Relatively simple question I've been unable to resolve

[Dubslow]

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

[Batalov]

Then define the function f as f(n)=^{[FONT=MathJax_Math][I]x[/I]2}[/FONT], and...

...Hmmmm...

[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.

[axn]

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).

[Dubslow]

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

[LaurV]

Originally Posted by Dubslow

I had previously noted that if l was even...

no way! you were odd...

[Dubslow]

Originally Posted by LaurV

no way! you were odd...

And that's why we use things like TeX to make the typesetting clear...


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