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Let us define $n \in \mathbb{N} \setminus \{0\}$ and let us define the digit sum function for the base $b$ as $F_b(n)$ (according to wikipedia).

Also as mentioned in the previous wikipedia link the base is $b \ge 2$.

What I am asking myself is if $b$ could be $1$, would this mean that $F_1(n) = n$?

If this should be true then we could say that the basic counting is similar (or equal?) to counting of $1$'s for the number $n$ in base $1$?

PiMathCLanguage
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1 Answers1

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Base $n$ has digits $0, 1, ..., n-1$

Base 1 would only have $0$ as a digit, so is impossible to use as a way to represent numbers.

Therefore $b\ge2$

tomi
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  • I guess we could define the base b representation of a number $n$ as the lexicographically first sequence of non-negative integers $c_i$ such that $\sum_i c_i b_i = n$. That would give usual base $b$ representations for $b > 1$ and tallies for $b = 1$. – Daniel McLaury Jan 12 '20 at 21:17
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    Oh that didn't quite work because you would prefer e.g. 111000 to 111. I wonder it can be fixed cleanly, though. – Daniel McLaury Jan 12 '20 at 21:59
  • My thoughts were that $n$ in base $1$ would be a concatenation of $1$'s $n$-time. Therefore I was suggesting that $F_1(n)=n$, but maybe I have a flaw in my understanding. Also does base $n$ has only $n-1$ digits? Should it not be digits $0, 1, ..., n-1$? – PiMathCLanguage Jan 13 '20 at 18:55
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    Well spotted. I’ve changed my answer. – tomi Jan 13 '20 at 20:12