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Q: What are the next two terms in the sequence 1 11 21 1112 3112 211213?

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11 21 1112 3112 111213 411213 31121314 41122314 31221324

No. All members of the sequence are in order and all fit the requirements of being a sequence.

no

It's the look-say sequence starting with 2. initially, starting with 2, you see that there is one 2, hence 12. looking at this, you see there is one 1 and one 2, giving you 1112, and so on.

This is a "Look and Say" sequence; you get the next term by looking at the previous one and saying what digits it contains. 2 contains "one 2", so the second term is 12; that is then "one 1, one 2", yielding 1112, and so on.

71422315

The answer to your sequence is 41122314 and then 31221324. The next number in the sequence is in the form of a1b2c3d4e5..., where "a" is the quantity of 1 in the previous line, "b" is the quantity of 2 in the previous line, etc etc. Another way I have seen this type of sequence is this: 1 11 21 1211 111221 312211 12112221 1112213211 The sequence is made by "reading" the line before it. The first line is One of the number One, so thus the second line is 1 1. The second line is Two Ones, so 2 1. The third line is One Two, One One, so 1211. In turn, this is One One, One Two, Two Ones, so 11 12 21.

3112 x 4 = 12448

There are 256. Some of them are: 1111, 1112, 1113, 1114, 1121, ... 2111, 2112, 2113, ... 3111, 3112, ... 4111, etc.

57.29%

3112

13211321322112 This line describes the last line. Each two digits describes how often a number repeats. One three, two one's, one three, two one's, three two's, three two's, two one's, one two. The problem with the sequence provided is the fourth set where there has been a typo (31112) should be (3112).

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