OFFSET
1,1
COMMENTS
If a(n) = 3 or 7 then a(n+1) = 2 or 6 respectively.
Conjecture: every term is < 10, i.e. for every n at least one of the numbers n+2, n+3, n+5, n+6 or n+7 is squarefree.
The conjecture is false. Here are 9 counterexamples, each of which is less than 10000: 1857, 2522, 3570, 4470, 6169, 6645, 7981, 9553, 9745. There are 16 counterexamples within the first 10000 squarefree numbers. - Harvey P. Dale, May 24 2014
EXAMPLE
a(12) = 2 as 2+12 = 14 is squarefree.
MATHEMATICA
With[{sqfree=Select[Range[2, 20], SquareFreeQ]}, Flatten[ Table[ Select[ sqfree+ n, SquareFreeQ, 1]-n, {n, 70}]]] (* Harvey P. Dale, May 21 2014 *)
PROG
(PARI) a(n) = {k = 2; while(!issquarefree(k) || !issquarefree(k+n), k++); k; } \\ Michel Marcus, May 24 2014
CROSSREFS
KEYWORD
nonn
AUTHOR
Amarnath Murthy, Nov 01 2002
EXTENSIONS
Corrected and extended by Harvey P. Dale, May 21 2014
STATUS
approved