Writing down consecutive natural numbers until a certain number of digits $k$ is reached.

A person starts writing consecutive natural numbers from $$5$$ until $$k$$ digits are reached. For some values of $$k$$, this will be impossible, for example $$6$$ or $$8$$ are impossible as then after writing the single digit numbers, we will have $$1$$ and $$3$$ left over respectively, but then we can only write double digit numbers.

An example of this would be if the person wanted to write $$9$$ digits, they write $$5, 6, 7, 8, 9, 10, 11$$ for a total of $$5 + 2(2) = 9$$ digits.

Question: Prove that if $$A$$, $$B$$, and $$n$$ are positive integers, then it is not possible to write $$An^{2}$$ + $$B$$ digits for each possible $$n$$ i.e for each pair of $$(A,B)$$, there exists an $$n$$ such that the person cannot write $$An^{2}$$ + $$B$$ digits.

I proved that the statement is true for $$An$$ + $$B$$, and think that it is true in this case as well, but I can't come up with a proof.

Edit: A sketch of the proof for the above case would be to note that as we go from $$n$$ to $$n + 1$$, the initial $$An + B$$ part is identical, and we only add $$A$$ digits after. Note that the smallest number we use in this "$$A$$ " digits interval is ever increasing, so there will come a point when even the smallest number in this interval shall contain more than $$A$$ digits.

Some perhaps useful facts:

$$1)$$ The differences between consecutive iterations is $$a(2n + 1)$$, a linear function in $$n$$.

$$2)$$ For sufficiently large $$n$$, if the $$n$$-th iteration uses numbers with at most $$P$$ digits, then the next iteration will use numbers with at most $$P + 1$$ digits.

$$3)$$ I imagine the proof is something which can be generalized to higher order polynomials as well.

• This is not clear. What does "until $k$ digits are reached" mean? I assumed you meant "natural numbers of length $k$" but that's not consistent with the rest of what you wrote. – lulu Jan 15 at 20:35
• For example, until $9$ digits would mean writing 5, 6, 7, 8, 9, 10, 11. So the total number of digits is 5 + 2(2) = 9. – Saad Jan 15 at 20:37
• I don't think that interpretation would occur to very many people. Please edit your post for clarity. – lulu Jan 15 at 20:39
• No, @lulu. The question is to show that for each pair $(A,B)$, you cannot write $A(n^{2}) + B$ digits for each $n$. For instance, if $A$ = 2, $B$ = 3, the question is whether you can write out consecutive natural numbers to hit $5, 11, 21, 35, ..$ digits – Saad Jan 15 at 21:05
• Sure...but if I could change the start each time, isn't it obvious that I could achieve whatever digit count I liked? I understand it if the starting point is fixed, but you indicated that you thought that irrelevant. – lulu Jan 15 at 21:06