Find the pattern of the following sequence and determine the next few terms:

2, 10, 12, 38, 42, 52, 56, 142, 150, 170

Note:  My solution requires you to replace 52 with 54 and 142 with 136.

The new sequence is:
2, 10, 12, 38, 42, 54, 56, 136, 150, 170.

Write each number in the sequence in binary code.

n           #        Binary Code

1           2                     10
2         10                  1010
3         12                  1100
4         38               100100
5         42               101010
6         54               110110
7         56               111000
8       136            10001000
9       150            10010110
10     170            10101010

Now, notice that each string of binary digits is of even length.  Divide the strings in half.

n      1st half    2nd half

1             1           0
2           10          10
3           11          00
4         100         100
5         101         010
6         110         110
7         111         000
8       1000        1000
9       1001        0110
10     1010         1010

Aha!  For the nth term, the first half of binary digits equals n when converted back to base 10!

1 = 1 in base ten
10 = 2 in base ten
.
.
.
1001 = 9  in base ten
1010 = 10 in base ten

For the second half of digits, notice that when n is even, the first and second half of binary digits are the same.  However, when n is odd, the first and second half are different.  Specifically, the first and second half are opposite in that 1 becomes 0 and 0 becomes 1. 

So, to find the 11th and 12th terms:

n = 11

11 = 1011 in binary code = first half of binary string.

11 is odd.  0100 = second half of binary string.

11th term in binary code = 1011 0100

10110100 = 180 in base ten.

n = 12

12 = 1100 in binary code

12 is even.

12th term in binary code = 1100 1100

11001100 = 204 in base ten.

So, I think the next two numbers are 180 and 204.

Posted by Reid on 2004-12-07 20:49:20