20170225, 17:07  #1 
"Rashid Naimi"
Oct 2015
Out of my Body
11100010001_{2} Posts 
Largest Primorial that would Fit on a Terabyte Drive.
Hi all,
What would be a good estimate for the largest prime of the largest primorial that would fit on a terabyte dive? Thank you in advance. 
20170225, 17:24  #2 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
probably would depend on format for example 2*3*5= 11110 in binary which could be brought down to 1(4)0 which isn't any shorter in this case but in larger cases if this happened to be more than 4 it would start to be more efficient to compress it.

20170225, 17:35  #3 
"Rashid Naimi"
Oct 2015
Out of my Body
3^{3}×67 Posts 
Hi SM,
I am looking for a rough estimate. The base of the number should not make much of a difference. Compression will, but only by a few orders of magnitude at its best. So let's say decimal digits stored without compression. I will be happy with a reasonable upper bound for the largest prime factor of the primorial that would fit on a terabyte dive. 
20170225, 17:52  #4 
"Rashid Naimi"
Oct 2015
Out of my Body
3^{3}·67 Posts 
Let's break it down:
How many decimal digits can be written to a terabyte dive without compression? 
20170225, 17:56  #5  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20170225, 18:04  #6  
Bamboozled!
May 2003
Down not across
3^{2}×11×101 Posts 
Quote:
Now write yourself a little program which sums the decimal logarithm of the successive prime numbers and stop when the sum exceeds 1e12. Barring rounding errors, which I will leave you to think about, you then have your answer. Rounding errors will be severe, so you are still going to have to do some thinking to get an accurate answer. I'm prepared to give you a hint but not do all the work for you. Note to real mathematicians: there are much better approaches but those take more thinking ability than a1call has demonstrated so far in public. Last fiddled with by xilman on 20170225 at 18:09 Reason: Add note. 

20170225, 19:13  #7 
Feb 2017
Nowhere
3,251 Posts 
xilman said: If you store one digit per byte you are asking, in effect, for the largest primorial with not more than 1e12 digits. The decimal logarithm of this number is, of course, 1e12
Taking 1e12 as a ballpark figure for the baseten logarithm makes sense. Of course, given that kind of bound, figuring out (at least approximately) the size of the largest prime whose primorial isn't any larger is one thing, and actually determining all the primes up to that point and then multiplying them together, is quite another. How long might that take? My guess is, "WAY too long to worry about writing the answer to a drive." 
20170225, 20:09  #8 
"Rashid Naimi"
Oct 2015
Out of my Body
3421_{8} Posts 
Thank you for all the replies.
I have always disliked the e in scientific and engineering notations. So let's say the primorial can have 1T dd. Now roughly for what size of p would p# have 1T dd? Now for a very rough estimate (that suits me just fine) I get. p^x=10^(T1) p^x=10^T And I don't know how to go further with 2 variables and one equation. Last fiddled with by a1call on 20170225 at 20:10 
20170225, 20:23  #9 
"Rashid Naimi"
Oct 2015
Out of my Body
3421_{8} Posts 
I think the next equation I need is the number of primes less than n.
Isn't that ln(n)? Corrections are appreciated. p^x=10^T p^ln (p)=10^T Anyway this can be calculated without programming? Last fiddled with by a1call on 20170225 at 20:30 
20170225, 20:25  #10  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
my logic was that an n digit number times an m digit number can give at most a n+m digit number ( or roughly like that) so if you knew the number of primes with a certain number of digits ( in PARI/GP primepi(10^x)primepi(10^(x1)) ) you could add x that many times for an upper bound on the number of digits added when you multiply by the product of all the primes in that range. ( my math has up to 10 digits only giving about 4.5 *10^9 so a long way to go potentially.) 

20170225, 20:28  #11  
Aug 2006
2^{2}·5·293 Posts 
Quote:
I don't know how long these multiplications would take; they are huge compared to any benchmarks I've seen. 

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