Question 1208787
<pre>
0.3450543003450005430000

Here is the problem assuming the introductory 0 on the left of the decimal is the 1st digit

The 4's occur in positions 3, 7, 12, 18, ...

Its sequence of first differences 4, 5, 6,... is linear, so the
sequence of positions of 4s is quadratic.

Let the general term of 3, 7, 12, 18, ... be {{{a[n]=An^2+Bn+C}}}

{{{system(A(1)^2+B(1)+C=3, A(2)^2+B(2)+C=7, A(3)^2+B(3)+C=12)}}}

{{{system(A+B+C=3, 4A+2B+C=7, 9A+3B+C=12)}}}

Solve that and get A=0.5, B=2.5, C=0

So the general term for the positions of 4's is {{{a[n]=0.5n^2+2.5n}}}

So let's find the position of the 4 nearest (or 'at') the 81403rd term by 
setting

{{{0.5n^2+2.5n}}}{{{""=""}}}{{{81403}}}

{{{0.5n^2+2.5n-81403}}}{{{""=""}}}{{{0}}}

Solving that with a quadratic program on my TI-84 graphing calculator:

n=401 and n=-406 

That has solution 401 (Ignore the negative), so the 81403rd digit is a 4.

Answer = 4

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It's easier when we consider the 0 to the left of the decimal as the
first digit.

In the olden days, people didn't, as a rule, put a 0 before the decimal
in decimal fractions between -1 and 1, as they do today.  We would just write,
say, .5, not 0.5, and -.75, not -0.75 and:
  
.3450543003450005430000... not 0.3450543003450005430000...
 
I'm an old fogy, so that's why I didn't consider the initial 0 before
the decimal as the first digit in my other answer. LOL

Edwin</pre>