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While digging through some old notebooks today, I found a problem from a long time ago that I was never able to solve. It involves a sequence of positive integers called the “special numbers” defined：

A number $m$ is called special if there is no $k$ such that $$k + S(k) = m$$

Prove that there are infinitely many special numbers of the form $10^n+b$ if and only if $b-1$ is special.

the problem is from (problems from the book) the chapter the digit sum of a positive integer and created by:Christopher D. Long

I have found sequence A003052 on OEIS.

markvs
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math110
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    So $S$ is digitale sum, should be added to the question body – Hagen von Eitzen Mar 09 '22 at 06:56
  • in base $10$ and will denote the decimal sum of the digits of the positive integer $x$ by $s(x)$ – math110 Mar 09 '22 at 12:05
  • The q was asked and answered more than a month ago. You either accept the answer, post your own or add a comment saying what else do you need. Right now the Q is listed as having no good answer. – markvs Apr 09 '22 at 01:52

1 Answers1

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One way is obvious: if $b=k+S(k)+1$ then for all sufficiently large $n$, $10^n+b=(10^n+k)+S(10^n+k)$, so $10^n+k$ is special provided $b-1$ is special.

Suppose that $10^n+b=k+S(k)$ for some $n$ with $10^n\gg b$ and some $k$ (such an $n$ surely exists in an infinite set).

Note that $S(k)\le 9\log_{10} k<k/10$ if $n$ is large enough. Then $k=10^n+c$ for some $c\le 10^{n-1}$. Then $c+S(c)+1+1=b+1$, so $b-1$ is special.

QED

markvs
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