A280721 a(n) is the n-th b > 1 such that p = prime(n) satisfies b^(p-1) == 1 (mod p^2).

5, 10, 24, 31, 81, 89, 134, 127, 255, 267, 430, 476, 744, 424, 629, 895, 1105, 1079, 1301, 1331, 1440, 2092, 1451, 2466, 2488, 2140, 3326, 2638, 2815, 3517, 4345, 4138, 4505, 4659, 5571, 6021, 5383, 5460, 6071, 6844, 8434, 7076, 8961, 10215, 7522, 9817, 11499

Offset: 1

Author

Felix Fröhlich, Jan 07 2017

Keywords

• nonn

Comments

Main diagonal of A244249.

There are exactly p-1 residue classes b mod p^2 for which b^(p-1) == 1 (mod p^2), of which p-2 will be greater than 1. p-2 > n for n > 3, so a(n) < prime(n)^2 for n > 2 (case n = 3 is fortuitous). - Charles R Greathouse IV, Jun 23 2021

Examples

			For n=2, prime(2)=3, 10^1 == 1 (mod 9), 10^2 == 1 (mod 9), a(2) = 10. - _N. J. A. Sloane_, Jan 14 2017
		

Links

• Robert Israel, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A244249, A259909.

Programs

Maple

f:= proc(n) local p,S,nS,r;
   p:= ithprime(n);
   S:= sort(map(t -> rhs(op(t)), [msolve(b^(p-1)=1, p^2)]));
   nS:= nops(S);
   r:= (n mod nS)+1;
   S[r] + (n+1-r)/nS*p^2;
end proc:
map(f, [$1..100]); # Robert Israel, Jan 09 2017

Mathematica

Table[Function[p, Select[Range[p^2 + 1], PowerMod[#, p - 1, p^2] == 1 &][[n + 1]]]@ Prime@ n, {n, 47}] (* Michael De Vlieger, Jan 09 2017 *)

PARI

base(p, n) = my(b=2, i=0); while(1, if(Mod(b, p^2)^(p-1)==1, i++); if(i==n, return(b)); b++)
a(n) = base(prime(n), n)

Formula

a(n) < prime(n)^2 for n > 2, see comments. - Charles R Greathouse IV, Jun 23 2021