TITLE: Given sequence of 7-digit positive numbers: $a_{1},...,a_{2017}$ with $a_{1}<a_{2}<...<a_{2017}$. Determine $a_{2017}$.
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: Hint: I don't think there  is   an  essentially better  way  than   suggested  by  OP and  the nicely elaborated answer by @antkam.

Note the sequence $a_1,\ldots,a_{2017}$  is a  subsequence  of  the Numbers with digits in nonincreasing order  stored  as   A009996 in OEIS.
The formula which  is presented there to calculate sequence elements  is  essentially the same that we can see in the answer by @antkam indicating that no simpler way is known.