Fermat number factorization expressed as bits

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 F₁₂ and F₁₃, the final terms are still composite, 3942 and 7940 bits respectively. The .006 fractional part of the composite cofactor of F₁₂ 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

F₁₄ is still completely unfactored.

log(2ⁿ+1) = log(2ⁿ·(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 = log₁₀(2^-n/ln(2)) = log₁₀(2^-n)-log₁₀(ln(2)) = -n·log₁₀(2)-log₁₀(ln(2)) = -n·ln(2)/ln(10)-ln(ln(2))/ln(10). For n=2¹⁴=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(2¹⁶³⁸⁴+1) = 16384 + 1.213×10^-4932 (The calculator bc only has ln and exp functions.)