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

1,2

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.

Links

Table of n, a(n) for n=1..52.

Prog

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

Crossrefs

Cf. A001653, A002315.

Sequence in context: A370597 A296387 A302658 * A358428 A084909 A038619

Adjacent sequences: A324172 A324173 A324174 * A324176 A324177 A324178

Keyword

nonn

Author

Jinyuan Wang, Feb 23 2019

Status

approved