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 A324175 Integers k such that floor(sqrt(k)) + floor(sqrt(k/2)) divides k. 4
 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 (list; graph; refs; listen; history; text; internal format)
 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 PROG (PARI) is(n) = n%(floor(sqrt(n)) + floor(sqrt(n/2))) == 0; CROSSREFS Cf. A001653, A002315. Sequence in context: A260400 A296387 A302658 * A084909 A038619 A118257 Adjacent sequences:  A324172 A324173 A324174 * A324176 A324177 A324178 KEYWORD nonn AUTHOR Jinyuan Wang, Feb 23 2019 STATUS approved

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Last modified May 18 22:08 EDT 2022. Contains 353825 sequences. (Running on oeis4.)