A372971 a(1)=1, then a(n) = floor(n/min(a(n-1),a(floor(n/2)))).

1, 2, 3, 2, 2, 3, 2, 4, 4, 5, 5, 4, 4, 7, 7, 4, 4, 4, 4, 5, 4, 5, 4, 6, 6, 6, 6, 4, 7, 4, 7, 8, 8, 8, 8, 9, 9, 9, 9, 8, 8, 10, 10, 8, 9, 11, 11, 8, 8, 8, 8, 8, 8, 9, 9, 14, 14, 8, 8, 15, 15, 8, 9, 8, 8, 8, 8, 8, 8, 8, 8, 9, 8, 9, 8, 9, 8, 9, 8, 10

Offset

1,2

Comments

It seems that limsup and liminf of a(n)/sqrt(n) exist (see link).

Links

Hugo Pfoertner, Table of n, a(n) for n = 1..10000

Benoit Cloitre, Plot of a(n)/sqrt(n) for n=1 up to 400000

Hugo Pfoertner, Plot of a(n)/sqrt(n), n=1..400000, zoom into pdf to see details.

Mathematica

a[1]=1; a[n_]:=Floor[n/Min[a[n-1], a[Floor[n/2]]]]; Array[a, 80] (* Stefano Spezia, May 18 2024 *)

Prog

(PARI) a(n)=if(n<2, 1, floor(n/min(a(n-1), a(n\2))))

Crossrefs

Cf. A372970, A097051.

Sequence in context: A374369 A242872 A329362 * A241604 A282900 A126014

Adjacent sequences: A372968 A372969 A372970 * A372972 A372973 A372974

Keyword

nonn,look

Author

Benoit Cloitre, May 18 2024

Status

approved