2007-11-26 20:51:37 · 11 answers · asked by mayank_jain05 2 in Science & Mathematics ➔ Mathematics
1 , 11 ,12 ,1121, 122111
there are
number 1 with amount 2
number 2 with amount 1
number 1 with amount 1
so next number will be 122111
the next next number is
number 1 with amount 1
number 2 with amount 2
number 1 with amount 3
112213
2007-11-26 21:48:59 · answer #1 · answered by safrodin 3 · 0⤊ 0⤋
Just to throw my hat into ring:
y = 1117n³/3 - 2243n²/2 + 6214n/3 - 1135
Will give those values for n = 1, 2, 3, 4
Also any solution in the form:
y = 1117n³/3 - 2243n²/2 + 6214n/3 - 1135 + A(n-1)(n-2)(n-3)(n-4) will work where A is any number you like! If I pick A = -102 then the fifth number is 2007!
2007-11-26 21:59:38 · answer #2 · answered by Anonymous · 0⤊ 0⤋
122111
2016-07-03 18:14:56 · answer #3 · answered by ashok 1 · 0⤊ 0⤋
Describe what you see with the prior number:
The term after 1121 is obtained by saying "1 twice, 2 once, 1 once", which gives 122111.
Continuing the pattern:
.... 122111, 112213, 12221131, 1123123111, etc.
For example, the term after 122111 is 1(2 times), 2(2 times), 1(3 times) --> 122213.
A good source for integer sequences is the "Online Encyclopedia of Integer Sequences". The link to this specific sequence (A007651) is given below.
2007-11-26 21:11:46 · answer #4 · answered by Puzzling 7 · 2⤊ 1⤋
22
2007-11-26 20:55:18 · answer #5 · answered by ? 3 · 0⤊ 2⤋
1, 11, 12, 1121
1
1 1
1+11=12
11 21
1121+12=1133
2007-11-26 20:57:35 · answer #6 · answered by Raju_89$ 2 · 0⤊ 2⤋
1122
2007-11-26 20:59:40 · answer #7 · answered by dlt 3 · 0⤊ 1⤋
1133
2007-11-26 21:04:48 · answer #8 · answered by Anonymous · 0⤊ 2⤋
Good luck, you alread have many choices, and which one is correct, well I have no clue.
2007-11-26 21:07:54 · answer #9 · answered by Anonymous · 1⤊ 1⤋
111221, 312211, 13112221, 1113213211, ...
OR
1231, 131221, 132231, 232221, 134211, ...
OR
3112, 211213, 312213, 212223, 114213, ...
2007-11-26 20:55:29 · answer #10 · answered by Anonymous · 0⤊ 1⤋