OFFSET
1,2
COMMENTS
If an odd prime p divides a(n) then it has the form 4k+1.
Conjecture. For every n>=1 there exist infinitely many primes p of the form 4k+1 for which for a(n) > 1 we have s*p-(floor(sqrt(s*p)))^2 is not a perfect square for s=1,...,a(n)-1 while a(n)*p-(floor(sqrt(a(n)p))^2 is a perfect square. (See A145016(s=1) and A145022, A145023, A145047, A145048, A145149, A145050 correspondingly for s=2, s=5, s=10, s=13, s=17, s=26.) - Vladimir Shevelev, Sep 30 2008
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1000
MATHEMATICA
Select[Range@ 500, And[SquareFreeQ@ #, IntegerQ@ Sqrt[# - Floor[Sqrt@ #]^2]] &] (* Michael De Vlieger, Jan 12 2020 *)
PROG
(PARI) is(n)={issquarefree(n) && issquare(n-sqrtint(n)^2)} \\ Andrew Howroyd, Jan 12 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Sep 29 2008
EXTENSIONS
Missing a(40) inserted and terms a(42) and beyond from Andrew Howroyd, Jan 12 2020
STATUS
approved