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.
PROG
(PARI) is(n) = n%(floor(sqrt(n)) + floor(sqrt(n/2))) == 0;
CROSSREFS
KEYWORD
nonn
AUTHOR
Jinyuan Wang, Feb 23 2019
STATUS
approved