OFFSET
1,1
COMMENTS
First prime p such that the expansion of 1/p has period (p-1)/(2*n) in base 4. Also the first prime p such that {k/p : 1 <= k <= p-1} has 2*n different cycles when written out in base 4.
Since ord(a^m,k) = ord(a,k)/gcd(m,ord(a,k)) for gcd(a,k) = 1, we have that (p-1)/ord(4,p) = ((p-1)/ord(2,p)) * gcd(2,ord(2,p)) is always even. Here ord(a,k) is the multiplicative order of a modulo k.
EXAMPLE
In the following examples let () denote the reptend. The prime numbers themselves and the fractions are written out in decimal.
The base-4 expansion of 1/3 is 0.(1), so the reptend has length 1 = (3-1)/2. Also, the base-4 expansions of 1/3 = 0.(1) and 2/3 = 0.(2) have two cycles 1 and 2. 3 is the smallest such prime, so a(1) = 3.
The base-4 expansion of 1/17 is 0.(0033), so the reptend has length 4 = (17-1)/4. Also, the base-4 expansions of 1/17, 2/17, ..., 16/17 have four cycles 0033, 0132, 1023 and 1122. 17 is the smallest such prime, so a(2) = 17.
The base-4 expansion of 1/31 is 0.(00133), so the reptend has length 5 = (31-1)/6. Also, the base-4 expansions of 1/31, 2/31, ..., 30/31 have three cycles 00201, 01203, 02211, 03213, 11223 and 13233. 13 is the smallest such prime, so a(3) = 13.
PROG
(PARI) a(n, {base=4}) = forprime(p=2, oo, if((base%p) && znorder(Mod(base, p)) == (p-1)/(n * if(issquare(base), 2, 1)), return(p)))
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, May 13 2024
STATUS
approved