A324175 Integers k such that floor(sqrt(k)) + floor(sqrt(k/2)) divides k.

1, 2, 6, 8, 10, 15, 21, 40, 60, 65, 90, 102, 119, 126, 133, 160, 168, 176, 216, 225, 270, 290, 319, 330, 341, 384, 396, 408, 468, 546, 615, 630, 704, 736, 782, 799, 816, 918, 1007, 1026, 1120, 1160, 1218, 1239, 1260, 1342, 1364, 1386, 1495, 1632, 1750, 1775

Offset: 1

Author

Jinyuan Wang, Feb 23 2019

Keywords

• nonn

Comments

This sequence is infinite. Proof: if x > y > 1 satisfies x^2 - 2*y^2 = -1 (x=A002315(j), y=A001653(j+1), j>0), then x < 2*y. Let k = 2*y^2 + m; then 0 <= m <= 2*x - 1, because x^2 < x^2 + my + 1 < (x+1)^2 and y^2 <= y^2 + m/2 < y^2 + 2*y, floor(sqrt(k)) = floor(sqrt(x^2+m+1)) = x and floor(sqrt(k/5)) = floor(sqrt(y^2+m/2)) = y. x + y < 2*x, so by the pigeonhole principle there exists a number m belonging to [0, 2*x - 1] such that x + y | 2*y^2 + m, so such k is a term.

Crossrefs

Cf. A001653, A002315.

Programs

PARI

is(n) = n%(floor(sqrt(n)) + floor(sqrt(n/2))) == 0;