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drummerslowrise

Message boards :
Sophie Germain Prime Search :
SGS primes decimal length
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This may be a silly question, but I was wondering why so many (if not all) SGS found primes have exactly 200700 digits. Could there be as many with, say, 200701 or 200699?
How come so many different exponents (and at least four different powers) generate results with the same length in decimal representation?
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676754^262144+1 is prime  

HonzaVolunteer moderator Volunteer tester Project scientist Send message
Joined: 15 Aug 05 Posts: 1844 ID: 352 Credit: 2,624,974,039 RAC: 931,024

2^30 = 1 073 741 824
2^31 = 2 147 483 648
2^32 = 4 294 967 296
2^33 = 8 589 934 592
All with exactly the same number of digits.
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My stats
Badge score: 1*1 + 5*1 + 7*1 + 8*8 + 9*7 + 11*1 + 12*3 = 187  


Ok, then here is a question that TheDawgz have been trying to find an answer for for a very long time.
How does one predict the decimal length for a given prime or potential prime?
Or for that matter the number of digits in the result of any calculation in the form of: A*B^C?
For example: Honza's prime from the PRPNet GFN32768 subproject 1*4108672^32768+1 which TheDawgz know to be 216,718 digits long (we looked it up).
Or in the case of one of TheDawgz primes: 789*2^1114779+1 which TheDawgz know to be 335,585 digits long (we were told).
TheDawgz are fairly sure that the answer is obvious, but our furry little brains just can't see it.
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There's someone in our head but it's not us.  


You simply take the logarithm with base 10 of the number n.
log[10](n) = log(n) / log(10)
(log is the natural logarithm)
 


Thanks Honza, you've answered my last question. Your example shows that a large number of digits should allow a proportionally large number of results.
The answers to the other questions are also clearer now: the power is much more important to determine the length of a given number than the exponent. If the power is the same or very close to each other as they are on SGS the result will be very close. I think this could help answering the TheDawgz question, or at least an approach to it: if you now the length of another number with the same power, your result should not be very different. If you compare two different a*b^c the one with the largest c will produce a result with the highest number of digits.
I now this must be trivial to those who deal with numbers and number theory. I'm not one of them. Just crunching for fun and thinking out loud right now...
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676754^262144+1 is prime  


Another hint if you want to do a quick calculation (without calculator) for getting an approximation (of base 2 numbers):
2^10 ~ 10^3 (1024 vs 1000)
Now you can convert e.g.
2^200000 into 10^60000 (divide by 10, multiply by 3)
If you have base 10, the exponent gives you the number of zeros (or, the number of digits).
If you have something like 357 * xxxx then it has 23 digits more (as many digits as the number  here 357  has). The 2/3 depends on whether it just 'flips' to the next digit or not. (I might be inaccurate here  anyway, it's only about few digits so doesn't really count on the large scale).
Coming now to the Sophies, and give an example here (which is currently on the homepage):
26059520697525*2^6666661
2^666666 ~ 10^200000
+ 14 digits from the multiplier
So you see, I'm a bit off here, but it gives you the right scale quickly.
Greetings!
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SGS primes decimal length 