# Seven Terms Periodic Sequence

A simple sequence of arithmetic operations leads to a curious result. First, to demonstrate the idea, I'll show an example, and afterwards prove a general result.

Take two numbers, say, 3 and 5. Perform the following sequence of operations:

Add 1 to the second number and divide the sum by the first | (5 + 1)/3 = 2 |

Add 1 to the third number and divide the sum by the second | (2 + 1)/5 = 3/5 |

Add 1 to the fourth number and divide the sum by the third | (3/5 + 1)/2 = 4/5 |

Add 1 to the fifth number and divide the sum by the fourth | (4/5 + 1)/(3/5) = 3 |

Add 1 to the sixth number and divide the sum by the fifth | (3 + 1)/(4/5) = 5 |

Note that the sixth and the seventh numbers coincide respectively with the first and the second, such that, should we continue the process, we would repeat the previous steps. This is a general result: for any starting numbers a and b, the sequence repeats itself from the sixth term on:

(1) | a | ||

(2) | b | ||

(3) | (b + 1)/a | ||

(4) | ((b + 1)/a + 1)/b | = (a + b + 1)/ab | |

(5) | ((a + b + 1)/ab + 1)/((b + 1)/a) | = (a + 1)(b + 1)/ab/((b + 1)/a) | = (a + 1)/b |

(6) | (a + 1)/b + 1)/((a + b + 1)/ab) | = (a + b + 1)/b/((a + b + 1)/ab) | = a |

(7) | (a + 1)/((a + 1)/b) | = b |

The derivation was greatly simplified on step (5) by an observation that

If one does not use algebra, but repeats the computations with different pairs of two numbers, how long would it take to notice that the same period always turns up? Meanwhile, one gets a good exercise practicing addition and multiplication.

It may be instructive to present the problem in a slightly different form. Define sequence U_{n}, n = 0, 1, 2, ... recursively

U_{0} = a,

U_{1} = b,

U_{n+1} = (1 + U_{n})/U_{n-1}, n > 0.

The gist of the discussion then is reduced to the claim that

For any selection of positive a and b, U_{n+5} = U_{n}.

### References

- W. W. Rouse Ball and H. S. M. Coxeter,
*Mathematical Recreations and Essays*, Dover, 1987

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