Wrong three digits

It's helpful to see the fractions in decimals. We can also change the fractions into sequences:

$\frac{1}{999}$ = 0.001001001001001001....

Since $\frac{x}{999}$ = $\frac{x}{1000}$ + $\frac{x}{(1000 \times 999)}$ , we can get..

$\frac{3}{(999 \times 999)}$ = $\frac{3}{(999 \times 1000)}$ + $\frac{3}{(999 \times 999 \times 1000)}$ = $\frac{3}{(999 \times 1000)}$ + $\frac{3}{(999 \times 1000 \times 1000)}$ + $\frac{3}{(999 \times 999 \times 1000 \times 1000)}$ =

$\frac{3}{(999 \times 1000)}$ + $\frac{3}{(999 \times 1000 \times 1000)}$ + $\frac{3}{(999 \times 1000 \times 1000 \times 1000)}$ + $\frac{3}{(999 \times 1000 \times 1000 \times 1000 \times 1000)}$ + .....

Since $\frac{3}{999}$ = 0.003003003003003003... then $\frac{3}{(999 \times 999 )}$ = 0.000003006009012015018...

With the same way, we can see that $\frac{2}{(999 \times 999 \times 999)}$ = $\frac{1}{1000}$ X $\frac{2}{(999 \times 999 )}$ + $\frac{1}{1000}$ X $\frac{1}{1000}$ X $\frac{2}{(999 \times 999 )}$ + $\frac{1}{1000}$ X $\frac{1}{1000}$ X $\frac{1}{1000}$ X $\frac{2}{(999 \times 999 )}$ + ... =

0.000000002004006008010012014.. +

0.000000000002004006008010012014.. +

0.000000000000002004006008010012014.. +

0.000000000000000002004006008010012014.. + .....

Now we know that the first three digits of $\frac{2}{(999 \times 999 \times 999)}$ is 000, the second is 000, the third is 002, the 4th is 006 or 002+004, the 5th is 012 or 002+004+006 , so now we can see that starting from the third three-digited number until a certain digits, the nth 3-digited number after decimal follows the arithmetic sequence of (n-2)(n-1) The following digits should still follow the pattern with the mistakes caused by the resulting sum being larger than 1000, and thus, give the excess to the 3-digit preceding it.

Now we can see that the sum of the three fraction caused the nth 3-digited number after decimal to follow the formula of: 1 + 3 (n-1) + (n-2)(n-1) = $n^{2}$

This method of predicting digit becomes wrong after n >= 31, such that the next n satisfies $n^{2}$ > 1000 such that the three digited number starting from digit 94th, 95th, and 96th, gives the excess sum to the digit 93rd, causing the 31st three-digited number becomes 962 instead.

With this logic, the 409th, 410th, and 411th digit is the last three digit of $137^{2}$ and the excess given by the sum of the following three digits, which is $138^{2}$

$137^{2}$ = 18769

$138^{2}$ = 19044

And thus, the 409th, 410th, and 411th digit, when concatenated, has the value of 769 + 19 = 788

position $411$ contains the square of $\frac{411}{3} = 137$ , $137^2=18769$ ,this concatinates with the first 2 digits of next number $138^2=19044$

so taking the numbers at positions 409,410,411 we get $769$ from $18769$ and $19$ from $19044$ wich add to $769+19=788$

For simplicity, let $q=10^{-3}$ . Then we can do partial fraction decomposition (PFD) to reorder Brian's new number $X$ into generalized geometric series $(*)$ : $\begin{aligned} X &= \frac{q}{1-q} + \frac{3q^2}{(1-q)^2} + \frac{2q^3}{(1-q)^3} = \frac{ q^2+q }{(1-q)^3} \underset{\text{PFD}}{=} \frac{1}{(1-q)} - \frac{3}{(1-q)^2} + \frac{2}{(1-q)^3}\\[1em] &\underset{(*)}{=}\sum_{k=0}^\infty\left[ \binom{k+0}{0}-3\binom{k+1}{1}+2\binom{k+2}{2} \right]q^k=\sum_{k=0}^\infty [1 -3(k+1) + (k+1)(k+2)]q^k = \sum_{k=0}^\infty k^2q^k=:\sum_{k=0}^\infty a_k \end{aligned}$ The only elements that can influence the digits $d_{409};\:d_{410};\:d_{411}$ are $a_{137}$ and $a_{138}$ . Let's write down their digits: $\begin{aligned}a_{137} &= 18769\cdot 10^{-411}, & a_{138} &= 19044\cdot 10^{-414} &&&\Rightarrow &&&& \begin{array}{r|rrrrrrrr} k & 407 & 408 & 409 & 410 & 411 & 412 & 413 & 414\\\hline a_{137} & 1 & 8 & 7 & 6 & 9 & &\\ a_{138} & & & & 1 & 9 & 0 & 4 & 4\\\hline & & & \red{7} & \red{8} & \red{8} & 0 & & \end{array}\end{aligned}$ Assuming the remainder of the series stays small enough that it does not influence these digits, the answer are the red digits $\boxed{788}$

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The generalized geometric series:

$\begin{aligned} \sum_{k=0}^\infty \binom{k+n}{k}q^k&=\frac{1}{(1-q)^{n+1}},&&&|q|&<1, &n&\in\mathbb{N}&(*) \end{aligned}$

Answer
= (411 ÷ 3)² mod 10³ + floor{ [(411 ÷ 3) + 1]² / 10³ }
= 137² mod 10³ + floor{ 138² / 10³ }
= 18769 mod 1000 + floor{ 19044 / 1000 }
= 769 + 19
= 788

Lucky 8, 88.8889%