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).

batystowy2b

batystowy2b

Answered question

2022-09-11

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 ) ) .

Answer & Explanation

letovanjelm

letovanjelm

Beginner2022-09-12Added 13 answers

Step 1
Write 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 > 1 , and if y i = 1 ,   f ( i ) = 1 i 1 for i > 1 . The only exception is i = 1 , as f ( i ) = 1 > 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 ) )

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