## Josephus Flavius Problem

Recursive Solution

A recent letter from a math teacher reminded my that the recursive solution I have been planning to describe for a while now is long overdue.

This solution applies to the case where every other person is executed *m* = 2)*r* = 1,

After the first go-round we essentially come up with the same problem but for a different number of individuals. We have to distinguish between two cases: *n* is even and *n* is odd. When *n* is even, *n* = 2*k*,*k* fellows remain and on the second round we start with the one who was initially number 1. When *n* is odd, *n* = 2*k* + 1,*k* + 1*k* people remain for the second round. The second round is very much like the first one, except that each person's number has been doubled and, in the first case, decreased by one, while, in the second, it was increased by one. Let J(*n*) denote the survivor's number. Obviously, *n*, we can write

J(2k) = 2J(k) - 1, and |

J(2k + 1) = 2J(k) + 1. |

There is a way to solve these recursive formulas but the result is simple enough for us just to guess the right expression. The formulas are very easy to apply. Let's build a table for the first few values of *n*:

n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | ||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

J(n) | 1 | 1 | 3 | 1 | 3 | 5 | 7 | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 | 1 |

We may now notice that J(*n*) is always odd and starts anew from 1 with every power of 2. It increases through all successive odd numbers until it is about to reach the next power of 2 when it again drops to 1. For every number *n*, there exists an integer *a* such that ^{a} ≤ *n* < 2^{a + 1}.*t* < 2^{a}*n* = 2^{a} + *t*.

J(2^{a} + *t*) = 2*t* + 1.

We prove the formula by induction on *a*. For *a* = 1*t* is 0 and we only have to verify that *a* up to a certain value. Now consider this value of *a*. The induction step has two parts, depending on whether *n* (and thus *t*) is even or odd. If ^{a} + *t* = 2*k*,

J(2^{a} + *t*) = 2J(2^{a -1} + *t*/2) - 1 = 2(2*t*/2 + 1) - 1 = 2*t* + 1

by the inductive assumption. Similarly, if 2^{a} + *t* = 2*k* + 1, then

J(2^{a} + *t*) = 2J(2^{a -1} + (*t*-1)/2) + 1 = 2(2(*t*-1)/2 + 1) + 1 = 2*t* + 1

This completes the induction step.

### Reference

- R. Graham, D. Knuth, O. Patashnik,
*Concrete Mathematics*, 2nd edition, Addison-Wesley, 1994.

- Josephus Flavius problem
- Two ancient variants
- A simple solution
- Partial recursive solution
- A problem from USAMTS

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