A145017 Squarefree positive integers k for which k-(floor(sqrt(k)))^2 is a perfect square.

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

Author

Vladimir Shevelev, Sep 29 2008

Keywords

• nonn

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, A145049, 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

Crossrefs

Cf. A005117, A020893, A145016, A145022, A145023, A145047, A145048, A145049, A145050.

Programs

Mathematica

Select[Range@ 500, And[SquareFreeQ@ #, IntegerQ@ Sqrt[# - Floor[Sqrt@ #]^2]] &] (* Michael De Vlieger, Jan 12 2020 *)

PARI

is(n)={issquarefree(n) && issquare(n-sqrtint(n)^2)} \\ Andrew Howroyd, Jan 12 2020

Extensions

Missing a(40) inserted and terms a(42) and beyond from Andrew Howroyd, Jan 12 2020