k such that L(H(k,2)) = 2*L(H(k,1)) where L(x) is the number of terms in the continued fraction of x and H(k,r) = Sum_{u=1..k} 1/u^r.

A336088

k such that L(H(k,2)) = 2*L(H(k,1)) where L(x) is the number of terms in the continued fraction of x and H(k,r) = Sum_{u=1..k} 1/u^r.