Dave Mullen wrote:
>Sorry, I'm no mathematician, and new to the Mersenne field.
>
>> No, in the x-y bit range (remember that n bit integers are
>> about >2^n) the first factor could be x/2 to y/2 bits long
>> (powers of a power >multiply).
>
>What I was trying to say in my disjointed way was ...
>
>(Example) M11 = 2047 (11 bits long). Now 2047 has only 2 factors
>(23 >x 89) and the square root of 2047 is approx 45. 45 is 6 bits
>long, therefore the factor lower than the square root must have
><= 6 bits, >and the factor higher than the square root must have
>>=6 bits.
>
>23. is 5 bits long, and 89 is 7 bits long.
>
>Thus for the exponent 11650000 bits long, if it only has 2 factors , >then the first factor must be between 2 and 3413 bits long, and
>the second factor must be between 3413 and 11649999 bits long.
>Note that the bit lengths of the 2 factors added together must
>equal the bit length of the Prime (or bit length of the
>Prime + 1) !!
There only a slight error with your logic... For the exponent
11650000, the root is *not* 3413 bits long, but more like 5825000
bits long. Perhaps you forgot exponents add, not multiply.
For simplication:
2^3 * 2^3 = 2^6
8 * 8 = 64
Therefore:
2^11650000 = 2^5825000 * 2^5825000
Eric Hahn
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