Sequence Problem

This one comes from puzzle/problem master Lee Bradley.

0, 3, 18, 69, 228, 711, next?

1. 1974

2. This is incorrect but I’d be interested in what makes you think it is correct.

3. It can’t be 1974. 1974 isn’t divisible by 3…

4. I think I got it. Is the next number 2166?

I got this answer using a recursive function.

Xn = 3[ Xn-1 + 2n - 1]

Let X0 = 0, Then
X1 = 3[0 + 1] = 3
X2 = 3[3 + 3] = 18
X3 = 3[18 + 5] = 69
and so on…

Essentially take the previous number, add the next consecutive odd number and multiply by 3.

5. Hmmmm … so being divisible by 3 is a common feature. Getting warmer.

6. Nice, Nick. You are correct. (I posted my “getting warmer” before seeing your last comment). I have one more challenge for you (and anyone else):

Express the nth term as a function of n. That is, give a closed form (non-recursive) formula.

7. Thanks Lee, another puzzle. Just as I was about to get back to work…

8. I. J. Kennedy

Also known as: 3ⁿ − 3n.

9. Well done everybody! I have been tempted to tack on yet another challenge but will spare you and point you to

http://primepuzzle.com/tc/sequence.html

This talks about a general method for “integrating” difference equations of this type and also gets a bit carried away discussing how you might go about determining things like the 999th and 99999th term in this sequence.

Thanks for participating!

Lee

10. Great problem, Lee! I’m sorry I didn’t have a chance to jump in there. These youngsters are too smart and fast. I look forward to reading up on things at PrimePuzzle.

11. Quick update in case anybody’s watching.

http://primepuzzle.com/tunxis/large-numbers.html