will repeat periodicially. For example powers of 2 end in 2, 4, 8, 6, 2, 4, 8, 6, ...

mark-r 9y ago

Yes they do. Python is a good resource for quick back-of-the-envelope calculations like this, I just tested the first digit of the powers of 2 from 0 to 9999. The percentages were 30.10, 17.61, 12.49, 9.70, 7.91, 6.70, 5.79, 5.12, and 4.58. These agree almost precisely with the Wikipedia numbers. Ditto for powers of 5.

madcaptenor 9y ago
I agree. But for powers from 1 to 50 those same percentages are 30, 20, 10, 10, 10, 8, 2, 10, 0. In particular there is only one power of 2 in this range starting in 7 and none that ends in 9. This might make someone doing hand computation suspicious. (But really you can't tell much from such a small sample.)