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 Strange sequence (Posted on 2005-11-15)
If f(0)=0 and for n>0, f(n)=n-f(f(n-1)),
what is f(1,000,000,000,000)?

Computer solutions welcome,
though a proof would be better!

 See The Solution Submitted by e.g. Rating: 3.1667 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: solution - no computer | Comment 4 of 9 |
(In reply to solution - no computer by pcbouhid)

You derive the correct formula:

"Examining the values of k*n for integer values of n, we realize that f(n) = [k*(n+1)], being [x] the smallest integer less than or equal to x."

But then add 1 after the multiplication, rather than adding 1 to n before the multiplication:

"Thus f(1,000,000,000,000) = [618,033,988,749.894...+ 1] = [618,033,988,750.894...] = 618,033,988,750."

This does not affect this particular answer, but does not illustrate the general formula, as 1,000,000,000,001*k comes out to 618033988750.51288..., and its integer part is the same as that for 618,033,988,750.894....

 Posted by Charlie on 2005-11-15 14:25:49

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