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.
posted by Wabulon @ 6:40 PM
5 Comments
5 Comments:
Six?
Er, that's not the answer the problem comes with...
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...
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.
In the above comment, p = 8223 approximately.
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