cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A306073 Bases in which 3 is a unique-period prime.

Original entry on oeis.org

2, 4, 5, 8, 10, 11, 17, 23, 26, 28, 35, 47, 53, 71, 80, 82, 95, 107, 143, 161, 191, 215, 242, 244, 287, 323, 383, 431, 485, 575, 647, 728, 730, 767, 863, 971, 1151, 1295, 1457, 1535, 1727, 1943, 2186, 2188, 2303, 2591, 2915, 3071, 3455, 3887
Offset: 1

Views

Author

Jianing Song, Jun 19 2018

Keywords

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 = 3^t + 1, t >= 1; (b) b = 2^s*3^t - 1, s >= 0, t >= 1.
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 = 3, there are no nontrivial bases, since ord(3,b) <= 2.

Examples

			If b = 3^t + 1, t >= 1, then b - 1 only has prime factor 3, so 3 is a unique-period prime in base b.
If b = 2^s*3^t - 1, t >= 1, then the prime factors of b^2 - 1 are 3 and prime factors of b - 1 = 2^s*3^t - 2, 3 is the only new prime factor so 3 is a unique-period prime in base b.

Links

Crossrefs

Cf. A040017 (unique primes in base 10), A144755 (unique primes in base 2).
Bases in which p is a unique prime: A000051 (p=2), this sequence (p=3), A306074 (p=5), A306075 (p=7), A306076 (p=11), A306077 (p=13).

Programs

  • PARI
    p = 3;
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, ", "))));

A306075 Bases in which 7 is a unique-period prime.

Original entry on oeis.org

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
Offset: 1

Views

Author

Jianing Song, Jun 19 2018

Keywords

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.

Examples

			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.)

Links

Crossrefs

Cf. A040017 (unique-period primes in base 10), A144755 (unique-period primes in base 2).
Bases in which p is a unique-period prime: A000051 (p=2), A306073 (p=3), A306074 (p=5), this sequence (p=7), A306076 (p=11), A306077 (p=13).

Programs

  • 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, ", "))));

A306076 Bases in which 11 is a unique-period prime.

Original entry on oeis.org

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
Offset: 1

Views

Author

Jianing Song, Jun 19 2018

Keywords

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.

Examples

			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.

Links

Crossrefs

Cf. A040017 (unique-period primes in base 10), A144755 (unique-period primes in base 2).
Bases in which p is a unique-period prime: A000051 (p=2), A306073 (p=3), A306074 (p=5), A306075 (p=7), this sequence (p=11), A306077 (p=13).

Programs

  • 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, ", "))));

A306077 Bases in which 13 is a unique-period prime.

Original entry on oeis.org

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
Offset: 1

Views

Author

Jianing Song, Jun 19 2018

Keywords

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.

Examples

			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.

Links

Crossrefs

Cf. A040017 (unique-period primes in base 10), A144755 (unique-period primes in base 2).
Bases in which p is a unique-period prime: A000051 (p=2), A306073 (p=3), A306074 (p=5), A306075 (p=7), A306076 (p=11), this sequence (p=13).

Programs

  • 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, ", "))));
Showing 1-4 of 4 results.

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