SOLUTION: Find the first 40 terms of the sequence defined by A(n+1) = {(1/2)An if An is an odd number and 3An + 1 if An is an even number} and A1 = 67, A2= 134.

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Question 969587: Find the first 40 terms of the sequence defined by A(n+1) = {(1/2)An if An is an odd number and 3An + 1 if An is an even number} and A1 = 67, A2= 134.
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
A1 = 67
A2 = 134 
134 is even, so A3 = ½·A2 = ½·(134) = 67
67 is odd, so A4 = 3A3+1 = 3·67+1 = 202
202 is even, so A5 = ½·A4 = ½·(202) = 101
101 is odd, so A6 = 3A5+1 = 3·101+1 = 304
304 is even, so A7 = ½·A6 = ½·(304) = 152
152 is even, so A8 = ½·A7 = ½·(152) = 76
76 is even, so A9 = ½·A8 = ½·(76) = 38
38 is even, so A10 = ½·A9 = ½·(38) = 19
19 is odd, so A11 = 3A10+1 = 3·19+1 = 58
58 is even, so A12 = ½·A11 = ½·(58) = 29
29 is odd, so A13 = 3A12+1 = 3·29+1 = 88
88 is even, so A14 = ½·A13 = ½·(88) = 44
44 is even, so A15 = ½·A14 = ½·(44) = 22
22 is even, so A16 = ½·A15 = ½·(22) = 11
11 is odd, so A17 = 3A16+1 = 3·11+1 = 34
34 is even, so A18 = ½·A17 = ½·(34) = 17
17 is odd, so A19 = 3A18+1 = 3·17+1 = 52
52 is even, so A20 = ½·A19 = ½·(52) = 26
26 is even, so A21 = ½·A20 = ½·(26) = 13
13 is odd, so A22 = 3A21+1 = 3·13+1 = 40
40 is even, so A23 = ½·A22 = ½·(40) = 20
20 is even, so A24 = ½·A23 = ½·(20) = 10
10 is even, so A25 = ½·A24 = ½·(10) = 5
5 is odd, so A26 = 3A25+1 = 3·5+1 = 16
16 is even, so A27 = ½·A26 = ½·(16) = 8
8 is even, so A28 = ½·A27 = ½·(8) = 4
4 is even, so A29 = ½·A28 = ½·(4) = 2
2 is even, so A30 = ½·A29 = ½·(2) = 1
1 is odd, so A31 = 3A30+1 = 3·1+1 = 4
4 is even, so A32 = ½·A31 = ½·(4) = 2
2 is even, so A33 = ½·A32 = ½·(2) = 1
1 is odd, so A34 = 3A33+1 = 3·1+1 = 4
4 is even, so A35 = ½·A34 = ½·(4) = 2
2 is even, so A36 = ½·A35 = ½·(2) = 1
1 is odd, so A37 = 3A36+1 = 3·1+1 = 4
4 is even, so A38 = ½·A37 = ½·(4) = 2
2 is even, so A39 = ½·A38 = ½·(2) = 1
1 is odd, so A40 = 3A39+1 = 3·1+1 = 4

Edwin
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