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%I #12 Oct 05 2020 00:03:39
%S 28,61,90,105,121,321,339,382,408,466,602,1079,1121,1596,1782,2067,
%T 2104,2170,2220,2250,2435,2456,2884,3141,3242,3321,3328,3435,4195,
%U 4323,4348,4497,4766,4914,5241,5526,6290,6581,6597,9306,9734
%N 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.
%C Conjecture: this sequence is infinite. More generally for any fixed integers a,b,c,d >= 1, there are infinitely many k's such that c*d*L(H(k,a)^b) = a*b*L(H(k,c)^d) 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. Here, (a,b,c,d) = (2,1,1,1).
%t c[n_, r_] := Length @ ContinuedFraction @ HarmonicNumber[n, r]; Select[Range[10^4], c[#, 2] == 2 * c[#, 1] &] (* _Amiram Eldar_, Oct 04 2020 *)
%o (PARI) H1=H2=1;for(n=2,10000,H1=H1+1/n;H2=H2+1/n^2;if(length(contfrac(H2))==2*length(contfrac(H1)),print1(n,",")))
%Y Cf. A055573, A070985.
%K nonn
%O 1,1
%A _Benoit Cloitre_, Oct 04 2020
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