A001918 -id:A001918 - cp's OEIS frontend

A046146 Largest primitive root modulo n, or 0 if no root exists.

0, 0, 1, 2, 3, 3, 5, 5, 0, 5, 7, 8, 0, 11, 5, 0, 0, 14, 11, 15, 0, 0, 19, 21, 0, 23, 19, 23, 0, 27, 0, 24, 0, 0, 31, 0, 0, 35, 33, 0, 0, 35, 0, 34, 0, 0, 43, 45, 0, 47, 47, 0, 0, 51, 47, 0, 0, 0, 55, 56, 0, 59, 55, 0, 0, 0, 0, 63, 0, 0, 0, 69, 0, 68, 69, 0, 0, 0, 0, 77, 0, 77, 75, 80, 0, 0

Offset: 0

Eric W. Weisstein

nonn

The value 0 at index 0 says 0 has no primitive roots, but the 0 at index 1 says 1 has a primitive root of 0, the only real 0 in the sequence. - Initial terms corrected by Harry J. Smith, Jan 27 2005

a(n) is nonzero if and only if n is 2, 4, or of the form p^k, or 2*p^k where p is an odd prime and k>0. - Tom Edgar, Jun 02 2014

T. D. Noe, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Primitive Root.

Cf. A001918, A046144, A046145, A002233, A071894, A219027, A008330, A010554.

f[n_] := Block[{pr = PrimitiveRootList[n]}, If[pr == {}, 0, pr[[-1]]]]; Array[f, 86, 0] (* Robert G. Wilson v, Nov 03 2014 *)

for(i=0,100,p=0;for(q=1,i-1,if(gcd(q,i)==1&&znorder(Mod(q,i))==eulerphi(i),p=q));print1(p",")) /* V. Raman, Nov 22 2012 */

Initial terms corrected by Harry J. Smith, Jan 27 2005

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

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}]

A002233 a(1) = 1; for n > 1, a(n) = least positive prime primitive root of n-th prime.

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, 19, 5, 2, 3, 2, 3, 2, 7, 3, 7, 7, 11, 3, 5, 2, 43, 5, 3, 3, 2, 5, 17, 17, 2, 3, 19, 2, 2, 3, 7, 11, 2, 2, 5, 2, 5, 3, 29, 2, 2, 7, 5, 17, 2, 3, 13, 2, 3, 2, 13, 3, 2, 7, 5, 2, 3, 2, 2, 2

Offset: 1

N. J. A. Sloane

According to Section F9 in Guy's book "Unsolved Problems in Number Theory" (Springer, 2004), P. Erdős asked whether for any large prime p there is a prime q < p so that q is a primitive root modulo p. See also the comments on A223942 related to this sequence. - Zhi-Wei Sun, Mar 29 2013

For n >= 2 the Dirichlet characters modulo prime(n), {Chi_{prime n}{(r,m)}, for n >= 1, r=1..(prime(n)-1) and m = 2..prime(n)-1, are determined from those for m = a(n), i.e., Chi_{prime n}(r,a(n)) = exp(2*Pi*I*(r-1)/(prime(n)-1)) and the power sequence S(n) := {a(n)^k (mod prime(n)), k = 1..(prime(n)-2)} by the strong multiplicity of Chi as Chi_{prime n}(r,m) = (Chi_{prime n}(r,a(n)))^{pos(m,S(n))} where S(n){pos(m,S(n))} = m. For m=1 Chi is always 1. For m = prime(n) Chi is always 0. For n=1 (prime 2) the characters are 1, 0 for r = 1 and m = 1, 2, respectively. See the example for a(4) below. - _Wolfdieter Lang, Jan 19 2017

			n=4, a(4) = 3: Dirichlet characters for prime(4) = 7 from Chi_7(r,3) = exp(Pi*I*(r-1)/3) and the power sequence S(4) = [3, 2, 6, 4, 5]. Hence Chi_7(r,2) = Chi_7(r,3)^2 = exp(2*Pi*I*(r-1)/3), Chi_7(r,4) = Chi_7(r,3)^4, Chi_7(r,5) = Chi_7(r,3)^5, Chi_7(r,6) = Chi_7(r,3)^3. Chi_7(r,1) = 1 and Chi_7(r,7) = 0, for r=1..6. This produces the character modulo 7 table. See the Apostol reference, p. 139, with interchanged rows r = 2..6. - _Wolfdieter Lang_, Jan 19 2017

T. M. Apostol, An Introduction to Analytic Number Theory, Springer-Verlag, NY, 1976, 1986, p. 139.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
A. E. Western and J. C. P. Miller, Tables of Indices and Primitive Roots. Royal Society Mathematical Tables, Vol. 9, Cambridge Univ. Press, 1968, p. 2.

T. D. Noe, Table of n, a(n) for n = 1..10000
A. E. Western and J. C. P. Miller, Tables of Indices and Primitive Roots, Royal Society Mathematical Tables, Vol. 9, Cambridge Univ. Press, 1968 [Annotated scans of selected pages]

See A122028 (least primitive root that is prime), A001918 (least positive primitive root), A223942.

a[1] = 1; a[n_] := (p = Prime[n]; Select[Range[p], PrimeQ[#] && MultiplicativeOrder[#, p] == EulerPhi[p] &, 1]) // First; Table[a[n], {n, 100}] (* Jean-François Alcover, Mar 30 2011 *)
a[1] = 1; a[n_] := SelectFirst[PrimitiveRootList[Prime[n]], PrimeQ]; Array[a, 101] (* Jean-François Alcover, Sep 28 2016 *)

leastroot(p)=forprime(q=2,p,if(znorder(Mod(q,p))+1==p,return(q)))
a(n)=if(n>1,leastroot(prime(n)),1) \\ Charles R Greathouse IV, Mar 20 2013

a(n) = A122028(n) for n>1. - Jonathan Sondow, May 18 2017

A023048 Smallest prime having least positive primitive root n, or 0 if no such prime exists.

Original entry on oeis.org

2, 3, 7, 0, 23, 41, 71, 0, 0, 313, 643, 4111, 457, 1031, 439, 0, 311, 53173, 191, 107227, 409, 3361, 2161, 533821, 0, 12391, 0, 133321, 15791, 124153, 5881, 0, 268969, 48889, 64609, 0, 36721, 55441, 166031, 1373989, 156601, 2494381, 95471, 71761, 95525767

Offset: 1

David W. Wilson

nonn

a(n) = 0 iff n is a perfect power m^k, m >= 1, k >= 2 (i.e., a member of A001597).

Of course if n is a perfect power then a(n) = 0, but it seems that the other direction is true only assuming the generalized Artin's conjecture. See the link from Tomás Oliveira e Silva below. - Jianing Song, Jan 22 2019

			a(2) = 3, since 3 has 2 as smallest positive primitive root and no prime p < 3 has 2 as smallest positive primitive root.
a(24) = 533821, since prime 533821 has 24 as smallest positive primitive root and no prime p < 533821 has 24 as smallest positive primitive root.

A. E. Western and J. C. P. Miller, Tables of Indices and Primitive Roots. Royal Society Mathematical Tables, Vol. 9, Cambridge Univ. Press, 1968, p. XLIV.

N. J. A. Sloane, Table of n, a(n) for n = 1..107 (from the web page of Tomás Oliveira e Silva)
Wouter Meeussen, Smallest Primes with Specified Least Primitive Root
Tomás Oliveira e Silva, Least primitive root of prime numbers
Index entries for primes by primitive root

Cf. A001122-A001126, A061323-A061335, A061730-A061741.
Indices of the primes: A066529.
For records see A133433. See A133432 for a version without the 0's.

t = Table[0, {100}]; Do[a = PrimitiveRoot@Prime@n; If[a < 101 && t[[a]] == 0, t[[a]] = n], {n, 10^6}]; Unprotect[Prime]; Prime[0] = 0; Prime@t; Clear[Prime]; Protect[Prime] (* Robert G. Wilson v, Dec 15 2005 *)

from sympy import nextprime, perfect_power, primitive_root
def a(n):
    if perfect_power(n): return 0
    p = 2
    while primitive_root(p) != n: p = nextprime(p)
    return p
print([a(n) for n in range(1, 40)]) # Michael S. Branicky, Feb 13 2023

# faster version for initial segment of sequence
from itertools import count, islice
from sympy import nextprime, perfect_power, primitive_root
def agen(): # generator of terms
    p, adict, n = 2, {None: 0}, 1
    for k in count(1):
        v = primitive_root(p)
        if v not in adict:
            adict[v] = p
        if perfect_power(n): adict[n] = 0
        while n in adict: yield adict[n]; n += 1
        p = nextprime(p)
print(list(islice(agen(), 40))) # Michael S. Branicky, Feb 13 2023

a(n) = min { prime(k) | A001918(k) = n } U {0} = A000040(A066529(n)) (or zero). - M. F. Hasler, Jun 01 2018

Comment corrected by Christopher J. Smyth, Oct 16 2013

A143548 Irregular triangle of numbers k < p^2 such that p^2 divides k^(p-1)-1, with p=prime(n).

Original entry on oeis.org

1, 1, 8, 1, 7, 18, 24, 1, 18, 19, 30, 31, 48, 1, 3, 9, 27, 40, 81, 94, 112, 118, 120, 1, 19, 22, 23, 70, 80, 89, 99, 146, 147, 150, 168, 1, 38, 40, 65, 75, 110, 131, 134, 155, 158, 179, 214, 224, 249, 251, 288, 1, 28, 54, 62, 68, 69, 99, 116, 127, 234, 245, 262, 292, 293, 299, 307, 333, 360

Offset: 1

T. D. Noe, Aug 24 2008

Row n begins with 1 and has prime(n)-1 terms. The first differences of each row are symmetric. For k > p^2, the solutions are just shifted by m*p^2 for m > 0. An open question is whether every integer appears in this sequence. For instance, 2 does not appear until the prime 1093 and 5 does not appear until the prime 20771.
For row n > 1, the sum of the terms in row n is (p-1)*p^2*(p+1)/2, which is A138416. - T. D. Noe, Aug 24 2008, corrected by Robert Israel, Sep 27 2016
Max Alekseyev points out that there is a much faster method of computing these numbers. Let p=prime(n) and let r be a primitive root of p (see A001918 and A060749). Then the terms in row n are r^(k*p) (mod p^2) for k=0..p-2. - T. D. Noe, Aug 26 2008
The numbers in this sequence are the bases to Euler pseudoprimes q, which are squares of prime numbers, such that n^((q-1)/2) == +-1 mod q. An exception is the first number 9 = 3*3, which is, following the strict definition in Crandall and Pomerance, no Fermat pseudoprime and hence no Euler pseudoprime. - Karsten Meyer, Jan 08 2011
For row n > 1, the sum is zero modulo p^2 (rows are antisymmetric due to Binomial Theorem). - Peter A. Lawrence, Sep 11 2016

			(2)   1,
(3)   1, 8,
(5)   1, 7, 18, 24,
(7)   1, 18, 19, 30, 31, 48,
(11)  1, 3, 9, 27, 40, 81, 94, 112, 118, 120,
(13)  1, 19, 22, 23, 70, 80, 89, 99, 146, 147, 150, 168,
(17)  1, 38, 40, 65, 75, 110, 131, 134, 155, 158, 179, 214, 224, 249, 251, 288,

R. Crandall and C. Pomerance, Prime Numbers: A Computational Perspective, Springer, NY, 2005

T. D. Noe, Rows n=1..50 of triangle, flattened

Cf. A039678, A056020, A056021, A056022, A056024, A056025, A056027, A056028, A056031, A056034, A056035, A096082, A138416.

f:= proc(n) local p,j,x;
  p:= ithprime(n);
  x:= numtheory:-primroot(p);
  op(sort([seq(x^(i*p) mod p^2, i=0..p-2)]))
end proc:
map(f, [$1..20]); # Robert Israel, Sep 27 2016

Flatten[Table[p=Prime[n]; Select[Range[p^2], PowerMod[ #,p-1,p^2]==1&], {n,50}]] (* T. D. Noe, Aug 24 2008 *)
Flatten[Table[p=Prime[n]; r=PrimitiveRoot[p]; b=PowerMod[r,p,p^2]; Sort[NestList[Mod[b*#,p^2]&,1,p-2]], {n,50}]] (* Faster version from T. D. Noe, Aug 26 2008 *)

A055578 "Non-generous primes": primes p whose least positive primitive root is not a primitive root of p^2.

Original entry on oeis.org

2, 40487, 6692367337

Offset: 1

Bernard Leak (bernard(AT)brenda-arkle.demon.co.uk), Aug 24 2000

For r a primitive root of a prime p, r + qp is a primitive root of p: but r + qp is also a primitive root of p^2, except for q in some unique residue class modulo p. In the exceptional case, r + qp has order p-1 modulo p^2 (Burton, section 8.3).
No other terms below 10^12 (Paszkiewicz, 2009).
Each term p is a Wieferich prime to base A046145(p). For example, a(2) and a(3) are base-5 Wieferich (see A123692). - Jeppe Stig Nielsen, Mar 06 2020

David Burton, Elementary Number Theory, Allyn and Bacon, Boston, 1976, first edition (cf. Section 8.3).

Joerg Arndt, Matters Computational (The Fxtbook), section 39.7.2, p.780.
Stephen Glasby, Three questions about the density of certain primes, Posting to Number Theory List (NMBRTHRY(AT)LISTSERV.NODAK.EDU), Apr 22, 2001.
Bryce Kerr, Kevin McGown, Tim Trudgian, The least primitive root modulo p^2, arXiv:1908.11497 [math.NT], 2019.
A. Paszkiewicz, A new prime for which the least primitive root (mod p) and the least primitive root (mod p^2) are not equal, Math. Comp. 78 (2009), 1193-1195.

Cf. A060503, A060504.

Select[Prime@Range[7!], ! PrimitiveRoot[#] == PrimitiveRoot[#^2] &] (* Arkadiusz Wesolowski, Sep 06 2012 *)

Prime A000040(n) is in this sequence iff A001918(n)^(A000040(n)-1) == 1 (mod A000040(n)^2).

Prime A000040(n) is in this sequence iff A001918(n) differs from A127807(n).

a(3) from Stephen Glasby (Stephen.Glasby(AT)cwu.EDU), Apr 22 2001

Edited by Max Alekseyev, Nov 10 2011

A071894 Largest positive primitive root (

Original entry on oeis.org

1, 2, 3, 5, 8, 11, 14, 15, 21, 27, 24, 35, 35, 34, 45, 51, 56, 59, 63, 69, 68, 77, 80, 86, 92, 99, 101, 104, 103, 110, 118, 128, 134, 135, 147, 146, 152, 159, 165, 171, 176, 179, 189, 188, 195, 197, 207, 214, 224, 223, 230, 237, 234, 248, 254, 261, 267, 269, 272

Offset: 1

N. J. A. Sloane, Jun 11 2002

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 864.
R. Osborn, Tables of All Primitive Roots of Odd Primes Less Than 1000, Univ. Texas Press, 1961.

T. D. Noe, Table of n, a(n) for n=1..10000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

A diagonal of triangle in A060749.

Cf. A000040, A001918 (least positive primitive root), A002199.

f[n_] := Block[{k = Prime[n] - 1, p = Prime[n], t = Table[i, {i, 1, Prime[n] - 1}]}, While[ Union[ PowerMod[ k, t, p]] != t, k-- ]; k]; Table[ f[n], {n, 1, 60}]

a(n) = my(p=prime(n)); forstep(q=p-1, 1, -1, if(znorder(Mod(q, p))==eulerphi(p), return(q))); \\ Michel Marcus, Sep 28 2023

a(n) = prime(n) - A002199(n) - T. D. Noe, Oct 24 2005

More terms from Robert G. Wilson v, Jun 11 2002

A111076 Smallest positive number of maximal order mod n.

Original entry on oeis.org

1, 1, 2, 3, 2, 5, 3, 3, 2, 3, 2, 5, 2, 3, 2, 3, 3, 5, 2, 3, 2, 7, 5, 5, 2, 7, 2, 3, 2, 7, 3, 3, 2, 3, 2, 5, 2, 3, 2, 3, 6, 5, 3, 3, 2, 5, 5, 5, 3, 3, 5, 7, 2, 5, 2, 3, 2, 3, 2, 7, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 7, 5, 5, 5, 2, 3, 2, 7, 3, 3, 2, 7, 2, 5, 3, 3, 2, 3, 3, 7, 2, 3, 11, 5, 2, 5, 5, 3, 2, 3

Offset: 1

Franklin T. Adams-Watters, Oct 10 2005

			a(6)=5 because order of 1 is 1 and 2 through 4 are not relatively prime to 6, but 5 has order 2, which is the maximum possible.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A002322 (orders); same as A046145 for n with primitive roots; see also A001918 (for primes), A229708.

Table[Min[
  Select[Range[n],
   CoprimeQ[#, n] &&
     MultiplicativeOrder[#, n] == CarmichaelLambda[n] &]], {n, 1, 100}]
(* Geoffrey Critzer, Jan 04 2015 *)

a(n)=if(n==1, return(1)); if(n<5,return(n-1)); my(o=lcm(znstar(n)[2]),k=1); while(gcd(k++,n)>1 || znorder(Mod(k,n))Charles R Greathouse IV, Jul 31 2013

a(n) = A229708(n) if and only if a(n) is prime. - Jonathan Sondow, May 17 2017

A234972 Least prime p < prime(n) such that 2^p - 1 is a primitive root modulo prime(n), or 0 if such a prime p does not exist.

Original entry on oeis.org

0, 0, 2, 2, 3, 3, 2, 2, 3, 2, 2, 17, 3, 2, 5, 2, 5, 3, 3, 3, 5, 2, 11, 2, 3, 2, 13, 3, 7, 2, 2, 5, 2, 2, 2, 3, 11, 2, 11, 2, 3, 7, 7, 7, 2, 2, 2, 2, 5, 3, 2, 3, 3, 7, 2, 3, 2, 11, 5, 2, 2, 2, 5, 5, 5, 2, 2, 5, 3, 3, 2, 3, 7, 7, 2, 7, 2, 3, 2, 7, 5, 31, 3, 3, 5, 3, 2, 5, 2, 2, 5, 5, 2, 3, 3, 5, 2, 2, 7, 7

Offset: 1

Zhi-Wei Sun, Apr 20 2014

nonn

Conjecture: a(n) > 0 for all n > 2.

			a(3) = 2 since 2 is a prime smaller than prime(3) = 5 with 2^2 - 1 = 3 a primitive root modulo prime(3) = 5.

Zhi-Wei Sun, Table of n, a(n) for n = 1..2000

Cf. A000040, A001348, A001918.

gp[g_,p_]:=Mod[g,p]>0&&(Length[Union[Table[Mod[g^k, p],{k,1,p-1}]]]==p-1)
Do[Do[If[gp[2^(Prime[k])-1,Prime[n]],Print[n," ",Prime[k]];Goto[aa]],{k,1,n-1}];Print[n," ",0];Label[aa];Continue,{n,1,100}]

A046147 Triangle read by rows in which row n lists the primitive roots mod n (omitting numbers n without a primitive root).

Original entry on oeis.org

1, 2, 3, 2, 3, 5, 3, 5, 2, 5, 3, 7, 2, 6, 7, 8, 2, 6, 7, 11, 3, 5, 3, 5, 6, 7, 10, 11, 12, 14, 5, 11, 2, 3, 10, 13, 14, 15, 7, 13, 17, 19, 5, 7, 10, 11, 14, 15, 17, 19, 20, 21, 2, 3, 8, 12, 13, 17, 22, 23, 7, 11, 15, 19, 2, 5, 11, 14, 20, 23, 2, 3, 8, 10, 11, 14, 15, 18, 19, 21, 26

Offset: 2

Eric W. Weisstein

			n followed by primitive roots, if any:
1 -
2 1
3 2
4 3
5 2 3
6 5
7 3 5
8 -
9 2 5
10 3 7
11 2 6 7 8
12 -
13 2 6 7 11
...

T. D. Noe, Table of n, a(n) for n = 2..3119 (first 99 nonempty rows of triangle, flattened)
Eric Weisstein's World of Mathematics, Primitive Root.

Cf. A001918, A046144 (row lengths), A046145, A046146.

Cf. A060749, A306252 (1st column), A306253 (last/maximum element)

f:= proc(n) local p,k,m,R;
     p:= numtheory:-primroot(n);
     if p = FAIL then return NULL fi;
     m:= numtheory:-phi(n);
     k:= select(i -> igcd(i,m) = 1, [$1..m-1]);
     op(sort(map(t -> p&^t mod n, k)))
end proc:
f(2):= 1:
map(f, [$2..50]); # Robert Israel, Apr 28 2017

a[n_] := Select[Range[n-1], GCD[#, n] == 1 && MultiplicativeOrder[#, n] == EulerPhi[n]& ]; Table[a[n], {n, 1, 30}] // Flatten (* Jean-François Alcover, Oct 23 2012 *)
PrimitiveRootList[Range[Prime[10]]]//Flatten (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Sep 10 2016 *)

a_row(r) = my(v=[], phi=eulerphi(r)); for(i=1, r-1, if(1 == gcd(r, i) && phi == znorder(Mod(i, r)), v=concat(v, i))); v \\ Ruud H.G. van Tol, Oct 23 2023

Edited by Robert Israel, Apr 28 2017

cp's OEIS Frontend

A046146 Largest primitive root modulo n, or 0 if no root exists.

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

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A002233 a(1) = 1; for n > 1, a(n) = least positive prime primitive root of n-th prime.

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A023048 Smallest prime having least positive primitive root n, or 0 if no such prime exists.

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A143548 Irregular triangle of numbers k < p^2 such that p^2 divides k^(p-1)-1, with p=prime(n).

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A055578 "Non-generous primes": primes p whose least positive primitive root is not a primitive root of p^2.

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A071894 Largest positive primitive root (

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