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Old 2020-02-26, 11:04   #1
henryzz
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Default Strings and Loops within Pi - Numberphile



Thought this video might interest some on the forum. The sequences thing at the end reminded me of Aliquot sequences.

Is it possible to construct an irrational number with a high density of self-locating digits?
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Old 2020-02-26, 23:29   #2
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Quote:
Originally Posted by henryzz View Post
Is it possible to construct an irrational number with a high density of self-locating digits?
x=sum(n=1,inf,0.1^(10^(n!))) is a trancendental number(!) (see https://en.wikipedia.org/wiki/Liouville_number), and obviously it is self locating at k=10^(n!) positions. OK, this gives only O(loglog(n)) self locating positions in the first n digits of x, but it could be easy to modify the definition of x: make it more dense, set self locating position at many "consecutive" position, and make it wrong at infinitely many place using long sequence of zeroes (like for the Liouville number), it could give much more than log(n) self locating positions (for every large n).
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Old 2020-02-27, 02:22   #3
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If you're not too picky about the base, there is an irrational number called the "rabbit constant" which has the unusual property that both its binary expansion and its simple continued fraction are known.
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Old 2020-02-27, 07:01   #4
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Quote:
Originally Posted by Dr Sardonicus View Post
If you're not too picky about the base, there is an irrational number called the "rabbit constant" which has the unusual property that both its binary expansion and its simple continued fraction are known.
It's A014565 in the OEIS.
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Old 2020-02-27, 09:14   #5
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It is not that hard to see that in the first n digits there could be only O(n/log(n)) self-repeating positions.

And surprisingly this is reachable by a trancendental number:
let a[1]=1,..a[9]=9,a[10]=10,a[11]=12,a[12]=14,..., consecutive positions that are using different digits, note that you can't use 11 after 10, because the 11th digit would be used twice.

For every k in the a[2*k-1] or a[2*k] position set a self-repeating position, that is 3 possible cases. So it gives 3^N_0=c real numbers, but there is only N_0 algebraic numbers. So there would be a trancendental number (and c>0) that has c*n/log(n) self-repeating positions in the first n digits of x for every n>1.

Last fiddled with by R. Gerbicz on 2020-02-27 at 09:17 Reason: typo
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