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%I #33 Dec 09 2017 19:47:33
%S 3,7,2,110,3,7,3,40,2,2,3,110,3,7,2,392,3,7,3,110,2,7,3,40,3,5,2,110,
%T 3,2,3,894,2,4,3,110,3,7,2,40,3,7,3,110,2,7,3,107,3,2,2,110,3,7,2,40,
%U 2,7,3,110,3,7,2,315,3,7,3,2,2,2,3,40,3,6,2,110,3,2
%N Least integer b > 0 such that b^n + 1 is not squarefree.
%C If m is an odd multiple of n, so m=(2k+1)n, then a(m)=a((2k+1)n)<=a(n). This follows from raising the congruence b^n == -1 (mod p^2) to the (2k+1)th power. Because of this, for all k, a(2k+1) <= a(1)=3, a(2*(2k+1)) <= a(2)=7, a(4*(2k+1)) <= a(4)=110, a(8*(2k+1)) <= a(8)=40, a(16(2k+1)) <= a(16)=392, etc. Also a(3(2k+1)) <= a(3)=2.
%C To show that a(n) is finite for all n, it suffices to show that a(2^j) is finite for all j.
%C Correspondingly (to the first comment), large & in particular record values are obtained for powers of 2: a(1)=3, a(2)=7, a(4)=110, a(16)=392, a(32)=894, ... - _M. F. Hasler_, Oct 08 2014
%C See A248576 for the least prime p such that p^2 divides b^n+1. - _M. F. Hasler_, Oct 08 2014
%C For a criterion for a(n) to be finite when n is a power of two, see A261117. - _Jeppe Stig Nielsen_, Aug 08 2015
%e For n = 12, we have that 110^12 + 1 is divisible by a (nonunit) square (namely by 5^2), and since 110 is minimal with this property, a(12) = 110.
%e For n=32, we have that 894^32 + 1 is divisible by 193^2, and there is no b < 894 such that b^32 + 1 would be divisible by a square > 1. (Conjectural: no factor p^2 with p < 10^6 for any b < 894.) - _M. F. Hasler_, Oct 08 2014
%o (PARI) for(n=1,1000,b=1;while(issquarefree(b^n+1),b++);print1(b,","))
%o (PARI) a(n,bound=b->n*b*20)=for(b=1,9e9,forprime(p=1,bound(b),Mod(b,p^2)^n+1||return(b))) \\ The given default search bound is experimental; might yield only an upper bound as result. You may use, e.g., a(n,b->10^5), for a constant bound. - _M. F. Hasler_, Oct 08 2014
%Y Cf. A248576, A260824, A261117.
%K nonn
%O 1,1
%A _Jeppe Stig Nielsen_, Oct 04 2014
%E More terms from _M. F. Hasler_, Oct 08 2014
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