A248576 -id:A248576 - cp's OEIS frontend

A248214 Least integer b > 0 such that b^n + 1 is not squarefree.

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, 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, 2, 7, 3, 110, 3, 7, 2, 315, 3, 7, 3, 2, 2, 2, 3, 40, 3, 6, 2, 110, 3, 2

Offset: 1

Jeppe Stig Nielsen, Oct 04 2014

nonn

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.
To show that a(n) is finite for all n, it suffices to show that a(2^j) is finite for all j.
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
See A248576 for the least prime p such that p^2 divides b^n+1. - M. F. Hasler, Oct 08 2014
For a criterion for a(n) to be finite when n is a power of two, see A261117. - Jeppe Stig Nielsen, Aug 08 2015

			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.
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

Cf. A248576, A260824, A261117.

for(n=1,1000,b=1;while(issquarefree(b^n+1),b++);print1(b,","))

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

More terms from M. F. Hasler, Oct 08 2014

A283452 The smallest square referenced in A248214 (Least integer b > 0 such that b^n + 1 is not squarefree).

Original entry on oeis.org

4, 25, 9, 289, 4, 25, 4, 289, 9, 25, 4, 289, 4, 25, 9, 591361, 4, 25, 4, 289, 9, 25, 4, 289, 4, 169, 9, 289, 4, 25, 4, 37249, 9, 289, 4, 289, 4, 25, 9, 289, 4, 25, 4, 289, 9, 25, 4, 9409, 4, 25, 9, 289, 4, 25, 121, 289, 9, 25, 4, 289, 4, 25, 9, 66049, 4, 25, 4, 289, 9, 25, 4, 289, 4, 1369, 9, 289, 4, 169

Offset: 1

Robert Price, Mar 07 2017

nonn

			A248214(2)=7, 7^2 + 1 = 50 = 2*5*5, so 50 is not squarefree, the square being 5*5 = 25.

Cf. A248214, A248576.

cp's OEIS Frontend

A248214 Least integer b > 0 such that b^n + 1 is not squarefree.

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