The answer is not what it seems

As you would have noticed

$\frac { 1 }{ 9 }$ =0.11111......

$\frac { 1 }{ 99 }$ =0.010101...

$\frac { 1 }{ 999 }$ =0.001001001...

$\cfrac { 1 }{ { 10 }^{ n }-1 }$ will have a one at every ${ n }^{ th }$ position after decimal

So,every ${ n }^{ th }$ place will have as many 1 as it has factors, if n+1 has factors between 0-9

So,for n having 11,21,31....... factors it will have 1 present at that position

Writing prime factorization we get

11=11*1

21=3*7

31=31*1

41=41*1

51=3*17........

Taking base 2 will generate the smallest number required.

So,smallest number with 11 factors is

n= ${ 2 }^{ 10 }$ =1024>576

Similarly,

21 factors

n= ${ 2 }^{ 6 }*{ 3 }^{ 2 }$ =576

31 factors

n= ${ 2 }^{ 30 }$ >576

41 factors

n= ${ 2 }^{40}$ >576

51 factors

n= ${ 2 }^{ 16 }*{ 3 }^{ 2 }$ >576

So ,1 at second place occur at $\boxed{n=576}$

However,if n has exactly 10,20,30...... factors and n+1 has factors in between 10 and 19 (both included) then also 1 will be present at that position

Writing prime factorization of 10 we get 10=2*5

So,we`ll have to check for numbers of the form ${ 2 }^{ 4}*k$ , ${ 3 }^{ 4}*k$ ....where k is a prime number other than the one it is multiplied with and so on till number is less than 1024

The smallest number satisfying above condition is given for n= ${ 2 }^{ 4}*59$ = $\boxed{944}$ ,it has exactly 10 factors and n+1 =945 has 16 factors

The last step of the solution is quite time taking so if any one has a better method please post it Please notify me about any typing mistakes since it was a long solution

My computer program generates all the figures one by one.

For n and N:

T1(n) = Output(n MOD 1) {n >= 1}

T2(n) = Output(n MOD 2) {n>= 2}

T3(n) = Output(n MOD 3) {n>= 3}

T4(n) = Output(n MOD 4) {n >= 4}

...

TN(n) = Output(n MOD N) {n>= N}

Sn = T1(n) + T2(n) + .... + TN(n) {1<= N <=n}

IF n MOD N <> 0 THEN Output:=0 ELSE Output:=1; {To be realistic}

Decision onto "<>" is a potentially faster than "=". No better trick.

Example:

Chain 12 of more than enough for longer counts of guaranteed correctness:

 FOR nx:=1 TO 100011 DO
 BEGIN
      Sn:=0;
      FOR Ny:=1 TO nx DO
          IF nx>=Ny THEN
          BEGIN
             IF nx MOD Ny <> 0 THEN
                Output:=0
             ELSE
                 Output:=1;

             Sn:=Sn+Output
          END;

      {For next use only}
       S1[flip0]:=Sn;
       S2[flip0]:=S1[flip1];
       S3[flip0]:=S2[flip1];
       S4[flip0]:=S3[flip1];
       S5[flip0]:=S4[flip1];
       S6[flip0]:=S5[flip1];
       S7[flip0]:=S6[flip1];
       S8[flip0]:=S7[flip1];
       S9[flip0]:=S8[flip1];
      S10[flip0]:=S9[flip1];
      S11[flip0]:=S10[flip1];


      IF (((((((((((
          Sn DIV 10 +
          S1[flip1]) DIV 10 +  S2[flip1]) DIV 10 + S3[flip1]) DIV 10 +
          S4[flip1]) DIV 10 +  S5[flip1]) DIV 10 + S6[flip1]) DIV 10 +
          S7[flip1]) DIV 10 +  S8[flip1]) DIV 10 + S9[flip1]) DIV 10 +
         S10[flip1]) DIV 10 + S11[flip1])
                     MOD 10 = 1 THEN

          BEGIN
               INC(count);
               Found:=nx-11
          END
      ELSE
          Found:=0;

      IF Found<>0 THEN {Convenience to change}
      BEGIN
           WRITELN(S11[flip1]:3, '   ' , S10[flip1]:3, '   ' ,
                    S9[flip1]:3, '   '  , S8[flip1]:3, ' ...' ,
                    S1[flip1]:3, '   ' ,  Sn:3,
                   '   n= ', nx:5,
                   '   Count= ', count:4,
                   '   Found= ', Found);
      END;

      IF flip0=0 THEN
      BEGIN
           flip0:=1;
           flip1:=0
      END
      ELSE
      BEGIN
           flip0:=0;
           flip1:=1
      END
 END;
 WRITE('Count = ', count);

From 1 to 65535, there are 1003 counts of '1'. {1, 576, 944, 1024, 1424, 1600, 1616, 1875, 2015, 2349, 2375, 2384, 2511, 2672, 2916, 2960, 3059, 3136, 3184, 3248, 3311, 3483, 3824, 4304, 4779, 4784, ..., 64375, 64385, 64448, 64557, 64574, 64624, 64655, 64679, 64745, 64749, 64784, 64823, 64827, 64880, 64976, 65008, 65009, 65168, 65189, 65199, 65264, 65279, 65339, 65351, 65375 and 65456.} By WORD which is faster; border at 65535 checked with LONGINT. Only 3 term chain has been sufficient to be correct for this range. Checked with chain up to 12 terms and found agreed.

From 65535 to 100000, there are 726 counts of '1'. {65648, 65659, 65680, 65691, 65744, 65823, 65904, 66015, 66064, 66128, 66149, 66177, 66224, 66263, 66299, 66338, 66384, 66416, 66429, 66443, 66554, 66663, 66779, 66815, 66824, 66832, ..., 98739, 98747, 98768, 98783, 98903, 98999, 99014, 99024, 99056, 99152, 99197, 99245, 99344, 99375, 99383, 99395, 99440, 99449, 99539, 99664, 99665, 99728, 99755, 99827, 99899 and 99952.} By LONGINT which is slower. Only 4 term chain has been sufficient to be correct for this range. Checked with chain up to 12 terms and found agreed. 3 term chain surprisingly made 725 counts to total of 1728 counts!

There are a total of 1729 counts of '1' from 1st to 100,000th decimal place; 4 terms is the least required.

72333 [20, 8, 16, 40] 72336 as an only case found by a 4 terms instead of 12 terms above:

      IF
        ((((Sn DIV 10 +  S1[flip1]) DIV 10 +  S2[flip1]) DIV 10 + S3[flip1])
               MOD 10 = 1)

               AND NOT

        (((S1[flip1] DIV 10 +  S2[flip1]) DIV 10 +  S3[flip1]) MOD 10 = 1)

      THEN
          Found:=nx-3;

Other than '1', we can find all others.

943 [4] --> [5]

944 [10] --> [1]

945 [16?] --> [6]

{1, 576, 944, 1024...}

I used a computer program to count the numbers of factors including checking the number carried forward from the next number of factors.

After manually adding up the sequence and observing the resulting fraction, I figured that the manual method would be prone to errors. Unix provides an arbitrary-precision command-line calculator program called "bc". I wrote the following script and examined the output.

1
2
3
4
5
6
7
8
9

scale=1000;
frac=0;
lastpow=0;
for ( i=1; i<1000; i++) {
   newpow=lastpow*10+9;
   frac += 1/newpow;
   lastpow=newpow;
   frac;
}

Looking at the results in a text editor gave me the answer. I wish I had the maths to solve it more analytically via formulae. It was luck I picked 1000 iterations, to find it at the 944th iteration.

0.1223242434262445262644283446282644492448282664303646284844322467482648322466483054324448324644522669282884322850284664563446522848324488283244554466322868522452502684363866287044382848484688342489323244524470562644564452483294322888324648402666684848384489285264364846363084722476282688583846324884584450523248362526285448556448562664602652524884323292284684443472483068522888563244368466365244582495284708562846525248442468684688402886483698324454384644962492363084384848487264444489522852324930483244444866365084524856322724603872484844842452562672564486326084324508363248812652484924384850368648442866683288582496283284605486642844526876283288405286285068442488924648642529285648384850562724564458365484522868568644504646723678522532523288364472563048642496284724583646326884644895482648644486365864528848403248604532282928382854685284523446727052522928363284444472564868922460383284364906485684674888482676404486563644646472366645042852285168325256324684683866563288384530285864364892643248522496