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Next Numbers Please #4 (Posted on 2004-11-16) Difficulty: 5 of 5
What are the next two numbers in this sequence:

1, 2, 4, 16, 26, 42, 57, 512, 730, 1010, 1343, 1872, 2367, 2954

See The Solution Submitted by Brian Smith    
Rating: 3.0000 (6 votes)

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Solution Solution (if errors in problem) | Comment 8 of 13 |
I believe I have the solution, but it means if I do there are errors in the sequence as given.

The missing terms are 3616 and 65536

To find term n, first write n as the sum of powers of 2. The term is the sum of those same powers of n.

1 = 2^0, 1^0 = 1
2 = 2^1, 2^1 = 2
3 = 2^1 + 2^0, 3^1 + 3^0 = 4
4 = 2^2, 4^2 = 16
5 = 2^2 + 2^0, 5^2 + 5^2 = 26
6 = 2^2 + 2^1, 6^2 + 6^1 = 42
7 = 2^2 + 2^1 + 2^0, 7^2 + 7^1 + 7^0 = 57
8 = 2^3, 8^3 = 512
9 = 2^3 + 2^0, 9^3 + 9^0 = 730 [PUZZLE HAS 738 WHICH IS 9^3 + 9^1 - ERROR?]
10 = 2^3 + 2^1, 10^3 + 10^1 = 1010
11 = 2^3 + 2^1 + 2^0, 11^3 + 11^1 + 11^2 = 1343 [PUZZLE HAS 1345 WHICH MAY BE A TYPO?]
12 = 2^3 + 2^2, 12^3 + 12^2 = 1872
13 = 2^3 + 2^2 + 2^0, 13^3 + 13^2 + 13^0 = 2367
14 = 2^3 + 2^2 + 2^1, 14^3 + 14^2 + 14^1 = 2954
Solution:
15 = 2^3 + 2^2 + 2^1 + 2^0 = 15^3 + 15^2 + 15^1 + 15^0 = 3616
16 = 2^4, 16^4 = 65536

-Jer
  Posted by Jer on 2004-11-19 20:24:48
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