A commas sequence starting with any n > 0 will eventually include an element < 100. I'm sure I could prove this, it would be a very messy and distasteful proof. Once the sequence reaches a value < 100, it must shortly terminate at one of the values 18, 27, 36 or 45, or else enter the loop (6, 73, 104, 59, 150, 149, 54, 7, 85, 32) I doubt there is a simple way to determine the fate of a sequence starting with n. The starting number 1 leads to the terminating sequence 1, 12, 35, 94, 135, 78, 159, 63, 30, 28, 109, 18 ----- Original Message ----- From: "Eric Angelini" <Eric.Angelini@kntv.be> To: "math-fun" <math-fun@mailman.xmission.com>; "Sequence Fans" <seqfan@ext.jussieu.fr> Sent: Thursday, December 06, 2007 12:31 PM Subject: The "commas" sequence revisited
Hello Math-Fun and SeqFans,
I would like to come back to this, if you don't mind (this was one year ago): http://www.groupsrv.com/hobby/about405191.html
Ed Murphy had asked (quote):
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What if d = abs(a(n+1)+a(n)), i.e. the sequence is not necessarily monotonically increasing? Which choices for a(1) result in a block? Which choices result in a loop?
Sample sequence: 1, 12, 35, 94, 135, 78, 159, 251, 239, 148, 62, 91, 74, 115, 166, 105, 156, 88, 4, 48, 129, 221, 209, 118, 199,...
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[Me]: -- Could someone check if the sequence starting with a(1)=1 ever enters a loop (should we always keep the terms as low as possible -- although we would still avoid blocks)?
-- Is there a seq which could be a loop by itself? a-->b-->c-->d-->e-->f-->a
Best, Ã.
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