This week's was solved really quickly, so I'm posting another. The following are the first twenty terms of a sequence: 1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144, 120, 65536, 180, 262144, 240, . . . Same idea as the other sequences. Describe the pattern of the sequence. If you use a search engine, please don't submit it in the comments. If it hasn't been solved by Thursday, I'll give a hint. Also, if you ask a polar question (i.e., a yes/no question), I will probably answer it if it relates to this sequence. I'll post the solution on next week's sequence. This afternoon's sequence (1, 4, 8, 9, 16, 25, 27, 32, 36, 49, . . .) was the sequence of Powerful Numbers. That is to say, the list of positive numbers where if the number has a prime divisor, the square of that prime divides it. E.g., 2 divides 8, so, since 8 is in the list it follows that 22=4 divides 8, which is obviously true. T solved it. Update: I've added another ten terms to the sequence so that there are now twenty. Update: Hint: There's a relationship between n and the divisors of a(n), but probably not the category of relationship that first comes to mind. Since I put the hint up late, I'm not going to put the solution on next week's sequence (that of April 20th) until Wednesday, April 22nd.
Tuesday, April 14, 2009
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3 comments:
This solution works, but it's not elegant enough for me to think its right.. anyways, here it is:
each number (so far) is a multiple of 2 raised to a power, and 3 raised to a power.
3 is raised to the power of (the smallest prime factor of the placeholder - 1).
2 is raised to the power of (the placeholder divided by it's smallest prime factor - 1). Prime placeholders use 1 as their smallest prime factor
So, for the 14th number; 14's smallest prime factor is 2.
Resulting in [2^(14/2-1)] x [3^(2-1)] = 2^(6) x 3^(1) = 192.
If this is the case, the rest of the sequence is:
...1024, 96, 4096, 192, 144, 384, 65536, 768, 262144, 1536...
The sequence fits, but my solution seems rather messy and random.
T
Nope, not elegant enough. But it does fit. For that reason, I'll put another ten terms in the sequence on the post. Try again. You're close, by the way.
Actually, upon closer look at your solution, I realize that it doesn't even fit for the ten terms I listed. For example, by your formula
a(3)=[2^(3/1-1)][3^(3-1)]=(2^2)(3^2)=36,
where I listed a(3)=4.
Keep trying.
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