A281747 Smallest b > 1 such that p = prime(n) satisfies b^(p-1) == 1 (mod p^p).

5, 26, 1068, 82681, 5392282366, 11356596271444, 34451905517028761171, 340625514346676110671584, 308318432223607315018221180590, 8566187045843934976180705488213013173127, 1099862052702774330481800364074681495062836757, 8170421001593885871548404108552563632485969048059688187

Offset: 1

Author

Felix Fröhlich, Jan 29 2017

Keywords

• nonn
• hard

Comments

a(n) is the element in row prime(n), column n of the table in A257833.

Is the sequence always nondecreasing, or stronger, is it always increasing?

For odd primes p, if c is a primitive root mod p^p then b == c^(p^(p-1)) (mod p^p) satisfies this. Thus a(n) < prime(n)^prime(n) for n > 1. - Robert Israel, Jan 30 2017

Links

• Robert Israel, Table of n, a(n) for n = 1..76
• W. Keller and J. Richstein, Solutions of the congruence a^(p-1) == 1 (mod p^r), Math. Comp. 74 (2005), 927-936.

Crossrefs

Cf. A257833.

Programs

Maple

f:= proc(p) local c,j;
       c:= numtheory:-primroot(p^p);
       min(seq(c &^ (j*p^(p-1)) mod p^p, j=1..p-2))
end proc:
5, seq(f(ithprime(i)),i=2..15); # Robert Israel, Jan 30 2017

Mathematica

Table[b = 2; While[PowerMod[b, (# - 1), #^#] &@ Prime@ n != 1, b++]; b, {n, 4}] (* Michael De Vlieger, Jan 30 2017 *)

PARI

a(n) = my(p=prime(n), b=2); while(Mod(b, p^p)^(p-1)!=1, b++); b

Extensions

More terms from Robert Israel, Jan 30 2017