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 1-3 of 3 results.

A236966 Number of primes p < prime(n)/2 such that 2^p - 1 is a primitive root modulo prime(n).

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 3, 2, 1, 3, 2, 1, 2, 2, 5, 6, 3, 4, 3, 5, 4, 5, 7, 9, 3, 5, 2, 10, 7, 7, 7, 7, 9, 5, 10, 4, 5, 7, 12, 11, 14, 6, 7, 5, 10, 9, 8, 5, 12, 15, 14, 8, 12, 11, 16, 12, 16, 9, 12, 10, 10, 14, 15, 10, 12, 14, 9, 10, 21, 9, 22, 21, 11, 9, 18, 24, 20, 17, 17, 16
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 22 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 2. In other words, for any prime p > 3, there is a prime q < p/2 with the Mersenne number 2^q - 1 a primitive root modulo p.
We have verified this for all n = 3, ..., 530000.
See also the comment in A234972.

Examples

			a(12) = 1 since 17 is a prime smaller than prime(12)/2 = 37/2 with 2^(17) - 1 = 131071 a primitive root modulo prime(12) = 37.
		

Crossrefs

Programs

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

A236308 Number of primes q < prime(n)/2 such that the Catalan number C(q) is a primitive root modulo prime(n).

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 1, 1, 4, 2, 1, 4, 1, 3, 3, 5, 5, 5, 2, 4, 5, 4, 10, 4, 7, 7, 4, 7, 4, 9, 5, 6, 10, 9, 7, 5, 5, 12, 12, 13, 12, 4, 10, 7, 13, 4, 7, 10, 18, 9, 14, 13, 9, 9, 15, 17, 16, 8, 9, 12, 10, 19, 13, 10, 14, 14, 13, 6, 18, 18, 14, 24, 13, 16, 9, 22, 20, 12, 23, 15
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 21 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 2. In other words, for any prime p > 3, there exists a prime q < p/2 such that the Catalan number C(q) = binomial(2q, q)/(q+1) is a primitive root modulo p.
We have verified this for all n = 3, ..., 2*10^5.

Examples

			a(13) = 1 since C(7) = 429 is a primitive root modulo prime(13) = 41.
		

Crossrefs

Programs

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

A237112 Number of primes p < prime(n)/2 with p! a primitive root modulo prime(n).

Original entry on oeis.org

0, 0, 1, 0, 2, 2, 1, 1, 1, 2, 2, 2, 2, 2, 4, 4, 6, 4, 2, 3, 3, 4, 7, 9, 5, 7, 5, 8, 4, 7, 6, 7, 8, 7, 11, 8, 9, 7, 13, 10, 16, 4, 7, 8, 13, 9, 8, 12, 17, 10, 14, 12, 4, 10, 14, 15, 18, 8, 9, 8, 8, 18, 6, 8, 7, 16, 9, 11, 21, 15
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 22 2014

Keywords

Comments

Conjecture: a(n) > 0 for all n > 4. In other words, for any prime p > 7, there exists a prime q < p/2 such that q! is a primitive root modulo p.
See also A236306 for a similar conjecture.

Examples

			a(7) = 1 since 3 is a prime smaller than prime(7)/2 = 17/2 and 3! = 6 is a primitive root modulo prime(7) = 17.
a(9) = 1 since  5 is a prime smaller than prime(9)/2 = 23/2 and 5! = 120 is a primitive root modulo prime(9) = 23.
		

Crossrefs

Programs

  • Mathematica
    f[k_]:=(Prime[k])!
    dv[n_]:=Divisors[n]
    Do[m=0;Do[Do[If[Mod[f[k]^(Part[dv[Prime[n]-1],i]),Prime[n]]==1,Goto[aa]],{i,1,Length[dv[Prime[n]-1]]-1}];m=m+1;Label[aa];Continue,{k,1,PrimePi[(Prime[n]-1)/2]}];Print[n," ",m];Continue,{n,1,70}]
Showing 1-3 of 3 results.