OEIS entry - A316312 has a question: Is it true that if k is a term then 100 * k is a term?

Question

Refer https://oeis.org/A316312 - the sequence in OEIS.

The sequence says

Numbers n such that sum of the digits of the numbers 1, 2, 3, ... up to (n - 1) is divisible by n.

A few terms from the sequence are as follows: 1, 3, 5, 7, 9, 12, 15, 20, 27, 40, 45, 60, 63, 80, 81, 100, 180, 181, 300, 360, 363, 500, 540, 545, 700, 720, 727, 900, 909, 912, 915, 1137, 1140, 1200, 1500, 1560, 1563, 2000, 2700, 2720, 2727, 4000, 4500, 4540, 4545, 6000, 6300, 6360, 6363, 8000, 8100, 8180, ...

It is evident from the sequence that if k is a term then 100*k is also a term. We (Author of the sequence and another Math nerd from GanitCharcha) haven't been able to prove or disprove it and we also do not know whether this has been proved or disproved.

If anyone can help us in this regard by providing pointers or any solution.

Answer 1

Let $a(n)$ be the function in question. Then

$$a(100n) = 100a(n) +na(100)$$ Since $a(100)$ is a multiple of $100$, we have that if $a(n)$ is a multiple of $n$, then $a(100n)$ is also a multiple of $100n$.

Answer 2

Yes it seems to be true.

Suppose $n$ is in $A316312$ and the cumulative digit sum up to $n-1$ is $d$;

the cumulative digit sum up to $100-1$ is $900$ so the cumulative digit sum up to $100n-1$ is $900n+100d = 100(9n+d)$.

Since $d$ is divisible by $n$ in $A316312$, we have $100(9n+d)$ being divisible by $100n$.