%I
%S 8,7,110,40,1497,894,315,48,166107,95853,63609,71589,492348,209628,
%T 388440,48853,6118793,2684186,25787045,49643800,54302036,3969770538,
%U 17592956651,7347360617,991255542,8249087392,11518171450,51385581002,2268777293,21252616802,2822082710511
%N Smallest positive integer b such that b^(2^n)+1 is divisible by the square of A035089(n+1).
%C For given n, if A035089(n+1) exists (which is true by Dirichlet's theorem on arithmetic progressions), then a(n) exists. Proof: p := A035089(n+1) is a prime of the form p=k*2^(n+1)+1, then the group (Z/(p^2)Z)* is cyclic of order p*(p1) = p*k*2^(n+1). It therefore has an element b of order exactly 2^(n+1). For that b we have then b^(2^n) == 1 (mod p^2).
%C For given n, a(n) is not necessarily the smallest b such that b^(2^n)+1 is nonsquarefree; see A260824.
%e Consider n=4, hence generalized Fermat numbers b^16+1. The first prime (A035089(4+1)) of the form 32*k+1 is 97. It follows that 97 is the smallest prime whose square divides a number of the form b^16+1. The first time 97^2 divides b^16+1 is for b=1497. Hence a(4)=1497. However, A260824(4) is smaller, A260824(4)=392. This is because already 392^16+1 is nonsquarefree (but the prime with a square dividing it, 769, exceeds 97).
%o (PARI) a(n)=for(k=1,10^10,p=(k<<(n+1))+1;if(isprime(p),break()));for(b=1,p^2,b%p!=0&Mod(b,p^2)^(1<<n)==1&return(b))
%o (PARI) a(n)=for(k=1, 10^10, p=(k<<(n+1))+1; if(isprime(p), break())); e=p*(p1)/(1<<(n+1)); h=znprimroot(p^2)^e; g=h^2; m=p^2; for(i=1,1<<n,m=min(m,lift(h));h*=g); m
%Y Cf. A260824, A248214, A035089.
%K nonn
%O 0,1
%A _Jeppe Stig Nielsen_, Aug 08 2015
