## Sequence Problem

This one comes from puzzle/problem master Lee Bradley.

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

This entry was posted on Monday, March 5th, 2012 at 8:00 am and is filed under Algebra, problem.
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1974

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

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

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.

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

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.

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

Also known as: 3ⁿ − 3n.

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

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.

Quick update in case anybody’s watching.

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