

A372797


Smallest prime p such that the multiplicative order of 4 modulo p is 2*n, or 0 if no such prime exists.


3



3, 17, 31, 73, 151, 433, 631, 337, 127, 241, 331, 601, 4421, 673, 3061, 257, 1429, 1657, 1103, 3121, 2143, 1321, 18539, 1777, 2351, 37441, 2971, 2857, 3191, 17401, 683, 15809, 17029, 9929, 38431, 1801, 11471, 63689, 49999, 13121, 17467, 21169, 83077, 25609, 5581, 5153, 26227
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OFFSET

1,1


COMMENTS

First prime p such that the expansion of 1/p has period (p1)/(2*n) in base 4. Also the first prime p such that {k/p : 1 <= k <= p1} 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 (p1)/ord(4,p) = ((p1)/ord(2,p)) * gcd(2,ord(2,p)) is always even. Here ord(a,k) is the multiplicative order of a modulo k.


LINKS



EXAMPLE

In the following examples let () denote the reptend. The prime numbers themselves and the fractions are written out in decimal.
The base4 expansion of 1/3 is 0.(1), so the reptend has length 1 = (31)/2. Also, the base4 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 base4 expansion of 1/17 is 0.(0033), so the reptend has length 4 = (171)/4. Also, the base4 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 base4 expansion of 1/31 is 0.(00133), so the reptend has length 5 = (311)/6. Also, the base4 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)) == (p1)/(n * if(issquare(base), 2, 1)), return(p)))


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



