Project Euler

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# [[Problem 69]]

## Totient Maximum

Euler's totient function, $\phi(n)$ [sometimes called the phi function], is defined as the number of positive integers not exceeding $n$ which are relatively prime to $n$. For example, as $1$, $2$, $4$, $5$, $7$, and $8$, are all less than or equal to nine and relatively prime to nine, $\phi(9)=6$.

| **$n$** | **Relatively Prime** | **$\phi(n)$** | **$n/\phi(n)$** |

|--|--|--|--|

| 2 | 1 | 1 | 2 |

| 3 | 1,2 | 2 | 1.5 |

| 4 | 1,3 | 2 | 2 |

| 5 | 1,2,3,4 | 4 | 1.25 |

| 6 | 1,5 | 2 | 3 |

| 7 | 1,2,3,4,5,6 | 6 | 1.1666... |

| 8 | 1,3,5,7 | 4 | 2 |

| 9 | 1,2,4,5,7,8 | 6 | 1.5 |

| 10 | 1,3,7,9 | 4 | 2.5 |

It can be seen that $n = 6$ produces a maximum $n/\phi(n)$ for $n\leq 10$.

Find the value of $n\leq 1\,000\,000$ for which $n/\phi(n)$ is a maximum.

## トーティエント関数の最大値

オイラーのトーティエント関数, φ(n) [時々ファイ関数とも呼ばれる]は, n と互いに素な n 未満の数の数を定める. たとえば, 1, 2, 4, 5, 7, そして8はみな9未満で9と互いに素であり, φ(9)=6.

| n | 互いに素な数 | φ(n) | n/φ(n) |

|--|--|--|--|

| 2 | 1 | 1 | 2 |

| 3 | 1,2 | 2 | 1.5 |

| 4 | 1,3 | 2 | 2 |

| 5 | 1,2,3,4 | 4 | 1.25 |

| 6 | 1,5 | 2 | 3 |

| 7 | 1,2,3,4,5,6 | 6 | 1.1666... |

| 8 | 1,3,5,7 | 4 | 2 |

| 9 | 1,2,4,5,7,8 | 6 | 1.5 |

| 10 | 1,3,7,9 | 4 | 2.5 |

n ≤ 10 では n/φ(n) の最大値は n=6 であることがわかる.

n ≤ 1,000,000で n/φ(n) が最大となる値を見つけよ.

```scheme

(import (scheme base)

        (gauche base))

(define (factorize num)

  (assume (exact-integer? num))

  (assume (<= 2 num))

  (let loop ([current 2] [rest num] [result '()])

    (cond

      [(= rest 1) (reverse result)]

      [(and (< num (* current current))

            (null? result))

       '()]

      [(< num (* current current))

       (reverse (cons rest result))]

      [(and (zero? (mod rest current))

            (null? result))

       (loop current

             (div rest current)

             (cons current result))]

      [(and (zero? (mod rest current))

            (= current (car result)))

       (loop current

             (div rest current)

             result)]

      [(zero? (mod rest current))

       (loop current

             (div rest current)

             (cons current result))]

      [else

        (loop (+ current 1) rest result)])))

; https://manabitimes.jp/math/667

(define (totient num)

  (assume (exact-integer? num))

  (assume (<= 2 num))

  (let ([lis (factorize num)])

    (if (null? lis)

      (- num 1)

      (apply * num

             (map (^n (- 1 (/ 1 n)))

                  lis)))))

(define (totient-100_0000-list)

  (map (^n

         (let ([totient-n (totient n)])

           `(,n ,totient-n ,(/ n totient-n))))

       (iota (- #e1e6 1) 2)))

(define answer-69

  (car

    (fold (^[lis current]

            (if (< (caddr current) (caddr lis))

              lis

              current))

          '(0 0 0)

          (totient-100_0000-list))))

(format #t "69: ~d~%" answer-69)

```