|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
It seems that limsup and liminf of a(n)/sqrt(n) exist (see link).
|
|
LINKS
|
|
|
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
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|