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The first 15 terms of the sequence {a_i} = 2^i are 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768. All of the digits in base-10, i.e. {0, 1, 2, ..., 9}, each appear in at least one of these terms, and 32768 is the first term to contain the digit 7.

Define the digital potency (this is a term I came up with, I am unaware if there is a proper term for this) of n in base-10 to be the minimum value of x such that the digits {0, 1, 2, ..., 9} each appear in at least one of n, n^2, n^3, ..., n^x.

Thus, from the above example, the digital potency of 2 in base-10 is 15.

More generally, define the digital potency of n in base-b to be the minimum value of x such that the digits {0, 1, 2, ..., b-1} each appear in at least one of the base-b representations of n, n^2, n^3, ..., n^x.

Observation 1: For any numeral system base-b, we can construct a number with digital potency of 1 by simply concatenating the digits {0, 1, 2, ..., b-1}.

Observation 2: If n is a perfect power of b, then the digital potency of n in base-b is undefined or 'infinite' (i.e., there exists a digit in {0, 1, 2, ..., b-1} that never appears in any of the terms of the sequence {a_i} = m^i). I believe that this is actually true so long as n is a rational power of b.

Has this concept been studied before? I wrote some Python code to calculate the digital potency of various numbers in various bases, and it seems to have some interesting behavior. For instance, if you fix n and calculate the digital potency for various choices of b, the values seem rather unpredictable. Similarly, if you fix b and calculate the digital potency for various choices of n, the values also seem unpredictable.

Some specific questions I have:

  • Is there an easy way to calculate or at least approximate digital potency without explicitly calculating the powers of a number?
  • Is there a non-trivial example of a number and a base system where digital potency is undefined?
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  • $\begingroup$ For the second question, no. For any number $n$ and base $b$ with $log_b(n)$ irrational, $klog_b(n)$ for $k\geq1$ obeys the Weyl equidistribution law, hence all possible digits in $0...b-1$ occur as first digits. $\endgroup$
    – LeechLattice
    Commented Apr 9, 2019 at 23:45
  • $\begingroup$ @Bullet, I'd say zero never occurs as first digit. $\endgroup$
    – Gerry Myerson
    Commented Apr 10, 2019 at 3:23
  • $\begingroup$ A distantly related and, I believe, unsolved problem is to show that for all sufficiently large $n$ there is a zero somewhere in the decimal representation of $2^n$. See, e.g., oeis.org/A238938 $\endgroup$
    – Gerry Myerson
    Commented Apr 10, 2019 at 3:26
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    $\begingroup$ @GerryMyerson But zero could appear as second digit. The Weyl equidistribution law still applies. $\endgroup$
    – LeechLattice
    Commented Apr 10, 2019 at 4:26
  • $\begingroup$ @Bullet, sure, and if you had written, all possible digits in $0,\dots,b-1$ occur as second digits, I would have had no objection. $\endgroup$
    – Gerry Myerson
    Commented Apr 10, 2019 at 6:30

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