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.

28, 61, 90, 105, 121, 321, 339, 382, 408, 466, 602, 1079, 1121, 1596, 1782, 2067, 2104, 2170, 2220, 2250, 2435, 2456, 2884, 3141, 3242, 3321, 3328, 3435, 4195, 4323, 4348, 4497, 4766, 4914, 5241, 5526, 6290, 6581, 6597, 9306, 9734

Offset

1,1

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) = (2,1,1,1).

Links

Table of n, a(n) for n=1..41.

Mathematica

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

Prog

(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, ", ")))

Crossrefs

Cf. A055573, A070985.

Sequence in context: A132769 A329306 A329307 * A255159 A071750 A255152

Adjacent sequences: A336085 A336086 A336087 * A336089 A336090 A336091

Keyword

nonn

Author

Benoit Cloitre, Oct 04 2020

Status

approved