Over 2000 numbers are around a circle. Each number is the sum of its left and right neighbors.
Given that one of the numbers is a one, how many numbers (as a minimum) must there be?
(In reply to
re(4): What now ?  you're right again !! by pcbouhid)
But it is proven that such a cycle must be a multiple of 6. Each successive number around the circle is the previous number minus the one before that, so that starting with any two numbers, a and b, the remaining numbers are forced:
a, b, ba, a, b, ab, ...
The next number is forced to be (ab)  (b) = a; then comes a  (ab) = b, so after the 6 items shown, the same 6 appear again starting with a and b. When it comes full circle, it has to match, and so can only come in again at a multiple of 6.
The only way to have a cycle of three would be if a = a and b = b; but that's true only when a = b = 0.
The only way to have a cycle of two would be if a = ba = b; but in that instance 2*b would equal b, and b would be zero. A similar algebra on the other three terms of the 6cycle shows a would also have to be zero, so again it would be all zeroes.

Posted by Charlie
on 20050601 15:20:09 