I thought this changes the way series of numbers are looked at
1
1 1
2 1
1 2 1 1
Write down the next three lines
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3$\begingroup$ One of my favorites! $\endgroup$– Pierre BegonCommented Feb 6, 2015 at 12:16
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7$\begingroup$ you can't say there is a unique answer... as the answer given drovani is also valid... $\endgroup$– Omid GhayourCommented Jun 20, 2016 at 19:32
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1$\begingroup$ Go here at $5$:$33$, youtube.com/watch?v=r5P-f5arPXE $\endgroup$– Mr PieCommented May 4, 2018 at 7:08
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$\begingroup$ @user477343 you should leave this link in youtube really :) $\endgroup$– skvCommented May 4, 2018 at 11:54
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$\begingroup$ Also: oeis.org/A005150 $\endgroup$– MattAllegroCommented May 31, 2021 at 22:33
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4 Answers
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5
The answer is this:
111221
312211
13112221
The first number specifies the quantity of digits of the set above it and the second number specifies what the digit is. The second line is 11 it is saying that the line above it is one one. The third line states that the line above it is two ones. The fourth line is saying there is one two and one one.
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29$\begingroup$ Also known as the "look and say sequence." $\endgroup$– DoorknobCommented Sep 17, 2014 at 13:51
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1$\begingroup$ There you go :) thats the fun (and embarrassment) of interacting with intellectuals like you @Doorknob $\endgroup$– skvCommented Sep 17, 2014 at 14:03
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3$\begingroup$ Bonus points: Will a four ever occur, and how would you prove that? $\endgroup$ Commented Sep 17, 2014 at 23:23
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$\begingroup$ @MooingDuck there's a bit more info on this at A005150 in The On-Line Encyclopedia of Integer Sequences. $\endgroup$– user2322Commented Sep 18, 2014 at 1:15
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5$\begingroup$ You can never have a four, as this would imply that the previous line contained four of the same digit in a row, which in turn implies you've described the line before that incorrectly (1111 = "one 1 then one 1", which would actually be described as "21" or "two 1s" $\endgroup$– IanF1Commented Jan 3, 2015 at 19:19
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Another valid answer:
1231
131221
132231
232221
Using similar rules as the accept answer. The first number specified the quantity of digits of the entire set above it and the second number specifies what the digit is. The accepted answer is a reading of the prior sequence, this answer is a summary of the prior sequence.
Fun fact, coming back to this. This sequence will hit a point where it will output the same number forever.
1
11
21
1211
1231
131221
132231
232221
134211
14131231
14231241
24132231
14233221
14233221
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2$\begingroup$ Why is the next number 1231 and not 3112? I would think you order the pairs in the order of first appearance. This appears to be in descending order of the digit being counted. $\endgroup$ Commented Sep 17, 2014 at 18:27
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4$\begingroup$ It could be either way, really. The OP was a little vague, so there will be multiple answers. $\endgroup$– drovaniCommented Sep 17, 2014 at 19:03
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3$\begingroup$ @MichaelT the "look-and-summarize" sequence, we could christen it $\endgroup$– smciCommented Feb 1, 2015 at 20:47
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$\begingroup$ You start counting how many of the same digit there is, starting from the largest to the smallest. $\endgroup$– Mr PieCommented May 4, 2018 at 7:09
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The answer can also be this:
1231
211213
223113
222321
421311
14123113
The first number specifies the quantity of digits of the set above it and the second number specifies what the digit is. The second line is 11 it is saying that the line above it is one one. The third line states that the line above it is two ones. The fourth line is saying there is one two and one one.
Thanks user2314
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So reading the first sentence of the question made me see the solution differently.
"I thought this changes the way series of numbers are looked at"
1
1 1
2 1 1
1 2 1 1
Depending on how you define "lines", you can add a "1" to the third row to make the triangle full/symmetrical. The number "1" consists of three "lines" to draw it (depending on the font used).
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$\begingroup$ Sorry, but there's a 'pattern' tag, not a 'lateral-thinking'. Nice try tho $\endgroup$ Commented Jul 13, 2017 at 7:32
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$\begingroup$ Ahh ok didn't notice the tags haha :) $\endgroup$– OJ7Commented Jul 13, 2017 at 16:29