### What's the next term?

Here's one of those sequence questions that testers love so much:

What is the next term of this sequence?

0, 1, 2, 8.252 x 10^2466, ...

Here, "10^2466" means "ten to the 2466th power." The corresponding term is necessarily slightly approximate, since it contains 2467 decimal digits.

1. Anonymous9:46 PM

Six?

2. Er, that's not the answer the problem comes with...

3. Anonymous3:22 PM

Any chance that even the decimals shown for the huge number are wrong?

I have 0, 1, 2, 8.72598x10^2466, 1.02019x10^3910 ...

This is 1+n^8195, n=-1,0,1,2,3...

4. Congratulations, Andy! You have in effect solved it--at least, you have found a valid solution.

Indexing from zero:

t(n) = 1 + (n-1)^p

where p = (2466+log 8.252)/log 2.

However, p is not an integer, which gives a problem with (-1)^p.

The best kludge I can come up with:

t(n) = 1 + (n-1) |(n-1)^(p-1)|.

This is required, because "8.252" is right. But your solution is in principle entirely valid. Again, congratulations. The next term is then 1 + 3^p. This is a big jump, but nothing compared with the previous jump, relatively speaking. In the limit, the ratio of successive terms is one. The order of growth is polynomial.

I like my answer better, because it is much simpler, and the jumps get hugely huger every time.

5. In the above comment, p = 8223 approximately.

6. 7. Anonymous8:28 PM

The page is so wonderful that I want to write something about myself.
Under the help of my friend, I have got a lot of World of Kung fu Gold and the WoKf gold is sold very cheap on a famous website. If you also want to buy World of Kung fu Gold and own plenty of cheap World of Kung fu Gold, you can ask me for help to get the World of Kung fu money.