Relevant wiki: Integers - Problem Solving

The first thought that occurs is that we will have a string of consecutive digits (with different digits before and after) when switching from $k$ -digit numbers to $(k+1)$ -digit numbers: $98,99,100$ has a sub-string of two $9$ 's; $998,999,1000$ has a sub-string of three $9$ 's; and $9998,9999,10000$ has a sub-string of four $9$ 's, this last one potentially being the sub-string we seek. So if there is an earlier such sub-string, it must occur among $k$ -digit numbers, where $k=2, 3$ or $4$ .

We check each case separately. (In the following, $x$ is the repeating digit while $a$ and $b$ are any digits not equal to $x$ .)

Two-digit numbers:

$\begin{aligned} &\bullet \quad \_a,xx,xx,b\_ && \quad \small \text{Two of the same number; clearly } \color{#D61F06} \text{not possible} \\ &\bullet \quad ax,xx,xb && \quad \small \text{The ones digit of the first and second numbers are the same; } \color{#D61F06} \text{not possible} \end{aligned}$

Three-digit numbers:

$\begin{aligned} &\bullet \quad \_ax,xxx,b\_\_ && \quad \small \text{The ones digit of the two numbers are the same; } \color{#D61F06} \text{not possible} \\ &\bullet \quad axx,xxb && \quad \small \text{The hundreds digit changes but the tens digit does not; } \color{#D61F06} \text{not possible} \\ &\bullet \quad \_\_a,xxx,xb\_ && \quad \small \text{The tens digit changes, so the ones digit is 9, so the second number is 999; but then the third number has first digit 9, } \color{#D61F06} \text{not possible} \end{aligned}$

Four-digit numbers:

$\begin{aligned} &\bullet \quad \_\ \_\ \_a,xxxx,b\_\ \_\ \_ && \quad \small \text{The thousands digit of the second number changes, so x=9 and xxxx=9999, } \color{#3D99F6} \text{already mentioned above} \\ &\bullet \quad \_\ \_ax,xxxb && \quad \small \text{The tens digit changes, so x=9, then a=8 and b=0; } \color{#20A900} \text{possible} \color{#333333} \text{, first example is 9989,9990} \\ &\bullet \quad \_axx,xxb\ && \quad \small \text{The tens digit changes, so x=9, then a=8 and b=0; } \color{#20A900} \text{possible} \color{#333333} \text{, first example is 9899,9900} \\ &\bullet \quad axxx,xb\_\ \_ && \quad \small \text{The hundreds digit is changing, so x=9, then a=8 and b=0; } \color{#20A900} \text{possible} \color{#333333} \text{, first example is 8999,9000} \end{aligned}$

Of the four possible cases, the earliest is $8999,9000$ . We need to find the location of the second digit of $8999$ .

$\begin{aligned} &\bullet \quad 0\ -\ 9 && \quad \small \text{The first ten numbers are single-digit, for a total of 10 digits.} \\ &\bullet \quad 10\ -\ 99 && \quad \small \text{The next ninety numbers are two-digit, for a total of 180 digits.} \\ &\bullet \quad 100\ -\ 999 && \quad \small \text{The next nine hundred numbers are three-digit, for a total of 2700 digits.} \\ &\bullet \quad 1000\ -\ 8999 && \quad \small \text{The next eight thousand numbers are four-digit, for a total of 32000 digits.} \end{aligned}$

Thus after $10 + 180 + 2700 + 32000 = 34890$ digits we've reached the last digit of $8999$ , so the position of the second digit of $8999$ is $34890 - 2 = \boxed{34888}$

the substring we are searching will occur at the point ...8999,9000......
knowing this here is a $Mathematica$ code to count the digits

Length@Flatten[IntegerDigits/@Range[0,8998]]

this returns $34886$

the next number is $8999$ so we add $2$ (position where the sequence begins)
and the answer is $34888$

For two digit numbers or Between 1 to 100 we can have $\cdots10{\color{#3D99F6}111}21 \cdots 21{\color{#3D99F6}222}3\cdots 32{\color{#3D99F6}333}4 \cdots 43{\color{#3D99F6}444}5 \cdots 54{\color{#3D99F6}555}6 \cdots 87 {\color{#3D99F6}888} {\color{#D61F06}99}0\cdots 9899.$ We notice that as the identical digits $aaa$ starts to appear the preceding digits is $a-1$ and following digits is $a+1$ . With this notice we can be sure that there is no possibilities to appear such nth sub-string except $898 \phantom{c}899 \phantom{c}900$ for greater than 3 digits numbers and $1010\phantom{c}111\phantom{c}112$ . If $abc$ is three digit then on concatenation we find $\cdots (c-1)\phantom{c} abc \phantom{c}ab(c+1)\cdots$ . If $c+1=0$ then only we have chance to meet such number for which $c$ corresponds to $9$ .

On the same manner, there is no possibilities to appear other sub-string except $8998\phantom{c} 8999 \phantom{c} 9000$ . Therefore , the first required sub-string is $012345678910\cdots \cdots \cdots 8998 8999 9000$ Now to determine the position of sub-string we can use the fact $\text{from 0 to 9 digits } =9\times 1+1=10\\ \text{from 10 to 99 digits } =89\times 2 +2 =180\\ \text{from 100 to 999 digits } = 899\times 3 +3 = 2700 \\ \text{ from 1000 to 8999 digits } = 7999\times 4 +4 = 32000\\\ \text{or total digits } = (8999\times 4+4) -(1000×4) = 32000$ Totalling we find $34890$ . Hence deducing $-2$ (since we jump to last $9$ of $8999$ ) we find $32890-2=\boxed{34888}$ which is the required position of sub-string.

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# I create a string of concatenated integers
bigString = ''
for i in range(10000):
    bigString += str(i)

# I print the indices of the sub-strings with only 4 identical digits
for idx, item in enumerate(bigString):
    if (bigString[idx - 1] == bigString[idx] == bigString[idx + 1] == bigString[idx + 2]  # we want 4 identicals digits
            and bigString[idx + 3] != bigString[idx]  # the next digit needs to be different
            and bigString[idx - 2] != bigString[idx]):   # the previous digit needs to be different
        print(idx)

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34888   <===
38489
38850

$\boxed{34888}$ is the first one !

C++ code to do this:

include <stdio.h>

include <cstdlib>

include <cmath>

include <iostream>

include <string>

using namespace std;

int main(int argc, char **argv) {

int len = 4;
if ( argc > 1 )
    len = atoi( argv[1] ); 
int i = 0;
char prev = -1;
string number;
int pos = 0;
int sequence = 1;

do {

    if ( number.length() == 0 ){
        number = to_string(i++);
    }

    char next = number[0];
    cout << next;
    number.erase(0, 1);
    pos++;

    if ( next == prev ){
        sequence++;
    }
    else{
        if ( sequence == len ){
            cout << endl << len << " consecutive " << prev << "'s found at position " << pos-len << endl;
            return 0;
        }
        sequence = 1;
        prev = next;
    }

} while( true );

}

4 consecutive 9's found at position 34888

Python solution:

# all non-negative integers concatenated into a single infinite string
str_txt = ''
for i in range(10001):
    str_txt = str_txt + str(i)

print(str_txt)

# the function of the position of the first digit of the four
def str_search(str_txt,pattern):

    M = len(str_txt)
    N = len(pattern)

    for i in range(M-N+1):

        temp = str_txt[i:i+N]

        if i>0 and i+N < M:
            temp_pre = str_txt[i-1]
            temp_suf = str_txt[i+N]
        elif i == 0:
            temp_pre = ''
            temp_suf = str_txt[i+N]
        elif i+N == M:
            temp_pre = str_txt[i-1]
            temp_suf = ''

        if temp == pattern and temp_suf != pattern[0] and temp_pre != pattern[0]:
            return i+1

print(str_search(str_txt,'0000'))
print(str_search(str_txt,'1111'))
print(str_search(str_txt,'2222'))
print(str_search(str_txt,'3333'))
print(str_search(str_txt,'4444'))
print(str_search(str_txt,'5555'))
print(str_search(str_txt,'6666'))
print(str_search(str_txt,'7777'))
print(str_search(str_txt,'8888'))
print(str_search(str_txt,'9999'))

Output :

None
None
None
None
None
None
None
None
34888
38892

So the position of the first digit of the four is 34888 .