Question 1208787
<pre>
0.3450543003450005430000

I assume you meant digits AFTER the decimal, not including the introductory 0.

The 4's occur in positions 2, 6, 11, 17, ...

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

Also if 4 has occurred an even-number of times, the digit that follows it is a
3, and if 4 has occurred an odd-number of times, the digit that follows it is a
5.

Let the general term of 2, 6, 11, 17, ... be {{{a[n]=An^2+Bn+C}}}

{{{system(A(1)^2+B(1)+C=2, A(2)^2+B(2)+C=6, A(3)^2+B(3)+C=11)}}}

{{{system(A+B+C=2, 4A+2B+C=6, 9A+3B+C=11)}}}

Solve that and get A=0.5, B=2.5, C=-1

So the general term for the positions of 4's is {{{a[n]=0.5n^2+2.5n-1}}}

So let's find the position of the 4 nearest the 81403rd term by setting

{{{0.5n^2+2.5n-1}}}{{{""=""}}}{{{81403}}}

{{{0.5n^2+2.5n-81404}}}{{{""=""}}}{{{0}}}

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

n=401.0024783 and n=-406.0024783

So it is almost equivalent to a quadratic factorable as (n-401)(n+406) which
would have been this quadratic in n:

{{{n^2+5n-162806=0}}}

The first two terms of that are twice the first two terms of the general
equation for the positions of the 4's. So we divide through by 2

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

Adding 81403 to both sides

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

Adding -1 makes the left side the general term for digit positions of 
the 4's. So let's add -1 to the right side also:

{{{0.5n^2+2.5n-1=81402}}}

That has solutions 401 and -406, so the 81402nd digit is a 4 and since 
it is the 401st occurrence of a 4, and 401 is odd, the next digit, or the
81403rd digit, is a 5.

BTW, if you did include the introductory 0 left of the decimal point, then the
answer would have been 4, and the solution would have been a little easier.

Edwin</pre>