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

%I A281747 #34 Aug 08 2025 14:21:32
%S A281747 5,26,1068,82681,5392282366,11356596271444,34451905517028761171,
%T A281747 340625514346676110671584,308318432223607315018221180590,
%U A281747 8566187045843934976180705488213013173127,1099862052702774330481800364074681495062836757,8170421001593885871548404108552563632485969048059688187
%N A281747 Smallest b > 1 such that p = prime(n) satisfies b^(p-1) == 1 (mod p^p).
%C A281747 a(n) is the element in row prime(n), column n of the table in A257833.
%C A281747 Is the sequence always nondecreasing, or stronger, is it always increasing?
%C A281747 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
%H A281747 Robert Israel, <a href="/A281747/b281747.txt">Table of n, a(n) for n = 1..76</a>
%H A281747 W. Keller and J. Richstein, <a href="https://doi.org/10.1090/S0025-5718-04-01666-7">Solutions of the congruence a^(p-1) == 1 (mod p^r)</a>, Math. Comp. 74 (2005), 927-936.
%p A281747 f:= proc(p) local c,j;
%p A281747        c:= numtheory:-primroot(p^p);
%p A281747        min(seq(c &^ (j*p^(p-1)) mod p^p, j=1..p-2))
%p A281747 end proc:
%p A281747 5, seq(f(ithprime(i)),i=2..15); # _Robert Israel_, Jan 30 2017
%t A281747 Table[b = 2; While[PowerMod[b, (# - 1), #^#] &@ Prime@ n != 1, b++]; b, {n, 4}] (* _Michael De Vlieger_, Jan 30 2017 *)
%o A281747 (PARI) a(n) = my(p=prime(n), b=2); while(Mod(b, p^p)^(p-1)!=1, b++); b
%Y A281747 Cf. A257833.
%K A281747 nonn,hard
%O A281747 1,1
%A A281747 _Felix Fröhlich_, Jan 29 2017
%E A281747 More terms from _Robert Israel_, Jan 30 2017