1) For any natural i , f ( i ) ≤ ( i − 1 ) .2) If i divides a number of the form b n − 1 + 1 or b n − 1 (for applicable n) then f ( i ) ≤ n (I believe this to be an equality if n is minimal).3) If i = j 1 × j 2 with j 1 and j 2 coprime, then f ( i ) ≤ lcm ( f ( j 1 ) , f ( j 2 ) ) .

Question

1) For any natural   i  ,     f  (  i  )  ≤  (  i  −  1  ) .2) If i divides a number of the form       b          n      −      1        +  1 or       b          n        −  1 (for applicable n) then   f  (  i  )  ≤  n (I believe this to be an equality if n is minimal).3) If   i  =      j    1    ×      j    2   with       j    1   and       j    2   coprime, then   f  (  i  )  ≤  lcm  (  f  (      j    1    )  ,     f  (      j    2    )  ) .

letovanjelm · Accepted Answer

Step 1Write   i  =      x    1        y    1   , where every prime factor of       x    1   divides b, and   g  c  d  (      y    1    ,     b  )  =  1 . Then   f  (  i  )  =  f  (      y    1    )  =  o  r      d          y      i        (  b  ), where   o  r      d          y      i        (  b  ) is the multiplicative order of b(mod yi).This is because:      x    i   does not affect the minimal period f(i).If       y    1        |    (      b          n        −  1  ) for some n, then let       b    n    −  1  =  k      y    i   , then      1          y      i        =      k                  b        n            −      1        =      k                  b        n            (      1      −              b                  −          n                    )        =      k          n              n              +      k          b              2        n              +  …So       1          y              i             has n as a period. Since the smallest such n is given by       ord                  y                  i                      (  b  )  ,     f  (      y          i        )  ≤      ord                  y                  i                      (  b  ) .On the other hand, since f(i) is a period, let l be the number representing the repeating string, so we can write      1          y              i              =      l          b              f        (        i        )              +      l          b              2        f        (        i        )              +  ⋯  =      l                  b                  f          (          i          )                    (      1      −              b                  −          f          (          i          )                    )      Thus       b          f      (      i      )        −  1  =      l                  y                  i                     , so   f  (      y          i        )  ≥      ord                  y                  i                      (  b  ) .Now back to your results.1 is mostly correct, as we have   f  (  i  )  =  f  (      y          i        )  =      ord                  y                  i                      (  b  )  ≤  ϕ  (      y          i        )  ≤      y          i        −  1  ≤  i  −  1 for       y          i        &amp;gt;  1 , and if       y          i        =  1  ,     f  (  i  )  =  1  ≤  i  −  1 for   i  &amp;gt;  1 . The only exception is   i  =  1 , as   f  (  i  )  =  1  &amp;gt;  1  −  1 .2 is partially correct. The       b    n    −  1 part is correct, but not the       b          n      −      1       part.3 is correct.       y          i        =      y                  j                  1                          y                  j                  2                     so  f  (  i  )  =      ord                  y                  i                      (  b  )  =  l  c  m  (  o  r      d                  y                              j                          1                                            (  b  )  ,     o  r      d                  y                              j                          2                                            (  b  )  )    =  l  c  m  (  f  (      j          1        )  ,     f  (      j          2        )  )

1) For any natural i, f(i) <= (i−1). 2) If i divides a number of the form b^{n-1}+1 or b^n-1 (for applicable n) then f(i) <= n (I believe this to be an equality if n is minimal).

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