A347727 a(1)=2; then a(n) is the least integer > a(n-1) such that 2 is the largest element in the continued fraction for 1/a(1) + 1/a(2) + ... + 1/a(n).

2, 6, 18, 102, 40936, 4252528, 7112715120

Offset

1,1

Comments

3.5*10^15 < a(8) <= 8778368652367133280. - Jon E. Schoenfield, Sep 12 2021

Links

Table of n, a(n) for n=1..7.

Example

contfrac(1/2 + 1/6 + 1/18 + 1/102 + 1/40936) = [0, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2] and 1/2 + 1/6 + 1/18 + 1/102 + 1/40936 = sqrt(3) - 1.0000002354...

Mathematica

a[1] = 2; a[n_] := a[n] = Module[{k = a[n - 1] + 1, s = Sum[1/a[k], {k, 1, n - 1}]}, While[Max[ContinuedFraction[s + 1/k]] != 2, k++]; k]; Array[a, 6] (* Amiram Eldar, Sep 11 2021 *)

Crossrefs

Cf. A160390.

Sequence in context: A162063 A162064 A162056 * A308549 A212653 A323104

Adjacent sequences: A347724 A347725 A347726 * A347728 A347729 A347730

Keyword

nonn,more

Author

Benoit Cloitre, Sep 11 2021

Extensions

a(7) from Jon E. Schoenfield, Sep 11 2021

Status

approved