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%I #13 Oct 05 2020 00:04:09
%S 7,10,14,275,293,359,509,518,526,531,643,671,701,710,1081,1158,1318,
%T 1798,1836,2368,2441,2507,2591,2990,3477,3589,3818,4096,5582,5851,
%U 6968,7180,7523,8718,8745,8782,8817,8844,8946,9772,9905
%N k such that L(H(k,1)^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) =(1,2,1,1).
%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=1; for(n=2, 10000, H1=H1+1/n; if(length(contfrac(H1^2))==2*length(contfrac(H1)), print1(n, ", ")))
%Y Cf. A055573, A336088.
%K nonn
%O 1,1
%A _Benoit Cloitre_, Oct 04 2020