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By using WolframAlpha, I couldn't find any transcendental number without equal adjacent digits among the numbes $\tan(n)$, $\sin(n)$, $\cos(n)$, $\sec(n)$, $\cot(n)$, $\csc(n)$, $e^n$, and $ \log(n)$, where $n$ is an integer number.

How to find a transcendental number where no two adjacent decimal digits are equal?

Some results below

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  • $\begingroup$ by repeated do you mean consecutive? All transcendental numbers do not have a repetitive decimal expansion. This is a property of only rationals. $\endgroup$ – Eoin Jan 7 '15 at 19:56
  • $\begingroup$ yes, I mean the adjacent digits $\endgroup$ – user187581 Jan 7 '15 at 19:58
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    $\begingroup$ Probably, only specially constructed transcendental numbers have this property. $\endgroup$ – Peter Jan 7 '15 at 19:58
  • $\begingroup$ I imagine the best way to construct a transcendental number in this way is to write the decimal expansion yourself. A fairly easy way would be to write two sequences of digits with entirely different starting and ending points. Call them $a_1$ and $a_2$. Then string them together in a non-repeating fashion $a_1a_2a_1a_1a_2a_1a_1a_1a_2...$. This should be a non-repeating decimal expansion of a transcendental number. $\endgroup$ – Eoin Jan 7 '15 at 20:05
  • $\begingroup$ You can show the existence of such a number with cantor's diagonalization. List all algebraic numbers (there are only countably many), start with $0,$ and choose the firs digit different from $0$ and the first digit in the first line, the second digit different from the first digit and the second digit in the second line and so on. Then, the number must be transcendental and has no adjacent equal digits. $\endgroup$ – Peter Jan 7 '15 at 20:26
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Start with Liouville's constant $0.11000100000000000000000100 \dots$ which is known to be transcendental and add $\frac{2}{99} = 0.02020202 \dots$. The resulting number is transcendental (because it differs from Liouville's constant by a rational) and has no identical adjacent digits.

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  • $\begingroup$ Very good! This is a concrete example, so cantor's method became unnecessary. $\endgroup$ – Peter Jan 7 '15 at 20:27

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