Consider the powers of two: 1, 2, 4, 8, 16, 32, 64, 128... Their right-most digits follow a simple repetitive pattern: 2, 4, 8, 6, 2, 4, 8, 6, 2, 4, 8, 6...
What about their left-most digits, though?
Here are the first 40-odd.
Read left to right, line by line. The digits are jammed together, leaving out the courtesy commas, to save space.
124813612512481361251248136125
124813612512481371251249137125
124913712512491371361249137136
124913713612512481361251248136
125124813612512481361251248137...
There is a pattern there, but it keeps breaking down. When you get up into really big powers of 2, in fact, it breaks down altogether.
For the ten powers of 2 from the 30th to the 39th, for instance, we have lead digits:
1, 2, 4, 8, 1, 3, 6, 1, 2, 5.
For the ten powers from the billionth to the 1,000,000,009th, by contrast, we have lead digits:
4, 9, 1, 3, 7, 1, 2, 5, 1, 2.
Every digit (except, of course, zero) shows up, though.
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In the first N digits of the sequence, how many occurrences of 3 shall one find? Call the number T(N).
And how many occurrences of 4? Call the number F(N).
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The first few values of T(N) and F(N), for N from 1 to 15, look like this, as you can easily verify:
T(N) = 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2,...
F(N) = 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2,...
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1. Show that for sufficiently large N, T(N) will always be bigger than F(N).
2. Find the limit of the ratio T(N) to F(N) as N tends to infinity.