A241492 -id:A241492 - cp's OEIS frontend

A241504 a(n) = |{0 < g < prime(n): g is not only a primitive root modulo prime(n) but also a partition number given by A000041}|.

1, 1, 2, 2, 2, 3, 4, 3, 4, 4, 3, 4, 5, 3, 5, 4, 5, 3, 3, 5, 4, 4, 6, 5, 4, 6, 4, 6, 4, 3, 4, 4, 3, 7, 8, 5, 3, 6, 5, 8, 5, 2, 5, 7, 7, 6, 4, 7, 7, 2, 7, 5, 3, 6, 6, 10, 9, 5, 8, 7, 5, 10, 5, 5, 3, 8, 5, 5, 9, 4, 5, 5, 5, 8, 7, 10, 9, 6, 7, 4

Offset: 1

Zhi-Wei Sun, Apr 24 2014

nonn

Conjecture: (i) a(n) > 0 for all n > 0. In other words, any prime p has a primitive root g < p which is also a partition number.
(ii) Any prime p > 3 has a primitive root g < p which is also a strict partition number (i.e., a term of A000009).
We have checked part (i) for all primes p < 2*10^7, and part (ii) for all primes p < 5*10^6. See also A241516.

			a(92) = 1 since p(13) = 101 is a primitive root modulo prime(92) = 479, where p(.) is the partition function (A000041).
a(493) = 1 since p(20) = 627 is a primitive root modulo prime(493) = 3529.
a(541) = 1 since p(20) = 627 is a primitive root modulo prime(541) = 3911.
a(1146) = 1 since p(27) = 3010 is a primitive root modulo prime(1146) = 9241.
a(1951) = 1 since p(35) = 14883 is a primitive root modulo prime(1951) = 16921.
a(2380) = 1 since p(36) = 17977 is a primitive root modulo prime(2380) = 21169.
a(5629) = 1 since p(20) = 627 is a primitive root modulo prime(5629) = 55441.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.

Cf. A000009, A000040, A000041, A237121, A239957, A239963, A241476, A241492, A241516, A241568.

f[k_]:=PartitionsP[k]
dv[n_]:=Divisors[n]
Do[m=0;Do[If[f[k]>Prime[n]-1,Goto[bb]];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,Prime[n]-1}];Label[bb];Print[n," ",m];Continue,{n,1,80}]

A241516 Least positive primitive root g < prime(n) modulo prime(n) which is also a partition number given by A000041, or 0 if such a number g does not exist.

Original entry on oeis.org

1, 2, 2, 3, 2, 2, 3, 2, 5, 2, 3, 2, 7, 3, 5, 2, 2, 2, 2, 7, 5, 3, 2, 3, 5, 2, 5, 2, 11, 3, 3, 2, 3, 2, 2, 7, 5, 2, 5, 2, 2, 2, 22, 5, 2, 3, 2, 3, 2, 7, 3, 7, 7, 11, 3, 5, 2, 15, 5, 3, 3, 2, 5, 22, 15, 2, 3, 15, 2, 2, 3, 7, 11, 2, 2, 5, 2, 5, 3, 22

Offset: 1

Zhi-Wei Sun, Apr 24 2014

nonn

According to the conjecture in A241504, a(n) should be always positive.

			a(4) = 3 since 3 = A000041(3) is a primitive root modulo prime(4) = 7, but neither 1 = A000041(1) nor 2 = A000041(2) is.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014.

Cf. A000040, A000041, A237121, A239957, A239963, A241476, A241492, A241504.

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

A241568 a(n) = |{0 < k < prime(n)/2: k is not only a quadratic nonresidue modulo prime(n) but also a Fibonacci number}|.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Apr 25 2014

nonn

Conjecture: (i) a(n) > 0 for all n > 2. In other words, for any prime p > 3, there is a Fibonacci number among 1, ..., (p-1)/2 which is a quadratic nonresidue modulo p.
(ii) For any n > 2, there is a prime q < prime(n) such that the q-th Fibonacci number is a quadratic nonresidue modulo prime(n).
(iii) For any odd prime p, there is a Lucas number (i.e., a term of A000032) smaller than p which is a quadratic nonresidue modulo p.
We have checked part (i) for all primes p < 3*10^9, part (ii) for n up to 10^8, and part (iii) for the first 10^7 primes.
See also A241604 for a sequence related to part (i) of the conjecture.

			a(3) = 1 since F(3) = 2 is a quadratic nonresidue modulo prime(3) = 5, where F(n) denotes the n-th Fibonacci number.
a(4) = 1 since F(4) = 3 is a quadratic nonresidue modulo prime(4) = 7.
a(5) = 1 since F(3) = 2 is a quadratic nonresidue modulo prime(5) = 11.
a(9) = 1 since F(5) = 5 is a quadratic nonresidue modulo prime(9) = 23.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Z.-W. Sun, New observations on primitive roots modulo primes, arXiv preprint arXiv:1405.0290 [math.NT], 2014.

Cf. A000040, A000032, A000045, A236966, A239957, A239963, A241492, A241504, A241516, A241604.

f[k_]:=Fibonacci[k]
Do[m=0;Do[If[f[k]>Prime[n]/2,Goto[aa]];If[JacobiSymbol[f[k],Prime[n]]==-1,m=m+1];Continue,{k,2,(Prime[n]+1)/2}]; Label[aa];Print[n," ",m];Continue,{n,1,80}]

A241472 a(n) = |{0 < k < sqrt(prime(n)): k^2 + 1 is a quadratic residue modulo prime(n)}|.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 3, 2, 4, 2, 3, 3, 3, 1, 5, 5, 4, 6, 5, 5, 3, 5, 5, 3, 3, 4, 7, 3, 7, 4, 6, 7, 7, 4, 4, 3, 6, 8, 8, 6, 5, 8, 7, 8, 6, 6, 8, 11, 8, 6, 7, 7, 5, 9, 2, 8, 3, 11, 10, 8, 6, 8, 7, 10, 5, 8, 8, 9, 13, 10, 10, 6, 6, 10, 11, 10

Offset: 1

Zhi-Wei Sun, Apr 23 2014

nonn

a(n) > 0 for all n > 3. In fact, for any prime p > 5 one of 1^2 + 1 = 2, 2^2 + 1 = 5 and 3^2 + 1 = 10 is a quadratic residue modulo p. Similarly, if p > 7 is a prime not equal to 19, then k^2 - 1 is a quadratic residue modulo p for some positive integer k < sqrt(p). In fact, for any prime p > 7, one of 2^2 - 1 = 3, 3^2 - 1 = 8 and 5^2 - 1 = 24 is a quadratic residue modulo p.

See also A239957 for a similar conjecture involving primitive roots modulo primes. Note that a primitive root modulo an odd prime p must be a quadratic nonresidue modulo p.

			a(6) = 1 since 3^2 + 1 = 10 is a quadratic residue modulo prime(6) = 13.
a(7) = 1 since 1^2 + 1 = 2 is a quadratic residue modulo prime(7) = 17.
a(9) = 1 since 1^2 + 1 = 2 is a quadratic residue modulo prime(9) = 23.
a(10) = 1 since 2^2 + 1 = 5 is a quadratic residue modulo prime(10) = 29.
a(18) = 1 since 2^2 + 1 = 5 is a quadratic residue modulo prime(18) = 61.

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

Cf. A000040, A002522, A239957, A241492.

a[n_]:=Sum[If[JacobiSymbol[k^2+1,Prime[n]]==1,1,0],{k,1,Sqrt[Prime[n]-1]}]
Table[a[n],{n,1,80}]

cp's OEIS Frontend

A241504 a(n) = |{0 < g < prime(n): g is not only a primitive root modulo prime(n) but also a partition number given by A000041}|.

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A241516 Least positive primitive root g < prime(n) modulo prime(n) which is also a partition number given by A000041, or 0 if such a number g does not exist.

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A241568 a(n) = |{0 < k < prime(n)/2: k is not only a quadratic nonresidue modulo prime(n) but also a Fibonacci number}|.

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A241472 a(n) = |{0 < k < sqrt(prime(n)): k^2 + 1 is a quadratic residue modulo prime(n)}|.

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