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Sequence puzzle
#11
Posted 03 February 2008 - 03:00 AM
11
21
1211
111221
312211
13112221
1113213211
31131211131221
13211311123113112211
11131221133112122113212221
3113112221232112112221312113211
13213213211121312211221321112111221131221
1113121113123112111311222122111312311231222113112211
31131112311312132112311321321122311311121321121311322113212221
13211331121321131112131221121321131211131221221321133112111312211211132113222113
12113211
11131221232112111312211331121113112221121113122113111231131122112211131221232312
3113112221123113122113322113111221131221
#12
Posted 03 February 2008 - 05:21 PM
421
2
488
504
421
?
#13
Posted 03 February 2008 - 05:49 PM
I thought of a puzzle based on this topic, so I'm posting it here. What is the next number in this sequence?
421
2
488
504
421
?
probably 2 ... it looks circular.
#14
Posted 03 February 2008 - 05:56 PM
No it is not 2. It is not circular, though numbers can repeat obviously.probably 2 ... it looks circular.
#15
Posted 04 February 2008 - 12:19 PM
I thought of a puzzle based on this topic, so I'm posting it here. What is the next number in this sequence?
421
2
488
504
421
?
#16
Posted 06 February 2008 - 04:49 PM
#17
Posted 06 February 2008 - 07:22 PM
No to all three questions. The pattern in the sequence that I posted would be very hard to figure out without looking at any of the clues that I posted. I may post more clues later.Is it 1 3 1 1 2 2 1 1? The same as the other one? I did this without any thinking, just using the clues. Although i believe those were there just to confuse us right?
#18
Posted 07 February 2008 - 12:03 AM
#19
Posted 07 February 2008 - 12:48 AM
The following is a sequence that is based on this topic. What is the next number in this sequence?
421
2
488
504
421
113
421
113
3933
3375
4154
4000
1216
4000
4000
4219
4000
4219
4000
?
Sorry if this is a stupid sequence and the answer ends up making you groan, but I thought of it so I figured I'd give it a try.
#20
Posted 09 February 2008 - 05:55 AM
there is a slight alternative
1 1
2 1
1 2 1 1
3 1 1 2
....
go figure
I have not seen this version before and it intrigued me. (Actually I think there is a minor error in the second line: 1211 should be 1112) I have studied the other version and it goes on indefinitely and the largest digit is a three (3). However, this version, when continued, differs in two significant ways:
1. It is finite (i.e. it starts to repeat)
2. The largest digit is a four (4)
Here's the whole sequence:
1
11
21
1112
3112
211213
312213
212223
114213
31121314
41122314
31221324
21322314
21322314
The generic pattern is: a1b2c3d4, where a, b, c, d are the total number of occurrences of the digits 1, 2, 3, and 4 in the previous line.
Thanks for this new way of looking at an old problem!
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