A336089 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.

7, 10, 14, 275, 293, 359, 509, 518, 526, 531, 643, 671, 701, 710, 1081, 1158, 1318, 1798, 1836, 2368, 2441, 2507, 2591, 2990, 3477, 3589, 3818, 4096, 5582, 5851, 6968, 7180, 7523, 8718, 8745, 8782, 8817, 8844, 8946, 9772, 9905

Offset: 1

Author

Benoit Cloitre, Oct 04 2020

Keywords

• nonn

Comments

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).

Crossrefs

Cf. A055573, A336088.

Programs

Mathematica

c[n_, r_] := Length @ ContinuedFraction @ (HarmonicNumber[n]^r); Select[Range[10^4], c[#, 2] == 2 * c[#, 1] &] (* Amiram Eldar, Oct 04 2020*)

PARI

H1=1; for(n=2, 10000, H1=H1+1/n; if(length(contfrac(H1^2))==2*length(contfrac(H1)), print1(n, ", ")))