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A145017
Squarefree positive integers k for which k-(floor(sqrt(k)))^2 is a perfect square.
3
1, 2, 5, 10, 13, 17, 26, 29, 34, 37, 53, 58, 65, 73, 82, 85, 97, 101, 109, 122, 130, 137, 145, 170, 173, 178, 185, 194, 197, 205, 221, 226, 229, 241, 257, 265, 281, 290, 293, 298, 305, 314, 349, 362, 365, 370, 377, 386, 397, 401, 409, 442, 445, 457, 466, 485, 493
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
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
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