20180302, 13:57  #1 
Dec 2011
After milion nines:)
2^{3}×5×37 Posts 
Generating from 16 to 35 digit number in row (+1)
Hi!
What program I can use to generate such "big "numbers ? Excel round numbers if has above 16 digits, so it is not solution. And I dont need primes, I need all numbers. Thanks for any reply example 800000000000000000 800000000000000001 800000000000000002 800000000000000003 800000000000000004 etc etc Last fiddled with by pepi37 on 20180302 at 13:57 
20180302, 14:23  #2  
"Forget I exist"
Jul 2009
Dumbassville
8384_{10} Posts 
Quote:


20180302, 14:41  #3 
Dec 2011
After milion nines:)
2^{3}×5×37 Posts 

20180302, 14:50  #4 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
1110000110101_{2} Posts 
On the face of it, your question seems slightly insane. There isn't enough space on your hard drive to store 10^35 numbers.

20180302, 15:01  #5 
Dec 2011
After milion nines:)
2^{3}·5·37 Posts 

20180302, 15:06  #6 
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
[1*10^15..1*10^15+2499] would have potential to work.
Last fiddled with by science_man_88 on 20180302 at 15:20 Reason: Apparently e version doesn't work 
20180302, 15:14  #7 
Mar 2006
Germany
37×79 Posts 
Using Excel:
 column A (textformatted) with "80000000000" (or your length/number you need)  column B (special format with leading zeros") with "0001" etc.  save as "prn" 
20180302, 15:30  #8 
"Jacob"
Sep 2006
Brussels, Belgium
3316_{8} Posts 
You can use Excel, but treat the numbers as strings,: concatenate the string "80000000000000" with the numbers from 0 to 2499 modified by a mask.
Put the numbers from 0 to 2499 in the A column and in the B1 cell ="80000000000000"&TEXT(A1;"0000"), then copy the formula down. (The 4 0's in the "0000" format string ensures you have the initial 0's and that all numbers have 4 digits.) Then it is an easy task to copy the result in a text file... (Unfortunately the name of function changes according to the language of the interface so if your Excel is in another language you will have to change the name of the function, f.i. in Spanish it is TEXTO() and in Italian TESTO().) I just tested and it seems you can directly use the TEXT() function with the "800000000000000000" mask as in column C of the attached Excel spreadsheet. Jacob Last fiddled with by S485122 on 20180302 at 15:31 
20180302, 15:41  #9 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9562_{10} Posts 
Code:
ABC2 800000000000000000+$a a: from 0 to 25 
20180302, 16:14  #10 
Aug 2006
3·1,993 Posts 
It's not clear what you mean. I'll give some examples in PARI/GP and you can pick whichever you think seem helpful.
Loop from a 16digit number to a 35digit number, doing calculations along the way: Code:
for(n=10^15, 10^351, if(Mod(2, n^2)^(n1)==1, print(n))) \\ this won't finish before the sun swallows the earth Code:
v=[8*10^17..8*10^17+2499]; Code:
v=vector(2500,i, [10^15,10^351]); Code:
s=0; forstep(n=10^15,10^35, 40016006402561024409363745498199, s+=sqrt(n)); s 
20180302, 16:50  #11 
Dec 2011
After milion nines:)
2^{3}×5×37 Posts 
Thanks to all, Excel mask works perfectly!

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