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 1-5 of 5 results.

A242748 Number of ordered ways to write n = k + m with 0 < k <= m such that k is a primitive root modulo prime(k) and m is a primitive root modulo prime(m).

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 3, 3, 3, 1, 3, 3, 2, 2, 3, 3, 2, 1, 2, 3, 3, 2, 3, 3, 3, 3, 1, 2, 3, 3, 3, 2, 1, 4, 2, 3, 3, 3, 3, 2, 5, 3, 4, 2, 4, 6, 6, 1, 5, 4, 6, 7, 4, 6, 4, 6, 3, 6, 3, 7, 5, 5, 6, 7, 4, 6, 8, 5, 6, 4, 6, 4, 8, 3, 7
Offset: 1

Views

Author

Zhi-Wei Sun, May 21 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This implies that there are infinitely many positive integers k which is a primitive root modulo prime(k).

Examples

			a(6) = 1 since 6 = 3 + 3 with 3 a primitive root modulo prime(3) = 5.
a(7) = 1 since 7 = 1 + 6 with 1 a primitive root modulo prime(1) = 2 and 6 a primitive root modulo prime(6) = 13.
a(15) = 1 since 15 = 2 + 13 with 2 a primitive root modulo prime(2) = 3 and 13 a primitive root modulo prime(13) = 41.
a(38) = 1 since 38 = 10 + 28 with 10 a primitive root modulo prime(10) = 29 and 28 a primitive root modulo prime(28) = 107.
a(53) = 1 since 53 = 3 + 50 with 3 a primitive root modulo prime(3) = 5 and 50 a primitive root modulo prime(50) = 229.

Links

Crossrefs

Cf. A000040, A000720, A242750, A242752.

Programs

  • Mathematica
    dv[n_]:=Divisors[n]
Do[m=0;Do[Do[If[Mod[k^(Part[dv[Prime[k]-1],i]),Prime[k]]==1,Goto[aa]],{i,1,Length[dv[Prime[k]-1]]-1}];Do[If[Mod[(n-k)^(Part[dv[Prime[n-k]-1],j]),Prime[n-k]]==1,Goto[aa]],{j,1,Length[dv[Prime[n-k]-1]]-1}];m=m+1;Label[aa];Continue,{k,1,n/2}];Print[n," ",m];Continue,{n,1,80}]

A242750 Positive integers n with property that n is a primitive root modulo prime(n).

Original entry on oeis.org

1, 2, 3, 6, 7, 10, 11, 13, 15, 18, 24, 26, 28, 33, 39, 41, 44, 45, 48, 50, 54, 55, 56, 58, 62, 65, 68, 69, 71, 75, 79, 83, 85, 93, 95, 107, 108, 109, 110, 117, 118, 119, 120, 123, 126, 129, 130, 131, 133, 139, 142, 143, 145, 148, 157, 158, 160, 163, 164, 166, 170, 172, 173, 174, 179, 182, 186, 190, 191, 195
Offset: 1

Views

Author

Zhi-Wei Sun, May 21 2014

Keywords

Comments

According to the conjecture in A242748, this sequence should have infinitely many terms.

Examples

			6 is a member since 6 is a primitive root modulo prime(6) = 13, but 4 and 5 are not since 4 is not a primitive root modulo prime(4) = 7 and 5 is not a primitive root modulo prime(5) = 11.

Links

Crossrefs

Cf. A000040, A000720, A242748, A242752.

Programs

  • Mathematica
    dv[n_]:=Divisors[n]
n=0;Do[Do[If[Mod[k^(Part[dv[Prime[k]-1],j]),Prime[k]]==1,Goto[aa]],{j,1,Length[dv[Prime[k]-1]]-1}];n=n+1;Print[n," ",k];Label[aa];Continue,{k,1,195}]

A242753 Number of ordered ways to write n = k + m with 0 < k <= m such that the inverse of k mod prime(k) among 1, ..., prime(k) - 1 is prime and the inverse of m mod prime(m) among 1, ..., prime(m) - 1 is also prime.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 1, 2, 2, 2, 1, 2, 3, 3, 2, 2, 3, 1, 2, 4, 3, 2, 3, 4, 2, 1, 2, 3, 1, 1, 2, 1, 1, 3, 2, 1, 1, 2, 2, 1, 2, 3, 3, 4, 1, 1, 3, 4, 2, 4, 4, 5, 3, 4, 5, 4, 3, 5, 6, 3, 3, 6, 4, 4, 3, 5, 4, 4, 4, 6, 5, 3, 5, 6, 5, 5, 9, 5, 6, 4
Offset: 1

Views

Author

Zhi-Wei Sun, May 22 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 3.
This implies that there are infinitely many positive integers k such that k*q == 1 (mod prime(k)) for some prime q < prime(k).

Examples

			a(11) = 1 since 11 = 4 + 7, 4*2 == 1 (mod prime(4)=7) with 2 prime, and 7*5 == 1 (mod Prime(7)=17) with 5 prime.
a(36) = 1 since 36 = 18 + 18, and 18*17 == 1 (mod 61) with 17 prime.
a(46) = 1 since 46 = 6 + 40, 6*11 == 1 (mod prime(6)= 13) with 11 prime, and 40*13 == 1 (mod prime(40)=173) with 13 prime.

Links

Crossrefs

Cf. A000040, A000720, A242425, A242748, A242750, A242752, A242754, A242755.

Programs

  • Mathematica
    p[n_]:=PrimeQ[PowerMod[n,-1,Prime[n]]]
Do[m=0;Do[If[p[k]&&p[n-k],m=m+1],{k,1,n/2}];Print[n," ",m];Continue,{n,1,80}]

A242754 Positive integers k such that k*p == 1 (mod prime(k)) for some prime p < prime(k).

Original entry on oeis.org

2, 3, 4, 6, 7, 10, 11, 13, 17, 18, 21, 31, 37, 40, 41, 46, 48, 49, 52, 53, 58, 60, 64, 66, 70, 71, 72, 73, 75, 81, 85, 92, 93, 96, 100, 102, 109, 117, 119, 127, 136, 137, 140, 143, 145, 146, 149, 160, 162, 179, 189, 194, 200, 206, 215, 232, 233, 243, 246, 247
Offset: 1

Views

Author

Zhi-Wei Sun, May 22 2014

Keywords

Comments

According to the conjecture in A242753, this sequence should have infinitely many terms.
Conjecture: The number of terms not exceeding x > 1 has the main term x/(log x) as x tends to infinity.

Examples

			a(4) = 6 since 6*11 == 1 (mod prime(6)=13) with 11 prime, but 5*9 == 1 (mod prime(5)=11) with 9 composite.

Links

Crossrefs

Cf. A000040, A000720, A242425, A242748, A242750, A242752, A242753, A242755.

Programs

  • Mathematica
    p[n_]:=PrimeQ[PowerMod[n,-1,Prime[n]]]
n=0;Do[If[p[k],n=n+1;Print[n," ",k]];Continue,{k,1,247}]

A242755 Primes p such that pi(p)*q == 1 (mod p) for some prime q < p, where pi(p) is the number of primes not exceeding p.

Original entry on oeis.org

3, 5, 7, 13, 17, 29, 31, 41, 59, 61, 73, 127, 157, 173, 179, 199, 223, 227, 239, 241, 271, 281, 311, 317, 349, 353, 359, 367, 379, 419, 439, 479, 487, 503, 541, 557, 599, 643, 653, 709, 769, 773, 809, 823, 829, 839, 859, 941, 953, 1063
Offset: 1

Views

Author

Zhi-Wei Sun, May 22 2014

Keywords

Comments

According to the conjecture in A242753, this sequence should contain infinitely many primes.
Conjecture: The number of such primes not exceeding x > 1 has the main term x/(log x)^2 as x tends to infinity.

Examples

			a(4) = 13 since 13 is prime with pi(13) = 6, and 6*11 == 1 (mod 13) with 11 prime, but pi(11)*9 == 1 (mod 11) with 9 not prime.

Links

Crossrefs

Cf. A000040, A000720, A242748, A242750, A242752, A242753, A242754.

Programs

  • Mathematica
    p[n_]:=PrimeQ[PowerMod[n,-1,Prime[n]]]
n=0;Do[If[p[k],n=n+1;Print[n," ",Prime[k]]];Continue,{k,1,179}]
Showing 1-5 of 5 results.

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