OFFSET
0,1
COMMENTS
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
FORMULA
a(n) = A248214(2^n).
EXAMPLE
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.
PROG
(PARI) a(n) = for(b=1, 10^42, !issquarefree(b^(2^n)+1) & return(b) );
(Python)
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
CROSSREFS
KEYWORD
hard,nonn,more
AUTHOR
Jeppe Stig Nielsen, Aug 04 2015
EXTENSIONS
Edited and a(5)-a(7) added by Max Alekseyev, Feb 20 2018
STATUS
approved