`log()`

denotes logarithm base 2.

log(2^32 + 1) = 9 + 23 log(2^64 + 1) = 18 + 46 log(2^128 + 1) = 56 + 72 log(2^256 + 1) = 50 + 206 log(2^512 + 1) = 21 + 162 + 328 log(2^1024 + 1) = 25 + 33 + 132 + 834 log(2^2048 + 1) = 18 + 20 + 67 + 72 + 1871

For F_{12} and F_{13}, the final terms are still
composite, 3942 and 7940 bits respectively. The `.006`

fractional part of the composite cofactor of F_{12} means that
when it is written in binary it begins with seven zeros after the most
significant bit: `100000001...`

log(2^4096 + 1) = 17 + 25 + 26 + 37 + 50 + 3941.006 log(2^8192 + 1) = 41 + 61 + 62 + 88 + 7939.7997

F_{14} is still completely unfactored.

log(2^{n}+1) =
log(2^{n}·(1+2^{-n})) =
*n*+log(1+2^{-n}) =
*n*+ln(1+2^{-n})/ln(2) = (approximately, by Taylor
expansion) *n*+2^{-n}/ln(2).
In order to express this small value 2^{-n}/ln(2) in
scientific notation, let *x* = log_{10}(2^{-n}/ln(2)) =
log_{10}(2^{-n})-log_{10}(ln(2)) =
-*n*·log_{10}(2)-log_{10}(ln(2)) =
-*n*·ln(2)/ln(10)-ln(ln(2))/ln(10).
For *n*=2^{14}=16384, this value is *x* = -4931.9162.
Then, 10^{x}/10^{-4932} = 10^{x+4932} = 10^{0.0837} =
e^{(0.0837·ln(10))} = 1.213.

So finally, we have log(2^{16384}+1) = 16384 + 1.213×10^{-4932}
(The calculator `bc`

only has `ln`

and `exp`

functions.)

## No comments :

Post a Comment