Given a sequence of 7-digit positive numbers: $a_{1},a_{2},...,a_{2017}$ with $a_{1}<a_{2}<...<a_{2017}$. The digit is ordered such that it is nonincreasing. It is known that $a_{1}=1000000$ and $a_{n+1}$ is the smallest possible number that is greater than $a_{n}$. This means $a_{2}=1100000$ and $a_{3}=1110000$. Determine $a_{2017}$.
Attempt:
Let us define $a_{0}=0000000$ and present the terms for $n=1,2,...,20$, $$ a_{1} = 1000000, a_{2} = 1100000, a_{3} = 1110000, a_{4} = 1111000, a_{5} = 1111100$$ $$ a_{6} = 1111110, a_{7} = 1111111, a_{8} = 2000000, a_{9} = 2100000, a_{10} = 2110000$$ $$ a_{11} = 2111110, a_{12} = 2111111, a_{13} = 2200000, a_{14} = 2210000, a_{15} = 2211000$$ $$ a_{16} = 2211100, a_{17} = 2211110, a_{18} = 2211111, a_{19} = 2220000, a_{20} = 2221000$$
We can analyse that the number of sequence that has $1$ as leading digit is $\binom{7}{1} = \binom{6+1}{1}$. Also the number of sequence that has 2 as leading digit is $\binom{6+2}{2}$. In general, the number of sequence that has $j >= 1$ as leading digit is $\binom{6+j}{j}$.
The number of sequence that has leading digit $1,2,..,m$ is therefore: $$ \alpha_{m} = \sum_{j=1}^{m} \binom{6+j}{j}$$
Which means that $a_{\alpha_{m}}$ is the last sequence that has $m$ as leading digit. We can find the simplified form for each $\alpha_{m}$. We will prove that
$$ \alpha_{m} = \sum_{j=1}^{m} \binom{6+j}{j} = \sum_{j=1}^{m} \binom{6+j}{6} $$ $$ = \binom{7}{6} + \binom{8}{6} + \binom{9}{6} + ... + \binom{6+m}{6} = \binom{7+m}{7} - 1$$
We can see $\binom{7+m}{7}$ as the number of ways to pick 7 numbers from the set $\{1,2,...,7+m\}$. But the list can be grouped as, the choice(s) that has $1$ as minimum number, $2$ as minimum number, and so on. These means that
$$ \binom{7+m}{7} = \binom{7+m-1}{6} + \binom{7+m-2}{6} + ... + \binom{7+m-m}{6} + \binom{7+m-(m+1)}{6} $$ $$ \binom{7+m}{7} = 1 + \sum_{j=1}^{m} \binom{6 + j}{6} $$
Thus $\alpha_{m}= \sum_{j=1}^{m} \binom{6 + j}{6} = \binom{7+m}{7} - 1$.
Let us check what value of $m$ such that $\alpha_{m-1} \le 2017 \le \alpha_{m}$. $m$ is $7$, $$ \alpha_{m-1} = 1715 \le 2017 \le \alpha_{m}=3431 $$
So the leading digit of $a_{2017}$ must be $7$. The first sequence with leading 7 is $a_{1716}$. How about the 2nd digit? The number of 6 digits sequence leading 1,2,..,5 is $\binom{6+5}{6} - 1 = 461$, with leading 1,2,..,4 is $209$. Now because $1715 + 209 < 2017 < 1715 + 461$, then we can conclude that the 2nd digit must be a 5. We can continue doing this until we find that $a_{2017}=75$*****.
Are there better or quicker approaches?
I don't think there is a faster way than recursing down the digits, but at each step you could have found / reasoned about your result a bit faster.
Consider any $n$-digit number with leading digit $\in [0, m]$. (I.e. we allow $0000000$ for now.) By the rules given, this number is uniquely specified by the $(m+1)$-tuple $(a_0, a_1, \dots, a_m)$ where $a_j =$ how many times digit $j \in [0,m]$ appears, and $a_j \ge 0$. And we have:
$$a_0 + a_1 + a_2 + \dots + a_m = n$$
This is a standard stars-and-bars problem (Theorem 2 in the wikipedia article with $k=m+1$) and so the number of such numbers ($n$ digits, leading digit $\le m$) is:
$${n + (m+1) - 1 \choose (m+1)-1} = {n + m \choose m} = {n + m \choose n}$$
Now the above allowed $0000000$, but this just means you're looking for the $2018$th number instead of the $2017$th.
${7 + 6 \choose 7} = 1716 < 2018 \le {7 + 7 \choose 7} = 3432 \implies $ leading digit is $7$.
$2018 - 1716 = 302$ and now you're looking at $n=6$. So ${10 \choose 6} = 210 < 302 \le {11 \choose 6} = 462 \implies $ next digit is $5$.
$302 - 210 = 92$ and now you're looking at $n=5$. So ${8 \choose 5} = 56 < 92 \le {9 \choose 5} = 126 \implies$ next digit is $4$.
$92 - 56 = 36, n = 4$... ${7 \choose 4} = 35 < 36 \le {8 \choose 4} = 70 \implies$ next digit is $4$.
$36 - 35 = 1$ so we got lucky and the rest of the digits are $0$s.
Note: the math guarantees that the digits found by this process are non-increasing; but as a sanity check it's still good to see that they are indeed non-increasing. :)
Hence the answer is $7544000$.