A261117 -id:A261117 - 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

A260824 Least positive integer b such that b^(2^n)+1 is not squarefree.

Original entry on oeis.org

3, 7, 110, 40, 392, 894, 315, 48

Offset: 0

Jeppe Stig Nielsen, Aug 04 2015

For any n, a(n) <= A261117(n).
The smallest square in the factors of b^(2^n)+1 are 2^2, 5^2, 17^2, 17^2, 769^2. - Robert Price, Mar 07 2017; edited by Jeffrey Shallit, May 10 2017
a(8) <= 50104 (corresponding square 10753^2). - Jeffrey Shallit, May 10 2017
Some better bounds than A261117(n): a(9) <= 65863 (factor 13313^2), a(12) <= 265801 (factor 65537^2), a(16) <= 1493667 (factor 1179649^2), a(18) <= 15834352 (factor 7340033^2), a(19) <= 15786037 (factor 23068673^2), a(21) <= 78597313 (factor 230686721^2), a(22) <= 13753565041 (factor 469762049^2), a(23) <= 6276931961 (factor 469762049^2). - Max Alekseyev, Feb 20 2018

			a(1) = A049532(1) = 7.
For n=4, we consider b^16+1. The first time it is not squarefree is for b=392, where 392^16+1 is divisible by 769^2. So a(4)=392.

Subsequence of A248214.

Cf. A261117, A049532.

a(n) = for(b=1,10^42, !issquarefree(b^(2^n)+1) & return(b) );

from sympy.ntheory.factor_ import core
def a(n):
  b, pow2, t = 1, 2**n, 2
  while core(t, 2) == t:
    b += 1
    t = b**(pow2) + 1
  return b
print([a(n) for n in range(4)]) # Michael S. Branicky, Mar 07 2021

a(n) = A248214(2^n).

Edited and a(5)-a(7) added by Max Alekseyev, Feb 20 2018

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

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A260824 Least positive integer b such that b^(2^n)+1 is not squarefree.

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Extensions