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 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

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