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How do you read this correctly, where are the brackets if you set them?

2^2048 - 2^1984 - 1 + 2^64 * { [2^1918 pi] + 124476 }

Taken from: https://www.ietf.org/rfc/rfc3526.txt I have great problems solving this because I don't know how it is read correctly... There are some important brackets missing for me and I could write several different ways how it could be read and I don't know which one is correct. Maybe there is a rule for that which I don't know?

I would read it like that but because the numbers are that huge and there are several possibilities, it's very hard to know which result is correct:

$$2^{2048} - 2^{1984}-1+2^{64} \cdot (2^{1918 \pi} + 124476)$$

All I know from the link is that the hexadecimal of it is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    $\begingroup$ $2^{2048} - 2^{1984}-1+2^{64} \cdot (2^{1918 \pi} + 124476)$ seems okay. But maybe $[]$ are floor function? $\endgroup$ – Nyssa Jun 10 at 23:17
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It seems that the correct writing of your number is $$c=2^{2048}-2^{1984}-1+2^{64}\cdot\bigl(\,\lfloor2^{1918}\,\pi\rfloor +124476\bigr)\ .$$ This $c$ has $617$ decimal digits. Its conversion to hexadecimal is

enter image description here

which seems to coincide with your hexadecimal input.

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Your hex string, when converted to decimal, contains 617 decimal digits.

$2^{2048} - 2^{1984}-1+2^{64} \cdot (2^{1918\times 3} + 124476)$, when converted to decimal contains 1752 digits.

However, replacing $2^{1918\times 3}$ with $2^{1918}\times 3$ also contains 617 digits, so this interpretation may be the correct one.

Note that in my calculations I have replaced $\pi$ with $3$.

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