STATUS
proposed
approved
The OEIS is supported by the many generous donors to the OEIS Foundation.
proposed
approved
editing
proposed
Least integer b > 0 so such that b^n + 1 is not squarefree.
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.
approved
editing
reviewed
approved
proposed
reviewed
editing
proposed
Least integer b>0 so that b^n + 1 is not squarefree.
For n = 12, we have that 110^12 + 1 is divisible by a (non-unitnonunit) 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
approved
editing
proposed
approved
editing
proposed
For a criterion for a(n) to be finite when n is a power of two, see A261117. - Jeppe Stig Nielsen, Aug 08 2015
approved
editing