20151127, 09:17  #1 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3×29×83 Posts 
Relatively simple question I've been unable to resolve
I wrote up Clifford Stern's analysis of aliquot sequences in a much more thorough and pedagogical way that explains every step (basically the page I wish I had years ago when I first read his work).
I've found myself unable to prove a small side equality, which while not terribly important for the aliquot analysis, is still of some minor interest to me. It's no doubt fairly easy to prove, so I'd like to bring it to the attention of others on this forum who are far more sophisticated than I. For now it is available here, though it will be available in more permanent form here when the relevant site is updated with my latest changes. Please be kind with me 
20151127, 09:49  #2  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10,061 Posts 
Quote:
Quote:


20151127, 16:27  #3 
Jun 2003
2×2,719 Posts 
You want to prove that b(1+p^2+p^4+...p^(2(l1))) = 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(l1)) = (1+p^2)*(1+p^4+p^8+..p^(2(l2))). Note that p^(2(l2)) = p^(4(l/21)) (1+p^4+p^8+..p^(4(l/21))) 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(l1))) = b(l). 
20151127, 16:53  #4 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 
Brilliant. I had previously noted that if l was even, then we could break it down and I had wondered if there was some recursion  but this was before I had noted that I could pull out the (1+p) initially (I had done the pairing but not the factoring), so when I had the idea I was looking at (p^(2i)+p^(2i+1))/2^x rather than this recursive (!) pairing and factoring. Wish I had this most important brainwave on my own too  oh well. (You can see the previous hints in that direction at the second link in the OP, again until such time as it's updated to match the first.
Thanks for the help Last fiddled with by Dubslow on 20151127 at 16:55 
20151128, 01:48  #5 
Romulan Interpreter
"name field"
Jun 2011
Thailand
10100000101001_{2} Posts 

20151128, 06:24  #6 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
7221_{10} Posts 

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