A143548 -id:A143548 - cp's OEIS frontend

A134307 Primes p such that A^(p-1) == 1 (mod p^2) for some A in the range 2 <= A <= p-1.

11, 29, 37, 43, 59, 71, 79, 97, 103, 109, 113, 127, 131, 137, 151, 163, 181, 191, 197, 199, 211, 223, 229, 233, 241, 257, 263, 269, 281, 283, 293, 307, 313, 331, 347, 349, 353, 359, 367, 373, 379, 397, 401, 419, 421, 433, 439, 449, 461, 463, 487, 499, 509

Offset: 1

Joerg Arndt, Aug 27 2008

nonn

It's worth observing that there are p-1 elements of order dividing p-1 modulo p^2 that are of the form r^(k*p) mod p^2 where r is a primitive element modulo p and k=0,1,...,p-2. Heuristically, one can expect that at least one of them belongs to the interval [2,p-1] with probability about 1 - (1 - 1/p)^(p-1) ~= 1 - 1/e.
Numerically, among the primes below 1000 (out of the total number pi(1000)=168) there are 103 terms of the sequence, and the ratio 103/168 = 0.613 which is already somewhat close to 1-1/e ~= 0.632.
If we replace p^2 with p^3, heuristically it is likely that the sequence is finite (since 1 - (1 - 1/p^2)^(p-1) tends to 0 as p grows). - Max Alekseyev, Jan 09 2009
Replacing p^2 with p^3 gives just the one term (113) for p < 10^6. - Joerg Arndt, Jan 07 2011
If furthermore the number A can be taken to be a primitive root modulo p, i.e., A is a generator of (Z/pZ)*, then that p belongs to A060503. - Jeppe Stig Nielsen, Jul 31 2015
The number of terms not exceeding prime(10^k), for k=1,2,..., are 2, 55, 652, 6303, 63219, ... - Amiram Eldar, May 08 2021

			Examples (pairs [p, A]):
[11, 3]
[11, 9]
[29, 14]
[37, 18]
[43, 19]
[59, 53]
[71, 11]
[71, 26]
[79, 31]
[97, 53]

L. E. Dickson, History of the theory of numbers, vol. 1, p. 105.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)
Wilfrid Keller and Jörg Richstein Fermat quotients that are divisible by p.

Cf. A001220, A055578, A039678, A143548, A222184, A060503.

Select[ Prime[ Range[100]], Product[ (PowerMod[a, # - 1, #^2] - 1), {a, 2, # - 1}] == 0 &] (* Jonathan Sondow, Feb 11 2013 *)

{ forprime (p=2, 1000,
   for (a=2, p-1, p2 = p^2;
     if( Mod(a, p2)^(p-1) == Mod(1, p2), print1(p, ", ") ;break() );
  ); ); }

A222184 Primes p such that q^(p-1) == 1 (mod p^2) for some prime q < p.

Original entry on oeis.org

11, 43, 59, 71, 79, 97, 103, 137, 263, 331, 349, 359, 421, 433, 487, 523, 653, 659, 743, 859, 863, 907, 919, 983, 1069, 1087, 1091, 1093, 1163, 1223, 1229, 1279, 1381, 1483, 1499, 1549, 1657, 1663, 1667, 1697, 1747, 1777, 1787, 1789, 1877, 1993, 2011, 2213, 2221, 2251, 2281, 2309, 2371, 2393, 2473, 2671, 2719, 2777, 2791, 2803, 2833, 2861, 3037, 3079, 3163, 3251, 3257, 3463, 3511, 3557

Offset: 1

Jonathan Sondow, Feb 11 2013

nonn

Subsequence of A134307; see its interesting heuristics. (What is the analogous heuristic for the present sequence?)

The smallest corresponding primes q are A222185.

			3 is a prime < 11, and 11^2 divides 3^(11-1)-1 = 59048 = 121*488, so 11 is a member.

L. E. Dickson, History of the Theory of Numbers, vol. 1, chap. IV.

Giovanni Resta, Table of n, a(n) for n = 1..10000
W. Keller and J. Richstein, Fermat quotients that are divisible by p.

Cf. A001220, A039678, A134307, A143548, A222185.

Select[ Prime[ Range[500]], Product[ PowerMod[ Prime[k], # - 1, #^2] - 1, {k, 1, PrimePi[#] - 1}] == 0 &]

lista(nn) = {forprime (p=2, nn, ok = 0; forprime(q=2, p-1, if (Mod(q, p^2)^(p-1) == 1, ok=1; break);); if (ok, print1(p, ", ")););} \\ Michel Marcus, Nov 24 2014

A222185(n)^(a(n)-1) == 1 (mod a(n)^2).

A222185 Least prime q with q^(p-1) == 1 (mod p^2), where p = A222184(n).

Original entry on oeis.org

3, 19, 53, 11, 31, 53, 43, 19, 79, 71, 223, 257, 251, 349, 307, 241, 197, 503, 467, 643, 13, 127, 457, 419, 487, 617, 691, 2, 241, 997, 821, 683, 653, 421, 941, 1069, 1481, 709, 463, 461, 1153, 1381, 631, 449, 1091, 277, 1993, 367, 659, 151, 1657, 823, 1493, 431, 1787, 2063, 1487, 59, 2389, 2131, 479, 1907, 79, 173, 1151, 1831, 419, 1193, 2, 3319

Offset: 1

Jonathan Sondow, Feb 12 2013

nonn

			3 is the smallest prime < A222184(1) = 11 such that 11^2 divides 3^(11-1)-1 = 59048 = 121*488, so a(1) = 3.

L. E. Dickson, History of the Theory of Numbers, vol. 1, chap. IV.

Giovanni Resta, Table of n, a(n) for n = 1..10000
Keller and J. Richstein, Fermat quotients that are divisible by p

Cf. A001220, A039678, A134307, A143548, A222184.

L = Select[ Prime[ Range[500]], Product[ PowerMod[ Prime[k], # - 1, #^2] - 1, {k, 1, PrimePi[#] - 1}] == 0 &]; Table[  Min[ Select[ Prime[ Range[ PrimePi[L[[n]]] - 1]], PowerMod[#, L[[n]] - 1, L[[n]]^2] == 1 &]], {n, 1, Length[L]}]

A247072 Smallest Wieferich prime (> sqrt(n)) in base n.

Original entry on oeis.org

2, 1093, 11, 1093, 20771, 66161, 5, 3, 11, 487, 71, 2693, 863, 29, 29131, 1093, 46021, 5, 7, 281

Offset: 1

Eric Chen, Nov 16 2014

a(n) = Smallest prime such that n appears in A143548. - Eric Chen, Nov 26 2014
The square of a(n) is the smallest squared prime that is a pseudoprime (> n) in base n; for example, a(2) = 1093, and 1093^2 = 1194649 is the smallest squared prime that is pseudoprime in base 2. - Eric Chen, Nov 26 2014
Is a(n) defined for all n? - Eric Chen, Nov 26 2014
Does every prime appear in this sequence? - Eric Chen, Nov 26 2014
a(22)..a(28) = {13, 13, 5, 20771, 71, 11, 19}, a(30)..a(46) = {7, 7, 1093, 233, 46145917691, 1613, 66161, 77867, 17, 8039, 11, 29, 23, 103, 229, 1283, 829}, a(48)..a(49) = {7, 491531}, a(51)..a(60) = {41, 461, 47, 19, 30109, 647, 47699, 131, 2777, 29}, a(62)..a(71) = {19, 23, 1093, 17, 89351671, 47, 19, 19, 13, 47}, a(74)..a(81) = {1251922253819, 17, 37, 32687, 43, 263, 13, 11}, a(83)..a(100) = {4871, 163, 11779, 68239, 1999, 2535619637, 13, 6590291053, 293, 727, 509, 11, 2137, 109, 2914393, 28627, 13, 487}; a(n) is currently unknown for n = {21, 29, 47, 50, 61, 72, 73, 82, 126, 132, 154, 186, 187, 188, 200, 203, 222, 231, 237, 301, 304, 309, 311, 327, 335, 347, 351, 355, 357, 435, 441, 454, 458, 496, 541, 542, 546, 554, 570, 593, 609, 610, 639, 640, 654, 662, 668, 674, 692, 697, 698, 701, 718, 724, 725, 727, 733, 743, 760, 772, 775, 777, 784, 798, 807, 808, 812, 829, 841, 858, 860, 871, 883, 912, 919, 944, 980, 983, 986, ...}. - Eric Chen, Nov 26 2014
a(21) > 3.4 * 10^13. - Eric Chen, Nov 26 2014

			a(12) = 2693 because the Wieferich primes to base 12 are 2693, 123653, ..., and 2693 is greater than sqrt(12), so a(12) = 2693.
a(17) = 46021 because the Wieferich primes to base 17 are 2, 3, 46021, 48947, 478225523351, ..., but neither 2 nor 3 is greater than sqrt(17), so a(17) = 46021.

Eric Chen, Table of known a(n) up to a(1000)
Richard Fischer, Fermat quotients B^(P-1) == 1 (mod P^2)
Wikipedia, Wieferich prime

Cf. A039951, A096082, A174422, A178871.

Cf. A001220, A014127, A123692, A212583, A123693, A045616, A111027, A128667, A234810, A242741, A128668, A244260, A090968, A242982, A128669.

a247072[n_] := Block[{p = Int[Sqrt[n]]+1}, While[!PrimeQ[p] || [p < 10^8 && PowerMod[n, p - 1, p^2] != 1], p++]; If[p == 10^8, 0, p]]; Table[ a247072[n], {n, 100}] (* Eric Chen, Nov 27 2014 *)

a(n)=forprime(p=sqrtint(n)+1,,if(Mod(n^(p-1),p^2)==1,return(p)))
n=1; while(n<101, print1(a(n), ", "); n++) \\ Charles R Greathouse IV, Nov 16 2014

A320535 Number of solutions to (x+1)^p - x^p == 1 (mod p^2) in the range 1 <= x <= p - 2, p = prime(n).

Original entry on oeis.org

0, 0, 0, 2, 0, 2, 0, 2, 0, 0, 2, 2, 0, 2, 0, 0, 12, 2, 2, 0, 2, 8, 6, 0, 2, 0, 2, 0, 2, 0, 2, 0, 0, 2, 0, 2, 2, 2, 0, 0, 6, 2, 0, 8, 0, 2, 2, 2, 6, 2, 0, 0, 2, 0, 0, 0, 0, 2, 2, 0, 2, 0, 2, 0, 2, 0, 2, 8, 0, 2, 0, 0, 2, 2, 2, 0, 0, 2, 0, 2, 6, 8, 0, 2, 2, 6, 0

Offset: 1

Jianing Song, Oct 15 2018

nonn

For primes p > 2, (x+1)^p - x^p == 1 (mod p^2) has trivial solutions x == 0, -1 (mod p) so they are excluded.
Equivalently, a(n) is the number of solutions to x^(p-1) == (x+1)^(p-1) == 1 (mod p^2) in the range 1 <= x <= p^2 - 2, p = prime(n), that is, number of x such that both x and x + 1 occurs in the n-th row of A143548.
All terms shown here are even. The first odd terms are a(183) = 17 and a(490) = 5, with corresponding primes prime(183) = 1093 and prime(490) = 3511. a(n) is odd iff prime(n) is in A001220.
Let g be a primitive root modulo p^2, then (x+1)^p - x^p == 1 (mod p^2) has nontrivial solutions x == g^((p-1)/3) or g^(2*(p-1)/3) (mod p), and x^(p-1) == (x+1)^(p-1) == 1 (mod p^2) has nontrivial solutions x == g^(p*(p-1)/3) or g^(2*p*(p-1)/3) (mod p^2). As a result, if prime(n) == 1 (mod 6) then a(n) > 0. Primes p == 5 (mod 6) such that the equations have nontrivial solutions are listed in A068209.
a(17) = 12 is surprisingly large comparing with its nearby terms. Among the first 1000 terms there are only 7 that are larger than 12. They are a(183) = 17 and a(385) = a(552) = a(582) = a(593) = a(739) = a(922) = 14 (the corresponding primes are 1093, 2659, 4003, 4243, 4339, 5623 and 7213).

			The nontrivial solutions to (x+1)^7 - x^7 == 1 (mod 49) are x == 2, 4 (mod 7); the solutions to x^6 == (x+1)^6 == 1 (mod 49) are x == 18, 30 (mod 49), so a(4) = 2.

Robert Israel, Table of n, a(n) for n = 1..2500

Cf. A001220, A068209, A143548.

f:= proc(n) local p; p:= ithprime(n);
  nops(select(t -> (t+1)^p - t^p  mod p^2 = 1, [$1 .. p - 2]))
end proc:
map(f, [$1..100]); # Robert Israel, Mar 18 2024

a(n) = my(p=prime(n)); sum(x=1, p-2, Mod(x+1, p^2)^p-Mod(x, p^2)^p==1);

Name corrected by Robert Israel, Mar 18 2024

A222206 Least prime p such that q^(p-1) == 1 (mod p^2) for n primes q < p.

Original entry on oeis.org

2, 11, 349, 13691, 24329

Offset: 0

Jonathan Sondow, Feb 12 2013

I found no new terms < 5*10^6. - J. Stauduhar, Mar 23 2013

a(5) > 13*10^6, if it exists. Note that, up to 13*10^6, the only other prime p (apart 24329) such that the congruence is satisfied for 4 primes q < p is 9656869. - Giovanni Resta, May 23 2017

			For the prime p = 349, but for no smaller prime, there are 2 primes q = 223 and 317 < p with  q^(p-1) == 1 (mod p^2), so a(2) = 349.

L. E. Dickson, History of the Theory of Numbers, vol. 1, chap. IV.

W. Keller and J. Richstein, Fermat quotients that are divisible by p. [Broken link]
Wilfrid Keller and Jörg Richstein, Solutions of the congruence a^(p-1) == 1 (mod (p^r)), Math. Comp. 74 (2005), 927-936.

Cf. A001220, A039678, A134307, A143548, A222184, A222185.

f[n_] := Block[{p = 2, q = {}}, While[ Count[ PowerMod[ q, p - 1, p^2], 1] != n, q = Join[q, {p}]; p = NextPrime@ p]; p]; Array[f, 5, 0] (* Robert G. Wilson v, Mar 09 2015 *)

a(n) = {nb = 0; p = 2; while (nb != n, p = nextprime(p+1); nb = 0; forprime(q=2, p-1, if (Mod(q, p^2)^(p-1) == 1, nb ++); if (nb > n, break););); p;} \\ Michel Marcus, Mar 08 2015

A320161 Irregular triangle read by rows: row n lists 0 <= k < p^2 such that p^2 divides A316269(k, p-Kronecker(k^2-4, p)), p = prime(n).

Original entry on oeis.org

0, 1, 3, 0, 1, 8, 0, 1, 24, 0, 1, 10, 39, 48, 0, 1, 27, 36, 37, 84, 85, 94, 120, 0, 1, 6, 29, 34, 61, 108, 135, 140, 163, 168, 0, 1, 25, 45, 56, 75, 82, 132, 157, 207, 214, 233, 244, 264, 288, 0, 1, 42, 43, 73, 88, 106, 120, 161, 200, 241, 255, 273, 288, 318, 319, 360

Offset: 1

Jianing Song, Oct 06 2018

p always divides A316269(k, p-Kronecker(k^2-4, p)), so it's interesting to see when p^2 also divides A316269(k, p-Kronecker(k^2-4, p)).
In the following comments, let p = prime(n). Note that A316269(0, m) and A316269(1, m) is not defined, so here k must be understood as a remainder modulo p^2. because A316269(k+s*p^2, m) == A316269(k, m) (mod p^2).
Let p = prime(n). Every row contains 0, 1 and p^2 - 1. For n >= 3, the n-th row contains p - 2 numbers, whose remainders modulo p form a permutation of {0, 1, 3, 4, ..., p - 3, p - 1}.
Every row is antisymmetric, that is, k is a member iff p^2 - k is, k > 0. As a result, the sum of the n-th row is prime(n)^2*(prime(n) - 3)/2.
Equivalently, for n >= 2, row n lists 0 <= k < p^2 such that p^2 divides A316269(k, (p-Kronecker(k^2-4, p))/2), p = prime(n).

			Table starts
p = 2: 0, 1, 3,
p = 3: 0, 1, 8,
p = 5: 0, 1, 24,
p = 7: 0, 1, 10, 39, 48,
p = 11: 0, 1, 27, 36, 37, 84, 85, 94, 120,
p = 13: 0, 1, 6, 29, 34, 61, 108, 135, 140, 163, 168,
p = 17: 0, 1, 25, 45, 56, 75, 82, 132, 157, 207, 214, 233, 244, 264, 288,
p = 19: 0, 1, 42, 43, 73, 88, 106, 120, 161, 200, 241, 255, 273, 288, 318, 319, 360,
p = 23: 0, 1, 12, 15, 60, 86, 105, 141, 142, 156, 223, 306, 373, 387, 388, 424, 443, 469, 514, 517, 528,
p = 29: 0, 1, 42, 46, 80, 101, 107, 120, 226, 227, 327, 330, 358, 409, 432, 483, 511, 514, 614, 615, 721, 734, 740, 761, 795, 799, 840,
...

Jianing Song, Table of n, a(n) for n = 1..29083 (primes below 600)
Jianing Song, Table of n, a(n) for n = 1..75796 (primes below 1000)
Wikipedia, Wall-Sun-Sun prime

Cf. A143548, A316269, A320162 (discriminant k^2+4, a more studied case).

Cf. A238490 (primes p such that 4 occurs in the corresponding row), A238736 (primes p such that 6 occurs in the corresponding row).

B(k, p) = (([k, -1; 1, 0]^(p-kronecker(k^2-4,p)))[1,2])%(p^2)
forprime(p=2, 50, for(k=0, p^2-1, if(!B(k, p), print1(k, ", ")));print)

A320162 Irregular triangle read by rows: row n lists 0 <= k < p^2 such that p^2 divides A172236(k, p-Kronecker(k^2+4, p)), p = prime(n).

Original entry on oeis.org

0, 0, 4, 5, 0, 7, 18, 0, 12, 20, 24, 25, 29, 37, 0, 5, 18, 19, 24, 43, 78, 97, 102, 103, 116, 0, 2, 14, 45, 70, 82, 87, 99, 124, 155, 167, 0, 24, 38, 40, 63, 83, 103, 105, 184, 186, 206, 226, 249, 251, 265, 0, 31, 37, 63, 79, 100, 137, 144, 150, 180, 181, 211, 217, 224, 261, 282, 298, 324, 330

Offset: 1

Jianing Song, Oct 06 2018

p always divides A172236(k, p-Kronecker(k^2+4, p)), so it's interesting to see when p^2 also divides A172236(k, p-Kronecker(k^2+4, p)). If this holds, then p is called a k-Wall-Sun-Sun prime (and thus being a (k + s*p^2)-Wall-Sun-Sun prime for all integer s). Specially, there is no Wall-Sun-Sun prime below 10^14 implies that there is no 1 in the first pi(10^14) rows.
Note that A172236(0, m) is not defined, so here k must be understood as a remainder modulo p^2. because A172236(k+s*p^2, m) == A172236(k, m) (mod p^2).
Let p = prime(n). Every row contains 0. For n >= 2, if p == 3 (mod 4), then the n-th row contains p numbers, whose remainders modulo p form a permutation of {0, 1, 2, 3, ..., p - 2, p - 1}. If p == 1 (mod 4), then the n-th row contains p - 2 numbers, whose remainders modulo p form a permutation of {0, 1, 2, 3, ..., p - 2, p - 1} \ {+-2*((p - 1)/2)! mod p}.
Every row is antisymmetric, that is, k is a member iff p^2 - k is, k > 0. As a result, the sum of the n-th row is prime(n)^2*(prime(n) - 1)/2 if prime(n) == 3 (mod 4) and prime(n)^2*(prime(n) - 3)/2 if prime(n) == 1 (mod 4).
Equivalently, if p = prime(n) == 1 (mod 4), then row n lists 0 <= k < p^2 such that p^2 divides A172236(k, (p-Kronecker(k^2+4, p))/2). - Jianing Song, Jul 06 2019

			Table starts
p = 2: 0,
p = 3: 0, 4, 5,
p = 5: 0, 7, 18,
p = 7: 0, 12, 20, 24, 25, 29, 37,
p = 11: 0, 5, 18, 19, 24, 43, 78, 97, 102, 103, 116,
p = 13: 0, 2, 14, 45, 70, 82, 87, 99, 124, 155, 167,
p = 17: 0, 24, 38, 40, 63, 83, 103, 105, 184, 186, 206, 226, 249, 251, 265,
p = 19: 0, 31, 37, 63, 79, 100, 137, 144, 150, 180, 181, 211, 217, 224, 261, 282, 298, 324, 330,
p = 23: 0, 21, 30, 38, 40, 70, 79, 89, 111, 149, 198, 248, 281, 331, 380, 418, 440, 450, 459, 489, 491, 499, 508,
p = 29: 0, 15, 40, 41, 49, 51, 56, 64, 74, 84, 126, 182, 204, 381, 460, 637, 659, 715, 757, 767, 777, 785, 790, 792, 800, 801, 826,
...

Jianing Song, Table of n, a(n) for n = 1..29193 (primes below 600)
Jianing Song, Table of n, a(n) for n = 1..75966 (primes below 1000)
Wikipedia, Wall-Sun-Sun prime

Cf. A143548, A172236, A320161 (discriminant k^2-4).

Cf. A238736 (primes p such that 2 occurs in the corresponding row).

B(k, p) = (([k, 1; 1, 0]^(p-kronecker(k^2+4, p)))[1, 2])%(p^2)
forprime(p=2, 50, for(k=0, p^2-1, if(!B(k, p), print1(k, ", "))); print)

A143811 Number of numbers k

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 1, 2, 1, 2, 2, 3, 3, 2, 1, 1, 2, 1, 3, 1, 1, 1, 2, 2, 1, 2, 2, 3, 2, 1, 3, 2, 1, 2, 1, 2, 2, 4, 1, 1, 2, 2, 2, 2, 1, 3, 1, 4, 1, 3, 3, 3, 3, 3, 2, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 3, 1, 1, 5, 1, 2, 1, 3, 2, 2, 1, 2, 2, 2, 1, 4

Offset: 1

T. D. Noe, Sep 02 2008

nonn

Note that a(n)>0 because k=1 is always a solution. The primes for which a(n)>1 are given in A134307. The values of k are the terms

A143548. The largest known terms in this sequence are for the Wieferich primes 1093 and 3511, for which we have a(183)=11 and a(490)=12, respectively. It is not hard to show that k=p-1 is never a solution for odd prime p. In fact, (p-1)^(p-1)=p+1 (mod p^2) for odd prime p.

T. D. Noe, Table of n, a(n) for n=1..10000

Table[p=Prime[n]; s=Select[Range[p-1], PowerMod[ #,p-1,p^2]==1&]; Length[s], {n,100}]

A317706 Irregular triangle of numbers k < p^2 such that k is a primitive root modulo p but not p^2, p = prime(n).

Original entry on oeis.org

1, 8, 7, 18, 19, 31, 40, 94, 112, 118, 19, 80, 89, 150, 40, 65, 75, 131, 158, 214, 224, 249, 116, 127, 262, 299, 307, 333, 28, 42, 63, 130, 195, 263, 274, 352, 359, 411, 14, 60, 137, 221, 374, 416, 425, 467, 620, 704, 781, 827, 115, 117, 145, 229, 414, 513, 623, 726

Offset: 1

Jianing Song, Aug 05 2018

Also row n lists numbers k < p^2 such that the multiplicative order of k modulo p^2 is p - 1.
Row n has phi(prime(n) - 1) = A008330(n) terms.
Row sum is congruent to mu(prime(n) - 1) = A089451(n) modulo prime(n)^2, where mu is the Moebius function. For n >= 3, the product of n-th row is congruent to 1 modulo prime(n)^2.
Does every integer appear in this sequence? For example, 3 does not appear until the prime 1006003 and 5 does not appear until the prime 40487. Where does 2 first appear?

			(2)   1,
(3)   8,
(5)   7, 18,
(7)   19, 31,
(11)  40, 94, 112, 118,
(13)  19, 80, 89, 150,
(17)  40, 65, 75, 131, 158, 214, 224, 249,
(19)  116, 127, 262, 299, 307, 333,
(23)  28, 42, 63, 130, 195, 263, 274, 352, 359, 411,

Cf. A008330, A055578, A089451, A143548.

Table[Select[Range[p^2 - 1], MultiplicativeOrder[#, p^2] == p - 1 &], {p, Prime@ Range@ 11}] // Flatten (* Michael De Vlieger, Aug 05 2018 *)

forprime(p=2,100,for(i=1,p^2,if(Mod(i,p)!=0,if(znorder(Mod(i,p^2))==p-1,print1(i, ", ")))))

cp's OEIS Frontend

A134307 Primes p such that A^(p-1) == 1 (mod p^2) for some A in the range 2 <= A <= p-1.

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A222184 Primes p such that q^(p-1) == 1 (mod p^2) for some prime q < p.

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A222185 Least prime q with q^(p-1) == 1 (mod p^2), where p = A222184(n).

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A247072 Smallest Wieferich prime (> sqrt(n)) in base n.

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A320535 Number of solutions to (x+1)^p - x^p == 1 (mod p^2) in the range 1 <= x <= p - 2, p = prime(n).

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Maple

PARI

Extensions

A222206 Least prime p such that q^(p-1) == 1 (mod p^2) for n primes q < p.

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Mathematica

PARI

A320161 Irregular triangle read by rows: row n lists 0 <= k < p^2 such that p^2 divides A316269(k, p-Kronecker(k^2-4, p)), p = prime(n).

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