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Revision History for A242453 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
Showing entries 1-10 | older changes

(Conjectured) infinite nondecreasing sequence with a(1) = 1 such that the divisors of n appear a total of a(n) times in the sequence.
(history; published version)

#15 by Michel Marcus at Mon Dec 01 00:22:21 EST 2014

STATUS

reviewed

approved

#14 by Wesley Ivan Hurt at Sun Nov 30 21:40:00 EST 2014

STATUS

proposed

reviewed

#13 by Jon E. Schoenfield at Sun Nov 30 21:35:35 EST 2014

STATUS

editing

proposed

#12 by Jon E. Schoenfield at Sun Nov 30 21:35:33 EST 2014

EXAMPLE

Nor can a(2) be 3 or greater, because then a(2) would be greater than the number of terms in the sequence that divided 2 (there would be only 1 such term). Since, by definition, a(2) must equal the number of terms in the sequence that divide 2, this is a contradiction.

AUTHOR

Matthew Vandermast, July Jul 24 2014

STATUS

approved

editing

#11 by Michael Somos at Tue Aug 05 12:39:44 EDT 2014

STATUS

proposed

approved

#10 by Matthew Vandermast at Thu Jul 24 22:27:59 EDT 2014

STATUS

editing

proposed

Discussion

Fri Jul 25

18:25

Franklin T. Adams-Watters: It isn't clear in your definition if you are allowing backtracking. I.e., if it appears that a(n) might be k or k+1, you try it with k, and then setting a(n) = k get an impossibility for a(n+1). Do you then go back and try a(n) = k+1?

18:26

Franklin T. Adams-Watters: If you are not backtracking, take the "conjectured" and "infinite" out of the name, and add a comment that it is conjectured that this sequence is infinite.

Sat Jul 26

18:44

Matthew Vandermast: Reply to Franklin's comments:  The way this sequence is defined (including the "nondecreasing" condition), the issue you describe doesn't really come up. 
1) If this sequence as defined is problem-free, then the number of appearances in the sequence of any value of k must be 2*a(k) - {Sum_(d|k) a(d)}. So there's no future point where we could say of some k: "If k appears j times, as it should, there's a contradiction...let's make it appear j-1 times instead, and put a 'k+1' or some larger number where the last 'k' would have been." 
2) If this sequence *does* eventually have the problem I wrote about in the first comment, the sequence is not only finite; it's invalid past a(2.) (If the sequence is problem-free, every term here is accurate.)
Another comment coming up...

19:03

Matthew Vandermast: 1) If my last comment doesn't completely clarify why the concern in Franklin's first comment isn't an issue, working out the first several terms of this sequence might also help. 
2) If this sequence is valid, as conjectured, it is also definable as "Lexicographically earliest infinite sequence of positive integers such that a(n) of its terms divide n," or "...such that the divisors of n appear a total of a(n) times." 
3) Again, I propose that this sequence should be here for the convenience of the next person who wonders about it, even though it's only conjecturably a valid sequence.  If it's proved invalid, it can be marked "dead" at that time.  I hope the sequence's inclusion will help inspire someone to find the answer either way!

Mon Jul 28

21:24

Franklin T. Adams-Watters: OK. It would still be finite IMO, not "invalid". You can, in that case, make the sequence work up to some point, so it is a fine sequence up to that point. There are plenty of sequences like that. "Dead" is used for duplicate sequences where both have been there for a while, and for incorrect versions of sequences. Neither of those applies here.

Fri Aug 01

10:25

Matthew Vandermast: Sorry for the delay (I've had limited computer access and am busy moving); I'll reply soon.

13:07

Franklin T. Adams-Watters: A minor point: you don't have to make a(1) = 1 part of the definition; you can show that a(1) = 1 from the rest of the definition.

Tue Aug 05

12:39

Michael Somos: Franklin: You have to start somewhere and a(1)=1 does it.

#9 by Matthew Vandermast at Thu Jul 24 22:27:04 EDT 2014

COMMENTS

This Can it be proved or disproved that this sequence is valid? It is invalid if, for some value of n, it is impossible for a(n) to be as great as the total number of appearances that the aliquot divisors of n have already made in the sequence.

A near-example: The combination of 1) the sequence definition and 2) the values of terms a(1)-a(29) eliminates all numbers except 13 as possible values of a(30) except for the number 13. If the aliquot divisors of 30 (1, 2, 3, 5, 6, 10 and 15) appeared in the sequence a total 14 or more times, no value for a(30) 13 would also be eliminated as a possible, value, and this sequence would be invalid. Since 30's aliquot divisors appear in the sequence a total of only 12 times, no contradiction occurs.

Can it be proved or disproved that this is a valid sequence? The first 2500 terms appear to be free of any contradiction that would "annihilate" the sequence.

AUTHOR

Matthew Vandermast, May 14 July 24 2014

#8 by Matthew Vandermast at Thu Jul 24 21:51:57 EDT 2014

COMMENTS

This sequence is invalid if, for some value of n, it is impossible for a(n) to be as great as the total number of appearances that the aliquot divisors of n have already made in the sequence.

A near-example: The combination of 1) the sequence definition and 2) the values of terms a(1)-a(29) eliminates all numbers except 13 as possible values of a(30). If the aliquot divisors of 30 (1, 2, 3, 5, 6, 10 and 15) appeared in the sequence a total 14 or more times, no value for a(30) would be possible, and this sequence would be invalid. Since 30's aliquot divisors appear in the sequence a total of only 12 times, no contradiction occurs.

Can it be proved or disproved that this is a valid sequence? The first 2500 terms appear to be free of any contradiction that would "annihilate" the sequence.

CROSSREFS

Cf. A001462.

Discussion

Thu Jul 24

22:12

Matthew Vandermast: I propose this only-conjectured sequence for acceptance because, in my opinion, the sequence idea is simple enough that it will inevitably occur to other people, and some may also wonder whether the sequence is valid.  (Its validity or invalidity may well be obvious to many readers, but not to me.)   If the sequence proves to be invalid, I propose that it be retained in the database and ther keyword "dead" added, so such people can be saved the trouble of investigating it further - traditionally one of the major purposes of the OEIS. 
Sorry about the delays on this one, and thanks as always to the editors!

#7 by Matthew Vandermast at Thu Jul 17 22:24:38 EDT 2014

EXAMPLE

a(2) cannot be 1 , because then it would be the second term in the sequence to divide 1. (Since a(1) = 1, it is necessary that exactly 1 term in the sequence is a divisor of 1.) Therefore a(2) is the next integer, 2.

Analogously, Nor can a(32) cannot be 2 3 or greater, because it then a(2) would be greater than the number of terms in the third sequence that divided 2 (there would be only 1 such term to divide 2). (Since , by definition, a(2) = 2, it is necessary that exactly 2 must equal the number of terms in the sequence are divisors of that divide 2.) Therefore a(3) , this is the next integer, 3a contradiction.

Since 2 is the only possible value for a(2) that does not create a contradiction, a(2) = 2.

#6 by Matthew Vandermast at Thu Jul 10 23:16:26 EDT 2014

EXAMPLE

a(2) cannot be 1 because it would be the second term to divide 1. (Since a(1) = 1, it is necessary that exactly 1 term in the sequence is a divisor of 1.) Therefore a(2) is the next integer, 2. Analogously, a(3) cannot be 2 because it would be the third term to divide 2. (Since a(2) = 2, it is necessary that exactly 2 terms in the sequence are divisors of 2.) Therefore a(3) is the next integer, 3.

Analogously, a(3) cannot be 2 because it would be the third term to divide 2. (Since a(2) = 2, it is necessary that exactly 2 terms in the sequence are divisors of 2.) Therefore a(3) is the next integer, 3.