Must every sequence of squarefree positive integers satisfying the recurrence f(n) = SquarefreePart(f(n-1)+f(n-2)) eventually enter the cycle 1, 2, 3, 5, 2, 7, 1, 2, 3, 5, 2, 7, ...? E.g., 7, 4, 11, 15, 26, 41, 67, 3, 70, 73, 143, 6, 149, 155, 19, 174, 193, 367, 35, 402, 437, 839, 319, 1158, 1477, 2635, 257, 723, 5, 182, 187, 41, 57, 2, 59, 61, 30, 91, 1, 23, 6, 29, 35, 1, 1, 2, 3, 5, 2, 7, ... (This probably isn't new, but you never know.) Jim
On 5/4/2016 11:37 PM, James Propp wrote:
Must every sequence of squarefree positive integers satisfying the recurrence f(n) = SquarefreePart(f(n-1)+f(n-2)) eventually enter the cycle 1, 2, 3, 5, 2, 7, 1, 2, 3, 5, 2, 7, ...?
E.g., 7, 4, 11, 15, 26, 41, 67, 3, 70, 73, 143, 6, 149, 155, 19, 174, 193, 367, 35, 402, 437, 839, 319, 1158, 1477, 2635, 257, 723, 5, 182, 187, 41, 57, 2, 59, 61, 30, 91, 1, 23, 6, 29, 35, 1, 1, 2, 3, 5, 2, 7, ...
(This probably isn't new, but you never know.)
No, infinitely many other cycles are possible, including n, 2n, 3n, 5n, 2n, 7n (period 6) where n is any squarefree positive integer coprime to 210 = 2*3*5*7. Less trivially, there are also the cycles 1 83 21 26 47 73 30 103 133 59 3 62 65 127 3 130 133 263 11 274 285 559 211 770 109 879 247 1126 1373 51 89 35 31 66 97 163 65 57 122 179 301 30 331 (period 43) and 157 381 538 919 1457 66 1523 1589 778 263 1041 326 1367 1693 85 1778 23 1801 114 1915 2029 986 335 1321 46 1367 (period 26) and 15 2162 2177 4339 181 1130 1311 2441 938 3379 4317 481 4798 5279 10077 3839 71 3910 3981 7891 742 8633 (period 22) and also their multiples by suitable squarefree numbers. -- Fred W. Helenius fredh@ix.netcom.com
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. 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 Question: What is the probability that every (f[1], f[2]) in (Z+)^2 leads to a cycle: f[n] = f[n+p] for some p >= 1 and all sufficiently large n ? —Dan
On May 4, 2016, at 10:39 PM, Fred W. Helenius <fredh@ix.netcom.com> wrote:
On 5/4/2016 11:37 PM, James Propp wrote:
Must every sequence of squarefree positive integers satisfying the recurrence f(n) = SquarefreePart(f(n-1)+f(n-2)) eventually enter the cycle 1, 2, 3, 5, 2, 7, 1, 2, 3, 5, 2, 7, ...?
E.g., 7, 4, 11, 15, 26, 41, 67, 3, 70, 73, 143, 6, 149, 155, 19, 174, 193, 367, 35, 402, 437, 839, 319, 1158, 1477, 2635, 257, 723, 5, 182, 187, 41, 57, 2, 59, 61, 30, 91, 1, 23, 6, 29, 35, 1, 1, 2, 3, 5, 2, 7, ...
(This probably isn't new, but you never know.)
No, infinitely many other cycles are possible, including
n, 2n, 3n, 5n, 2n, 7n (period 6)
where n is any squarefree positive integer coprime to 210 = 2*3*5*7.
Less trivially, there are also the cycles
1 83 21 26 47 73 30 103 133 59 3 62 65 127 3 130 133 263 11 274 285 559 211 770 109 879 247 1126 1373 51 89 35 31 66 97 163 65 57 122 179 301 30 331 (period 43)
and
157 381 538 919 1457 66 1523 1589 778 263 1041 326 1367 1693 85 1778 23 1801 114 1915 2029 986 335 1321 46 1367 (period 26)
and
15 2162 2177 4339 181 1130 1311 2441 938 3379 4317 481 4798 5279 10077 3839 71 3910 3981 7891 742 8633 (period 22)
and also their multiples by suitable squarefree numbers.
-- Fred W. Helenius fredh@ix.netcom.com
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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? Andy
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
Question: What is the probability that every (f[1], f[2]) in (Z+)^2 leads to a cycle:
f[n] = f[n+p]
for some p >= 1 and all sufficiently large n ?
—Dan
On May 4, 2016, at 10:39 PM, Fred W. Helenius <fredh@ix.netcom.com> wrote:
On 5/4/2016 11:37 PM, James Propp wrote:
Must every sequence of squarefree positive integers satisfying the recurrence f(n) = SquarefreePart(f(n-1)+f(n-2)) eventually enter the cycle 1, 2, 3, 5, 2, 7, 1, 2, 3, 5, 2, 7, ...?
E.g., 7, 4, 11, 15, 26, 41, 67, 3, 70, 73, 143, 6, 149, 155, 19, 174, 193, 367, 35, 402, 437, 839, 319, 1158, 1477, 2635, 257, 723, 5, 182, 187, 41, 57, 2, 59, 61, 30, 91, 1, 23, 6, 29, 35, 1, 1, 2, 3, 5, 2, 7, ...
(This probably isn't new, but you never know.)
No, infinitely many other cycles are possible, including
n, 2n, 3n, 5n, 2n, 7n (period 6)
where n is any squarefree positive integer coprime to 210 = 2*3*5*7.
Less trivially, there are also the cycles
1 83 21 26 47 73 30 103 133 59 3 62 65 127 3 130 133 263 11 274 285 559 211 770 109 879 247 1126 1373 51 89 35 31 66 97 163 65 57 122 179 301 30 331 (period 43)
and
157 381 538 919 1457 66 1523 1589 778 263 1041 326 1367 1693 85 1778 23 1801 114 1915 2029 986 335 1321 46 1367 (period 26)
and
15 2162 2177 4339 181 1130 1311 2441 938 3379 4317 481 4798 5279 10077 3839 71 3910 3981 7891 742 8633 (period 22)
and also their multiples by suitable squarefree numbers.
-- Fred W. Helenius fredh@ix.netcom.com
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-- Andy.Latto@pobox.com
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
So Dan's version is https://oeis.org/A165911 as opposed to A272636 Best regards Neil Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com On Thu, May 5, 2016 at 9:45 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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Thanks, Fred! I'll amend my question to: does every such sequence eventually cycle? Despite the strong downward trend one typically sees in experiments with this recurrence, the conjecture could be extremely hard; compare with the infamous 3n+1 problem. Jim On Thursday, May 5, 2016, Fred W. Helenius <fredh@ix.netcom.com <javascript:_e(%7B%7D,'cvml','fredh@ix.netcom.com');>> wrote:
On 5/4/2016 11:37 PM, James Propp wrote:
Must every sequence of squarefree positive integers satisfying the recurrence f(n) = SquarefreePart(f(n-1)+f(n-2)) eventually enter the cycle 1, 2, 3, 5, 2, 7, 1, 2, 3, 5, 2, 7, ...?
E.g., 7, 4, 11, 15, 26, 41, 67, 3, 70, 73, 143, 6, 149, 155, 19, 174, 193, 367, 35, 402, 437, 839, 319, 1158, 1477, 2635, 257, 723, 5, 182, 187, 41, 57, 2, 59, 61, 30, 91, 1, 23, 6, 29, 35, 1, 1, 2, 3, 5, 2, 7, ...
(This probably isn't new, but you never know.)
No, infinitely many other cycles are possible, including
n, 2n, 3n, 5n, 2n, 7n (period 6)
where n is any squarefree positive integer coprime to 210 = 2*3*5*7.
Less trivially, there are also the cycles
1 83 21 26 47 73 30 103 133 59 3 62 65 127 3 130 133 263 11 274 285 559 211 770 109 879 247 1126 1373 51 89 35 31 66 97 163 65 57 122 179 301 30 331 (period 43)
and
157 381 538 919 1457 66 1523 1589 778 263 1041 326 1367 1693 85 1778 23 1801 114 1915 2029 986 335 1321 46 1367 (period 26)
and
15 2162 2177 4339 181 1130 1311 2441 938 3379 4317 481 4798 5279 10077 3839 71 3910 3981 7891 742 8633 (period 22)
and also their multiples by suitable squarefree numbers.
-- Fred W. Helenius fredh@ix.netcom.com
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Compare Conway's Subprime Fibonacci Sequences [article in Dec 2014 Math Mag]. R. On Thu, 5 May 2016, James Propp wrote:
Thanks, Fred!
I'll amend my question to: does every such sequence eventually cycle?
Despite the strong downward trend one typically sees in experiments with this recurrence, the conjecture could be extremely hard; compare with the infamous 3n+1 problem.
Jim
On Thursday, May 5, 2016, Fred W. Helenius <fredh@ix.netcom.com <javascript:_e(%7B%7D,'cvml','fredh@ix.netcom.com');>> wrote:
On 5/4/2016 11:37 PM, James Propp wrote:
Must every sequence of squarefree positive integers satisfying the recurrence f(n) = SquarefreePart(f(n-1)+f(n-2)) eventually enter the cycle 1, 2, 3, 5, 2, 7, 1, 2, 3, 5, 2, 7, ...?
E.g., 7, 4, 11, 15, 26, 41, 67, 3, 70, 73, 143, 6, 149, 155, 19, 174, 193, 367, 35, 402, 437, 839, 319, 1158, 1477, 2635, 257, 723, 5, 182, 187, 41, 57, 2, 59, 61, 30, 91, 1, 23, 6, 29, 35, 1, 1, 2, 3, 5, 2, 7, ...
(This probably isn't new, but you never know.)
No, infinitely many other cycles are possible, including
n, 2n, 3n, 5n, 2n, 7n (period 6)
where n is any squarefree positive integer coprime to 210 = 2*3*5*7.
Less trivially, there are also the cycles
1 83 21 26 47 73 30 103 133 59 3 62 65 127 3 130 133 263 11 274 285 559 211 770 109 879 247 1126 1373 51 89 35 31 66 97 163 65 57 122 179 301 30 331 (period 43)
and
157 381 538 919 1457 66 1523 1589 778 263 1041 326 1367 1693 85 1778 23 1801 114 1915 2029 986 335 1321 46 1367 (period 26)
and
15 2162 2177 4339 181 1130 1311 2441 938 3379 4317 481 4798 5279 10077 3839 71 3910 3981 7891 742 8633 (period 22)
and also their multiples by suitable squarefree numbers.
-- Fred W. Helenius fredh@ix.netcom.com
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In fact, subprime Fibonacci's are what inspired me to look at my recurrence. (It seemed capricious to treat the different prime factors differently, just because some of them are larger!) Jim On Thursday, May 5, 2016, rkg <rkg@ucalgary.ca> wrote:
Compare Conway's Subprime Fibonacci Sequences [article in Dec 2014 Math Mag]. R.
On Thu, 5 May 2016, James Propp wrote:
Thanks, Fred!
I'll amend my question to: does every such sequence eventually cycle?
Despite the strong downward trend one typically sees in experiments with this recurrence, the conjecture could be extremely hard; compare with the infamous 3n+1 problem.
Jim
On Thursday, May 5, 2016, Fred W. Helenius <fredh@ix.netcom.com <javascript:_e(%7B%7D,'cvml','fredh@ix.netcom.com');>> wrote:
On 5/4/2016 11:37 PM, James Propp wrote:
Must every sequence of squarefree positive integers satisfying the
recurrence f(n) = SquarefreePart(f(n-1)+f(n-2)) eventually enter the cycle 1, 2, 3, 5, 2, 7, 1, 2, 3, 5, 2, 7, ...?
E.g., 7, 4, 11, 15, 26, 41, 67, 3, 70, 73, 143, 6, 149, 155, 19, 174, 193, 367, 35, 402, 437, 839, 319, 1158, 1477, 2635, 257, 723, 5, 182, 187, 41, 57, 2, 59, 61, 30, 91, 1, 23, 6, 29, 35, 1, 1, 2, 3, 5, 2, 7, ...
(This probably isn't new, but you never know.)
No, infinitely many other cycles are possible, including
n, 2n, 3n, 5n, 2n, 7n (period 6)
where n is any squarefree positive integer coprime to 210 = 2*3*5*7.
Less trivially, there are also the cycles
1 83 21 26 47 73 30 103 133 59 3 62 65 127 3 130 133 263 11 274 285 559 211 770 109 879 247 1126 1373 51 89 35 31 66 97 163 65 57 122 179 301 30 331 (period 43)
and
157 381 538 919 1457 66 1523 1589 778 263 1041 326 1367 1693 85 1778 23 1801 114 1915 2029 986 335 1321 46 1367 (period 26)
and
15 2162 2177 4339 181 1130 1311 2441 938 3379 4317 481 4798 5279 10077 3839 71 3910 3981 7891 742 8633 (period 22)
and also their multiples by suitable squarefree numbers.
-- Fred W. Helenius fredh@ix.netcom.com
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_______________________________________________
math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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<< 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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participants (6)
-
Andy Latto -
Dan Asimov -
Fred W. Helenius -
James Propp -
Neil Sloane -
rkg