20130812, 04:22  #1 
May 2004
2^{2}·79 Posts 
"one step backward _two steps forward "
This refers to the Ramanujan  Nagell theorem. i.e. there are only 5 solutions to the Diaphontine eqn: x^2 + 7 = 2 ^n.
When x = 1, f(x) = 8 = 2^3 = 2^s, say. Then x = 1 + 2^(s1), when substituted in f(x) yields 32 = 2 ^5. The algo x_i+1 = x_i + 2^(s1) always yields a number such that f(x_i+1), whose power of 2 component is always greater than s. Here s is the power of 2 in f(x_i). 
20130812, 23:26  #2 
May 2013
East. Always East.
3277_{8} Posts 
The text editor on this forum offers super^{script} and sub_{script}, just by the by. Makes things a bit easier to read.

20130813, 06:04  #3 
Romulan Interpreter
"name field"
Jun 2011
Thailand
10,273 Posts 

20130814, 05:40  #4  
May 2004
2^{2}·79 Posts 
Quote:
A.K. Devaraj 

20130814, 10:25  #5 
May 2004
2^{2}×79 Posts 
onestep backward...

20130814, 10:30  #6 
May 2004
2^{2}×79 Posts 

20130814, 12:19  #7 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2A85_{16} Posts 
No, it is built in to the online interface.
You can use it by inserting the formatting markups within the [ TEX ][ /TEX ] (spaces inserted to keep the command from being interpreted) tags. 
20130814, 17:08  #8  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2743_{16} Posts 
Quote:
Code:
~/> gp ? f(x)=x^2+7 ? ilog2(t)={s=0;while(t%2==0,t/=2;s++);return(if(t==1,s,1))} # returns 1 if input is not a power of two ? f(1) %1 = 8 ? s=ilog2(8) %2 = 3 ? f(1+2^(s1)) %3 = 32 ? s=ilog2(32) %4 = 5 ? f(1+2^(s1)) %5 = 296 ? ilog2(296) %6 = 1 # not a power of two; the "algo" is invalid 

20130814, 17:28  #9 
"Gang aft agley"
Sep 2002
2×1,877 Posts 

20130820, 05:09  #10 
May 2004
100111100_{2} Posts 
onestep backward...
I am afraid some members have not understood the algorithm. I will clarify with a few examples relevant to the RN theorem.
The mother function is f(x)=x^2+7 The algo: x_i+1 = x_i + 2^(s1) where s is the power of two in (x_i)^2+7. result: f(x_i+1) is a number in which the power of 2 is always greater than that in f(x_i). f(1) = 2^3. x_i+1 = 1 + 2^(s1) = 5. f(5)= 2 ^5. next x_i+1 = 5 + 2^4 = 21. f(21) = 7*2^6. This goes on & on. A.K. Devaraj p.s I had presented an alternative proof of RN Theorem based on the above algo. at a con of Ramanujan mathematical society a few years ago. 
20130820, 12:57  #11  
May 2004
2^{2}×79 Posts 
One step backward....
Quote:


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