20110728, 18:00  #12  
Nov 2008
2·3^{3}·43 Posts 
Quote:
Now for a simple exercise: can you prove it? (Dr Silverman would probably say it was trivial ) 

20110728, 22:24  #13  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


20110729, 00:17  #14  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
237D_{16} Posts 
Quote:
Veuillez aller au "Relation de récurrence" et voilà! 

20110729, 06:17  #15 
May 2011
France
7×23 Posts 
Sum
For information it's the ssame with recurence
Thanks to search; John (It's the morning in France : I attack 966 from 0 to 122 :last 64 bits (19 digits)) 
20110729, 13:55  #16  
"Frank <^>"
Dec 2004
CDP Janesville
2×1,061 Posts 
Quote:
(You do realize that all the sequences with a starting number <1.000.000 have been worked to >100 digits, right?) 

20110729, 18:44  #17 
May 2011
France
7×23 Posts 
Aliquot
In fact I write personal tools: list of primes, factoring, primarily test
I don't want a generic code l my computer is a 64 bits so less 10^20 yous use the ROM after I write the add, div, mod,sort,sqrt... I extract a sqr of a number of 1 000 000 digits Just for the fun A modulo for 1500 digits is not useful but funny if it needs 2 or 3 seconds In fract Iuse 966 to verify py code. I have a little pb with the itrtations 70.... To be quiet I try value> 1.000.000 Is there better values to begin an Aliquot sequence? It's now evening My Aliquot search is write. Tomorrow I attack more 20 digits John 
20110730, 13:05  #18  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Code:
trying(x,y)=b=1;a=[x];until(b==y  isprime(x)sigma(x)x==x,if(isprime(x),break(),x=sigma(x)x;a=concat(a,x);b=b+1));a 

20110730, 14:31  #19  
"Frank <^>"
Dec 2004
CDP Janesville
2×1,061 Posts 
Quote:
As far as start values over 1.000.000, there is no organized project there, but we can suggest many values under that that would be owrth pursuing. 

20110730, 15:13  #20  
Nov 2008
2·3^{3}·43 Posts 
Quote:


20110730, 17:53  #21  
May 2011
France
7×23 Posts 
Quote:
I can see that I have a problem with iteration 65: i have the good sum but I find a bad factoring; I try 3630.....perdect,... or.... John 

20110731, 05:18  #22  
May 2011
France
7×23 Posts 
Quote:
and if you have a faster I'm interested!!!!! where :Dairo's factorization applet, John Last fiddled with by JohnFullspeed on 20110731 at 05:25 

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