[math-fun] A sequence puzzle
Here are the first 20 terms of a sequence: 1,2,6,12,60,3,21,168,504,2520,27720,4,52,364,5460,21840,371280,1113840,21 162960,5 The puzzle is to determine the rule of formation. Franklin T. Adams-Watters
On 7/31/06, franktaw@netscape.net <franktaw@netscape.net> wrote:
Here are the first 20 terms of a sequence:
1,2,6,12,60,3,21,168,504,2520,27720,4,52,364,5460,21840,371280,1113840, 21162960,5
The puzzle is to determine the rule of formation.
LCM with the next natural; if you get the same number twice, replace with the next unseen natural (1,2) = 2 (2,3) = 6 (6,4) = 12 (12,5) = 60 (60,6) = 60, so replace this with the first natural we haven't seen yet. We already have 1,2, so choose 3 (3,7)=21 (21,8)=168 etc.. (27720,12)=27720, so go to 4.
Franklin T. Adams-Watters -- Mike Stay metaweta@gmail.com http://math.ucr.edu/~mike
Franklin T. Adams-Watters asks:
Here are the first 20 terms of a sequence: 1,2,6,12,60,3,21,168,504,2520,27720,4,52,364,5460,21840,371280,1113840,21162960,5
The puzzle is to determine the rule of formation.
An OEIS search for the number 27720 will be enough to give it away, for anyone who's stumped. Well, I think you haven't gone quite far enough to give us the full rule.
From where you left off, I believe the next run-up in your sequence is 5, 105, 2310, 53130, 212520, 1062600, 13813800, 124324200 But is the next thing after that 6 or 7?
--Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
Apparently, I didn't give enough terms to really disambiguate the sequence. Here are 20 more. 105,2310,53130,212520,1062600,13813800,124324200,7,203,6090,188790,302064 0,33227040,564859680,35,1260,46620,885780,11515140,10 Franklin T. Adams-Watters -----Original Message----- From: franktaw@netscape.net Here are the first 20 terms of a sequence: 1,2,6,12,60,3,21,168,504,2520,27720,4,52,364,5460,21840,371280,1113840,21 162960,5 The puzzle is to determine the rule of formation. Franklin T. Adams-Watters
This sequence is now in the OEIS as A119862, if anybody wants to look at the answer. Franklin T. Adams-Watters
participants (3)
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franktaw＠netscape.net -
Michael Kleber -
Mike Stay