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Another question asked about the convergence of the sum of reciprocals of integers where digits appear only an odd number of times. (Digits which do not appear are ignored, so for example $14404$ is included since $1$ and $0$ appear once and $4$ appears three times.)

It would be possible to show that the sum of reciprocals diverges, using the number of $n$-digit integers where digits appear an odd number of times. For small $n$, the numbers of such integers appear to be

n  1  2   3    4     5      6       7        8        9        10         11          12
#  9 81 657 4860 33705 228096 1555857 10782720 75810825 543977856 3989477457 29758855680

and for large $n$

  • about $\frac{10-1}{10 \times 2^{10-1}} \times 10^n$ for even $n$
  • about $\frac{10-1}{ 2^{10-2}} \times (10-1)^n$ for odd $n$ (smaller because ten odd numbers cannot add up to an odd number)

These are empirical results. Something similar seems to work in other bases, and replacing all the $10$s with $2$s leads to an easy proof.

I was wondering whether a proof was possible justifying this $\frac{10-1}{10 \times 2^{10-1}}$, or at least a lower bound for the multiplicative factor for even $n$.

Henry
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    Here https://math.stackexchange.com/questions/1146548/number-of-strings-when-each-character-must-occur-even-times is a similar question which you might adapt – Empy2 Jun 21 '21 at 15:33
  • @Empy2 That looks useful - though the odd case likely to be more complicated since in your link zero occurrences can be used as an even number while here it is an ignored number – Henry Jun 21 '21 at 15:47
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    Choose your k digits, in how many ways, then do the generating function, then sum over k. – Empy2 Jun 21 '21 at 15:49

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