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29 Sep 2011 12:21:02 IST
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Hey guys.... Its time for teasing your brains a bit and answer the question... Here is a series of numbers. What is the next number in the sequence?
1
11
21
1211
111221
312211
13112221 Lets see what you come up with
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6 Oct 2011 00:23:17 IST
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1.
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6 Oct 2011 03:47:04 IST
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INDUCTION IS IMPOSSIBLE
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6 Oct 2011 18:39:40 IST
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??? How to do ??? Don't see any pattern
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7 Oct 2011 23:01:23 IST
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1 is the answer
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18 Oct 2011 20:07:26 IST
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3.
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18 Oct 2011 22:00:27 IST
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1113213211
The first row - 1 - contains one 1 -> 11
11 contains two 1's -> 21
21 contains one 2, one 1 ->1211
1211 contains one 1, one 2, two 1's ->111221
etcetera.
Building on that theory, it would go
111221 contains three 1's, two 2's, one 1-> 312211
312211 contains one 3,one 1,two 2's two 1's->13112221
13112221 contains one 1,one 3,three 2's, one 1,
therefore the next sequence of numbers would be ->1113213211
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If you think you cannot solve a question,then don't try it because you can't solve it |
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5 Feb 2012 13:49:51 IST
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1 ?
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8 Apr 2012 11:38:40 IST
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1
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8 Apr 2012 11:38:59 IST
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1
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8 Apr 2012 12:39:16 IST
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I think 1.....
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8 Apr 2012 13:15:46 IST
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I will tell you what this is first just write 1 now read (You can see a single(one) one) Hence read one one and write it . Now you get 11 Now again read there are two(ones) hence say two 1 write 21 Now read one two one 1 1211 now read 111221 go onnn continuing....... You wiil get this pLEASE CLICK LIKE
A small history of this sequence
A005150 as a simple table | n | | a(n) | | 1 | | 1 | | 2 | | 11 | | 3 | | 21 | | 4 | | 1211 | | 5 | | 111221 | | 6 | | 312211 | | 7 | | 13112221 | | 8 | | 1113213211 | | 9 | | 31131211131221 | | 10 | | 13211311123113112211 | | 11 | | 11131221133112132113212221 | | 12 | | 3113112221232112111312211312113211 | [1,11,21,1211,111221,312211,13112221,1113213211, 31131211131221,13211311123113112211, 11131221133112132113212221, 3113112221232112111312211312113211] | Look and Say Sequence The integer sequence beginning with a single digit in which the next term is obtained by describing the previous term. Starting with 1, the sequence would be defined by "1, one 1, two 1s, one 2 one 1," etc., and the result is 1, 11, 21, 1211, 111221, .... Similarly, starting the sequence instead with the digit  for  gives  , 1 , 111 , 311 , 13211 , 111312211 , 31131122211 , 1321132132211 , ..., as summarized in the following table.  | Sloane | sequence | | 1 | A005150 | 1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, ... | | 2 | A006751 | 2, 12, 1112, 3112, 132112, 1113122112, 311311222112, ... | | 3 | A006715 | 3, 13, 1113, 3113, 132113, 1113122113, 311311222113, ... |  The number of digits in the  th term of the sequence for  are 1, 2, 2, 4, 6, 6, 8, 10, 14, 20, 26, 34, 46, 62, ... (Sloane's A005341). Similarly, the numbers of digits for the  th term of the sequence for  , 3, ..., are 1, 2, 4, 4, 6, 10, 12, 14, 22, 26, ... (Sloane's A022471). These sequences are asymptotic to  , where  The quantity  is known as Conway's constant (Sloane's A014715), and amazingly is given by the unique positive real root of the polynomial  | (4) | all of whose roots are illustrated above. In fact, the constant is even more general than this, applying to all starting sequences (i.e., even those starting with arbitrary starting digits), with the exception of 22, a result which follows from the cosmological theorem. Conway discovered that strings sometimes factor as a concatenation of two strings whose descendants never interfere with one another. A string with no nontrivial splittings is called an "element," and other strings are called "compounds." It is postulated that every string of 1s, 2s, and 3s that does not contain four of the same number in succession eventually "decays" into a compound of 92 special elements, named after the chemical elements.
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Only two things are infinite, the universe and human stupidity, and I'm not sure about the former. |
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27 Apr 2012 12:53:50 IST
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are everybody okk with my solution
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