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a(1) = 3. For n > 1, a(n) = smallest prime q such that q^(a(n-1)-1) == 1 (mod a(n-1)^2).
5

%I #6 Jul 23 2023 18:59:39

%S 3,17,131,659,503,9833,49603,327317,13900147,144229223,5872276013

%N a(1) = 3. For n > 1, a(n) = smallest prime q such that q^(a(n-1)-1) == 1 (mod a(n-1)^2).

%C Is this overall an increasing sequence or does it enter a cycle?

%C The sequence decreases for the first time at n = 5.

%t sp[n_]:=Module[{p=2},While[PowerMod[p,n-1,n^2]!=1,p=NextPrime[p]];p]; NestList[sp,3,8] (* _Harvey P. Dale_, Jul 23 2023 *)

%o (PARI) seq(start, terms) = my(x=start, i=1); print1(start, ", "); while(1, forprime(q=1, , if(Mod(q, x^2)^(x-1)==1, print1(q, ", "); x=q; i++; if(i >= terms, break({2}), break))))

%o seq(3, 20) \\ Print initial 20 terms of sequence

%Y Row n = 2 of A249162.

%Y Cf. A355597, A355599, A355600, A355601, A355602.

%K nonn,hard,more

%O 1,1

%A _Felix Fröhlich_, Jul 09 2022