Let d(n) = f(n) - f(n-1). Re-writing the recurrence in terms of d(n) and simplifying yields ( d(n) + d(n-2) ) / d(n-1) = ( d(n-1) + d(n-3) ) / d(n-2) so in fact ( d(n) + d(n-2) ) / d(n-1) = C, a constant, and this is really a simpler recurrence in disguise: d(n) = C * d(n-1) - d(n-2) As long as C is an integer, you're all set (in your examples C = 1, 2 and -6, respectively). J.P. On Wed, Aug 20, 2014 at 10:09 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Consider the recurrence f(n) = f(n-3) + ( f(n-4) - f(n-1) ) * ( f(n-2) - f(n-1) ) / ( f(n-2) - f(n-3) ) . Initialised with integer 4-tuplets, this generates sequences which are variously periodic: 6, -3, 12, 36, 45, 30, 6, -3, 12, 36, 45, 30, 6, -3, 12, 36, 45, 30, 6, ...; quadratic: 3, 1, 1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, 133, 157, 183, 211, 241, ...; asymptotically exponential: 1, -1, 1, -9, 49, -289, 1681, -9801, 57121, -332929, 1940449, -11309769, ...; but remarkably remain integer despite the division operator.
Explain this behaviour. WFL
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