 mersenneforum.org Relatively simple question I've been unable to resolve
 Register FAQ Search Today's Posts Mark Forums Read 2015-11-27, 09:17 #1 Dubslow Basketry That Evening!   "Bunslow the Bold" Jun 2011 40 2015-11-27, 09:49   #2
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

9,901 Posts Quote:
 Then define the function f as f(n)=[FONT=MathJax_Math][I]x[/I]2[/FONT], and...
...Hmmmm...
Quote:
 [During a job interview] Dr. Lester: Which of these two letters comes first, this one or this one? Craig Schwartz: The symbol on the left is not a letter, sir? Dr. Lester: Damn, you're good. I was trying to trick you.   2015-11-27, 16:27 #3 axn   Jun 2003 5,387 Posts You want to prove that b(1+p^2+p^4+...p^(2(l-1))) = b(l) You can proceed as follows: If l is even, then pair up the terms and pull out (1+p^2) 1+p^2+p^4+...p^(2(l-1)) = (1+p^2)*(1+p^4+p^8+..p^(2(l-2))). Note that p^(2(l-2)) = p^(4(l/2-1)) (1+p^4+p^8+..p^(4(l/2-1))) has l/2 terms. We can proceed this way until we run into an expression with odd number of terms. The numbers of terms of the form (1+p^2^i) we can pull out depends on the power of two present in l, i.e. b(l). Next we note that b(1+p^2^i) = 1 (for i>=1). Proof: odd perfect squares are 1 (mod 8). So 1+p^2^i = 2 (mod 8). Hence b()=1 Final factor with odd number of terms will be odd, hence b()=0. Thus b(1+.p^2+...p^(2(l-1))) = b(l).   2015-11-27, 16:53 #4 Dubslow Basketry That Evening!   "Bunslow the Bold" Jun 2011 40 2015-11-28, 01:48   #5
LaurV
Romulan Interpreter

"name field"
Jun 2011
Thailand

17×19×31 Posts Quote:
 Originally Posted by Dubslow I had previously noted that if l was even...
no way! you were odd...    2015-11-28, 06:24   #6
Dubslow

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

3×29×83 Posts Quote:
 Originally Posted by LaurV no way! you were odd... And that's why we use things like TeX to make the typesetting clear... IlIlIlIlIlIl  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Unregistered Information & Answers 3 2012-11-26 02:55 nuggetprime No Prime Left Behind 16 2008-12-23 07:46 Bundu Math 11 2007-09-24 00:02 akruppa Puzzles 28 2006-02-04 03:40 Faraday Linux 13 2005-06-01 02:22

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