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Bases in which 5 is a unique-period prime.
+10
5
2, 3, 4, 6, 7, 9, 19, 24, 26, 39, 49, 79, 99, 124, 126, 159, 199, 249, 319, 399, 499, 624, 626, 639, 799, 999, 1249, 1279, 1599, 1999, 2499, 2559, 3124, 3126, 3199, 3999, 4999, 5119, 6249, 6399, 7999, 9999, 10239, 12499, 12799, 15624, 15626, 15999, 19999, 20479
COMMENTS
A prime p is called a unique-period prime in base b if there is no other prime q such that the period length of the base-b expansion of its reciprocal, 1/p, is equal to the period length of the reciprocal of q, 1/q.
A prime p is a unique-period prime in base b if and only if Zs(b, 1, ord(b,p)) = p^k, k >= 1. Here Zs(b, 1, d) is the greatest divisor of b^d - 1 that is coprime to b^m - 1 for all positive integers m < d, and ord(b,p) is the multiplicative order of b modulo p.
b is a term if and only if: (a) b = 5^t + 1, t >= 1; (b) b = 2^s*5^t - 1, s >= 0, t >= 1; (c) b = 2, 3, 7.
For every odd prime p, p is a a unique-period prime in base b if b = p^t + 1, t >= 1 or b = 2^s*p^t - 1, s >= 0, t >= 1. These are trivial bases in which p is a unique-period prime, with ord(b,p) = 1 or 2. By Faltings's theorem, there are only finitely many nontrivial bases in which p is also a unique-period prime, with ord(b,p) >= 3. For p = 5, the nontrivial bases are 2, 3, 7.
EXAMPLE
1/5 has period length 4 in base 2. Note that 3 and 5 are the only prime factors of 2^4 - 1 = 15, but 1/3 has period length 2, so 5 is a unique-period prime in base 2.
1/5 has period length 4 in base 3. Note that 2 and 5 are the only prime factors of 3^4 - 1 = 80, but 1/2 has period length 1, so 5 is a unique-period prime in base 3.
1/5 has period length 4 in base 7. Note that 2, 3 and 5 are the only prime factors of 7^4 - 1 = 2400, but 1/2 and 1/3 both have period length 1, so 5 is a unique-period prime in base 7.
PROG
(PARI)
p = 5;
gpf(n)=if(n>1, vecmax(factor(n)[, 1]), 1);
test(n, q)=while(n%p==0, n/=p); if(q>1, while(n%q==0, n/=q)); n==1;
for(n=2, 10^6, if(gcd(n, p)==1, if(test(polcyclo(znorder(Mod(n, p)), n), gpf(znorder(Mod(n, p)))), print1(n, ", "))));
Bases in which 7 is a unique-period prime.
+10
5
2, 3, 4, 5, 6, 8, 13, 18, 19, 27, 48, 50, 55, 97, 111, 195, 223, 342, 344, 391, 447, 685, 783, 895, 1371, 1567, 1791, 2400, 2402, 2743, 3135, 3583, 4801, 5487, 6271, 7167, 9603, 10975, 12543, 14335, 16806, 16808, 19207, 21951, 25087, 28671, 33613, 38415, 43903, 50175
COMMENTS
A prime p is called a unique-period prime in base b if there is no other prime q such that the period length of the base-b expansion of its reciprocal, 1/p, is equal to the period length of the reciprocal of q, 1/q.
A prime p is a unique-period prime in base b if and only if Zs(b, 1, ord(b,p)) = p^k, k >= 1. Here Zs(b, 1, d) is the greatest divisor of b^d - 1 that is coprime to b^m - 1 for all positive integers m < d, and ord(b,p) is the multiplicative order of b modulo p.
b is a term if and only if: (a) b = 7^t + 1, t >= 1; (b) b = 2^s*7^t - 1, s >= 0, t >= 1; (c) b = 2, 3, 4, 5, 18, 19.
For every odd prime p, p is a unique-period prime in base b if b = p^t + 1, t >= 1 or b = 2^s*p^t - 1, s >= 0, t >= 1. These are trivial bases in which p is a unique-period prime, with ord(b,p) = 1 or 2. By Faltings's theorem, there are only finitely many nontrivial bases in which p is also a unique-period prime, with ord(b,p) >= 3. For p = 7, the nontrivial bases are 2, 3, 4, 5, 18, 19.
EXAMPLE
1/7 has period length 3 in base 2. Note that 7 is the only prime factor of 2^3 - 1 = 7, so 7 is a unique-period prime in base 2.
1/7 has period length 3 in base 4. Note that 3, 7 are the only prime factors of 4^3 - 1 = 63, but 1/3 has period length 1, so 7 is a unique-period prime in base 4.
1/7 has period length 3 in base 18. Note that 7, 17 are the only prime factors of 18^3 - 1 = 5831, but 1/17 has period length 1, so 7 is a unique-period prime in base 18.
(1/7 has period length 6 in base 3, 5, 19. Similar demonstrations can be found.)
PROG
(PARI)
p = 7;
gpf(n)=if(n>1, vecmax(factor(n)[, 1]), 1);
test(n, q)=while(n%p==0, n/=p); if(q>1, while(n%q==0, n/=q)); n==1;
for(n=2, 10^6, if(gcd(n, p)==1, if(test(polcyclo(znorder(Mod(n, p)), n), gpf(znorder(Mod(n, p)))), print1(n, ", "))));
CROSSREFS
Cf. A040017 (unique-period primes in base 10), A144755 (unique-period primes in base 2).
Bases in which 11 is a unique-period prime.
+10
5
2, 3, 10, 12, 21, 43, 87, 120, 122, 175, 241, 351, 483, 703, 967, 1330, 1332, 1407, 1935, 2661, 2815, 3871, 5323, 5631, 7743, 10647, 11263, 14640, 14642, 15487, 21295, 22527, 29281, 30975, 42591, 45055, 58563, 61951, 85183, 90111, 117127, 123903, 161050, 161052, 170367, 180223, 234255, 247807, 322101, 340735
COMMENTS
A prime p is called a unique-period prime in base b if there is no other prime q such that the period length of the base-b expansion of its reciprocal, 1/p, is equal to the period length of the reciprocal of q, 1/q.
A prime p is a unique-period prime in base b if and only if Zs(b, 1, ord(b,p)) = p^k, k >= 1. Here Zs(b, 1, d) is the greatest divisor of b^d - 1 that is coprime to b^m - 1 for all positive integers m < d, and ord(b,p) is the multiplicative order of b modulo p.
b is a term if and only if: (a) b = 11^t + 1, t >= 1; (b) b = 2^s*11^t - 1, s >= 0, t >= 1; (c) b = 2, 3.
For every odd prime p, p is a unique-period prime in base b if b = p^t + 1, t >= 1 or b = 2^s*p^t - 1, s >= 0, t >= 1. These are trivial bases in which p is a unique-period prime, with ord(b,p) = 1 or 2. By Faltings's theorem, there are only finitely many nontrivial bases in which p is also a unique-period prime, with ord(b,p) >= 3. For p = 11, the nontrivial bases are 2, 3.
EXAMPLE
1/11 has period length 10 in base 2. Note that 3, 11, 31 are the only prime factors of 2^10 - 1 = 1023, but 1/3 has period length 2 and 1/31 has period length 5, so 11 is a unique-period prime in base 2.
1/11 has period length 5 in base 3. Note that 2, 11 are the only prime factors of 3^5 - 1 = 242, but 1/2 has period length 1, so 11 is a unique-period prime in base 3.
PROG
(PARI)
p = 11;
gpf(n)=if(n>1, vecmax(factor(n)[, 1]), 1);
test(n, q)=while(n%p==0, n/=p); if(q>1, while(n%q==0, n/=q)); n==1;
for(n=2, 10^6, if(gcd(n, p)==1, if(test(polcyclo(znorder(Mod(n, p)), n), gpf(znorder(Mod(n, p)))), print1(n, ", "))));
CROSSREFS
Cf. A040017 (unique-period primes in base 10), A144755 (unique-period primes in base 2).
Bases in which 13 is a unique-period prime.
+10
5
2, 3, 4, 5, 12, 14, 22, 23, 25, 51, 103, 168, 170, 207, 239, 337, 415, 675, 831, 1351, 1663, 2196, 2198, 2703, 3327, 4393, 5407, 6655, 8787, 10815, 13311, 17575, 21631, 26623, 28560, 28562, 35151, 43263, 53247, 57121, 70303, 86527, 106495, 114243, 140607, 173055, 212991, 228487, 281215, 346111
COMMENTS
A prime p is called a unique-period prime in base b if there is no other prime q such that the period of the base-b expansion of its reciprocal, 1/p, is equal to the period of the reciprocal of q, 1/q.
A prime p is a unique-period prime in base b if and only if Zs(b, 1, ord(b,p)) = p^k, k >= 1. Here Zs(b, 1, d) is the greatest divisor of b^d - 1 that is coprime to b^m - 1 for all positive integers m < d, and ord(b,p) is the multiplicative order of b modulo p.
b is a term if and only if: (a) b = 13^t + 1, t >= 1; (b) b = 2^s*13^t - 1, s >= 0, t >= 1; (c) b = 2, 3, 4, 5, 22, 23, 239.
For every odd prime p, p is a unique-period prime in base b if b = p^t + 1, t >= 1 or b = 2^s*p^t - 1, s >= 0, t >= 1. These are trivial bases in which p is a unique-period prime, with ord(b,p) = 1 or 2. By Faltings's theorem, there are only finitely many nontrivial bases in which p is also a unique-period prime, with ord(b,p) >= 3. For p = 13, the nontrivial bases are 2, 3, 4, 5, 22, 23, 239.
EXAMPLE
1/13 has period 12 in base 2. Note that 3, 5, 7, 13, 31 are the only prime factors of 2^12 - 1 = 4095, but 1/3 has period 2, 1/5 has period 4, 1/7 has period 3, 1/31 has period 5, so 13 is a unique-period prime in base 2. (For the same reason, 13 is a unique-period prime in base 4.)
1/13 has period 3 in base 3. Note that 2, 13 are the only prime factors of 3^3 - 1 = 26, but 1/2 has period 1, so 13 is a unique-period prime in base 3.
1/13 has period 3 in base 22. Note that 3, 7, 13 are the only prime factors of 22^3 - 1 = 10647, but 1/3 and 1/7 both have period 1, so 13 is a unique-period prime in base 22.
PROG
(PARI)
p = 13;
gpf(n)=if(n>1, vecmax(factor(n)[, 1]), 1);
test(n, q)=while(n%p==0, n/=p); if(q>1, while(n%q==0, n/=q)); n==1;
for(n=2, 10^6, if(gcd(n, p)==1, if(test(polcyclo(znorder(Mod(n, p)), n), gpf(znorder(Mod(n, p)))), print1(n, ", "))));
CROSSREFS
Cf. A040017 (unique-period primes in base 10), A144755 (unique-period primes in base 2).
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