A235712 -id:A235712 - cp's OEIS frontend

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

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

Zhi-Wei Sun, Apr 22 2014

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..1200
Z.-W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.

Cf. A000040, A001348, A001918, A234972, A235709, A235712, A236306, A236308.

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

Zhi-Wei Sun, Apr 21 2014

nonn

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.

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

Zhi-Wei Sun, Table of n, a(n) for n = 1..1000
Z.-W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.

Cf. A000040, A000108, A234972, A235709, A235712, A236306, A236966.

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

Zhi-Wei Sun, Apr 22 2014

nonn

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.

			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.

Zhi-Wei Sun, Table of n, a(n) for n = 1..600
Z.-W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.

Cf. A000040, A000142, A234972, A235709, A235712, A236306, A236308, A236966, A237121.

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}]

A236306 Least prime p < prime(n) for which both p and p! are primitive roots modulo prime(n), or 0 if such a prime does not exist.

Original entry on oeis.org

0, 2, 2, 0, 2, 2, 3, 2, 5, 2, 11, 2, 17, 5, 5, 2, 2, 2, 2, 13, 5, 3, 2, 3, 5, 2, 11, 2, 67, 3, 3, 2, 3, 2, 2, 13, 53, 2, 5, 2, 2, 2, 47, 5, 2, 3, 2, 3, 2, 29, 3, 7, 137, 11, 3, 5, 2, 59, 31, 13, 17, 2, 5, 23, 47, 2, 101, 23, 2, 2, 13, 7, 43, 2, 2, 5, 2, 109, 3, 127

Offset: 1

Zhi-Wei Sun, Apr 21 2014

nonn

Conjecture: a(n) > 0 for all n > 4. In other words, for any prime p > 7, there exists a prime q < p such that both q and q! are primitive roots modulo p.

P. Erdős asked whether for any sufficiently large prime p there exists a prime q < p which is a primitive root modulo p (see page 377 of Guy's book in the reference).

			a(7) = 3 since both 3 and 3! = 6 are primititive roots modulo prime(7) = 17, but 2 is not a primitive root modulo 17.

R. K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, New York, 2004.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000040, A000142, A234972, A235709, A235712.

f[k_]:=Prime[k]!
dv[n_]:=Divisors[n]
Do[Do[Do[If[Mod[Prime[k]^(Part[dv[Prime[n]-1],i]),Prime[n]]==1||Mod[f[k]^(Part[dv[Prime[n]-1],i]),Prime[n]]==1,Goto[aa]],{i,1,Length[dv[Prime[n]-1]]-1}];Print[n," ",Prime[k]];Goto[bb];Label[aa];Continue,{k,1,n-1}];Print[n," ",0];Label[bb];Continue,{n,1,80}]

cp's OEIS Frontend

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

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

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

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A236306 Least prime p < prime(n) for which both p and p! are primitive roots modulo prime(n), or 0 if such a prime does not exist.

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