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A248214
Least integer b > 0 such that b^n + 1 is not squarefree.
4
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
1,1
COMMENTS
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
EXAMPLE
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
PROG
(PARI) for(n=1, 1000, b=1; while(issquarefree(b^n+1), b++); print1(b, ", "))
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Jeppe Stig Nielsen, Oct 04 2014
EXTENSIONS
More terms from M. F. Hasler, Oct 08 2014
STATUS
approved