A collatz-inequality related to the waring-problem; a furtherquestion
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From: smallfrey on 8 Jul 2010 06:10 Halbeisen and Hungerbuhler's 1997 article "Optimal bounds for the length of rational Collatz cycles" is very useful; the parity vector of a loop can be computed from its length and the number of odd elements in it. In least-residue trees, certain limbs have the same parity vector except for the first and last elements. A lot can be learned from the study of these "near" cycles. Details are at "http://home.graysoncable.com/dkcox".
From: smallfrey on 10 Jul 2010 10:50 Hmmm. If I've got it right, you're trying to determine if 2^N is greater than 3^N - [3^N/2^N]*(2^N-1). I don't see any easy way to prove this.
From: smallfrey on 11 Jul 2010 06:34
It's easy to show that the integer portion makes a relatively small contribution to the sum since for a sufficiently large N, 2^(N-1) (or 2^(N-2), 2^(N-3), 2^(N-4), ...) is greater than [3^N/2^N]. The remainder appears to be just a "random" odd number between 1 and 2^N-1; there doesn't appear to be anything "special" about 3^N. I would say the proposition is false since the integer portion should tip the sum over 2^N occasionally. Let y=<3^(N-1)> where the "<>" denotes the remainder function and the modulus is understood to be 2^N. Then y is one of 1, 3, 5, ..., 2^(N-1). Then <3y> (illustrated for N=5) is one of 3, 9, 15, 21, 27, 1, 7, 13, 19, 25, 31, 5, 11, 17, 23, 29 (where large remainders are approached in a somewhat regular fashion). <3y>=<3^N>, so if there was something "special" about <3^N> there would have to be something "special" about <3^(N-1)>, <3^(N-2)>, <3<(N-3)>, .... I could be all wet though. |