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A372971
a(1)=1, then a(n) = floor(n/min(a(n-1),a(floor(n/2)))).
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, 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
Sequence in context: A374369 A242872 A329362 * A241604 A282900 A126014
KEYWORD
nonn,look
AUTHOR
Benoit Cloitre, May 18 2024
STATUS
approved