what are the next 2 terms of this series ... 1,11,21,1211,111221,.. ?
403722
1,11,21,1211,111221....\n\n312211,13112221,1113213211\n\n\nBTW i love this riddle, i use it on for my chess team for brain warm-ups
148653
[1, 11, 12, 1121, 122111, 112213, 12221131, 1123123111, \n 12 213 111 213 113,11 221 131 132 111 311 231, \n 12 221 231 123 121 133 112 213 111, \n 1 123 112 131 122 131 112 112 321 222 113 113]
972930
answer is 11,13,17,19,23,29
168953
121    since(1+4)+(1+4*2)+(1+4*4)
615524
May10\nJune 28/29 \nJuly NONE\nAugust8\nSeptember20\nOctober 24\nNovember NONE \nDecember 12
40461
well if you want to know what n1 is, i would say the best guess is 10 beacuse it is teh first term in the sequence, if you mean n subscript x then i would say you are looking for a general formula for terms
293052
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97\n\n101 is the next prime number ... \nso 24 prime numbers between 1 - 100
594189
34,\n\nYou are having the series as 2 times <prime numbers>\n2*2, 2*3, 2*5, 2*7, 2*11, 2*13, 2*17, 2*19, 2*23...\n = \n4, 6, 10, 14, 22, 26, 34, 38, 46 ...
869678
1 and 1\n9 and 3\n25 and 5\n49 and 7\n81 and 9\n121 and 11\n...
520658
#24,#99,#16,or #20\nWho knows?
140903
6250\n\nX    2    10    50    250   1250    6250\n\nEach term is 5x the previous.
653946
1 4\n2 15\n4 2\n5 100\n6 3\n7 15\n8 10\n9 10\n10 33\n11 -45\n12 2\n13 500\n14 5\n15 200
939729
Just "pair up" the terms.\n\n1 + (5-3) + (9-7) + (13-11) + ... + (2001-1999)\n\n1 + 2 + 2 + 2 + .... + 2 = 1 + 2 * 500 = 1001
417627
11 - No\n12 - yes\n13 - No
917530
21st\n\nIt starts off like this....\n0-99 is the 1st century..\n100-199 is the 2nd century...\n200-299..................3rd\n300-399..................4th\n400-499..................5th\netc etc etc....\n2000-2099.............21st century.\n2100-2199.............22nd\n\n\nhope that helps
1080683
1113213211\n\nAfter starting the sequence with 1, each term in the sequence consists of groups of two numbers based on the previous term - the first being the quantity and the second specifying which digit.
613494
Let n be the 1001th term of this series.\nNow, 7-3=11-7=15-11=4\n\nSo 4 is the common difference(d)\na is the first term=3\nNow the 1001th term is = a+(n-1)d\n                       = 3+(1001-1)4\n                       = 3+4000=4003\n4003 is the 1001th term