<< It looks like Dan is using the largest squarefree factor ... Yes, I foolishly assumed that is what "squarefree part" meant, instead of looking it up. I.e., I thought it meant the product of all distinct primes dividing the number. As I learned, "squarefree part" really means the number divided by its largest square factor. I.e., replace each exponent in the number's prime factorization by its remainder mod 2. —Dan
On May 5, 2016, at 6:45 AM, Fred W. Helenius <fredh@ix.netcom.com> wrote:
On 5/5/2016 8:28 AM, Andy Latto wrote:
On Thu, May 5, 2016 at 5:46 AM, Dan Asimov <dasimov@earthlink.net> wrote:
I just did a few Mma experiments on the sequence starting with f[1] = 0, f[2] = 1, and it doesn't appear to cycle.
The sequence starting with 0,1 continues
0, 1, 1, 2, 3, 5, 2, 7, 1, 2
and cycles from there. You must have some different sequence rule than the "add the last two elements, and divide by the largest square that divides this number" sequence rule that Jim and Fred are using.
Your rule might also be interesting to study, though; can you explain what it is?
It looks like Dan is using the largest squarefree factor (also called the "radical") of n instead of the squarefree part (called "core(n)" in PARI/GP). This makes a difference at f[9], since rad(2 + 7) = 3, but core(2 + 7) = 1. The sequence using rad looks like this:
0, 1, 1, 2, 3, 5, 2, 7, 3, 10, 13, 23, 6, 29, 35, 2, 37, 39, 38, ...
It struggles to get started (e.g., f[40] = f[58] = 6), but it does eventually get into big numbers.
Well, at least through the first 1000 terms it was still getting net larger, though the sequence is clearly decreases at times. The 1000th term has 39 decimal digits:
705105353768951034109967840494543269698
I get this same value using the radical, but as f[996]. This variant has cycles, too, including
2, 2 (period 1) 3, 3, 6, 3, 3, 6 (period 3) 5, 10, 15, 5, 10, 15 (period 3)
but they seem only to occur when the initial terms have a common factor. This is in contrast to the original question, where cycles seem to always happen.
-- Fred W. Helenius fredh@ix.netcom.com
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