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%I #29 May 20 2024 16:26:04
%S 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,
%T 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,
%U 8,8,8,8,8,8,8,9,8,9,8,9,8,9,8,10
%N a(1)=1, then a(n) = floor(n/min(a(n-1),a(floor(n/2)))).
%C It seems that limsup and liminf of a(n)/sqrt(n) exist (see link).
%H Hugo Pfoertner, <a href="/A372971/b372971.txt">Table of n, a(n) for n = 1..10000</a>
%H Benoit Cloitre, <a href="/A372971/a372971.png">Plot of a(n)/sqrt(n) for n=1 up to 400000</a>
%H Hugo Pfoertner, <a href="/A372971/a372971.pdf">Plot of a(n)/sqrt(n)</a>, n=1..400000, zoom into pdf to see details.
%t a[1]=1; a[n_]:=Floor[n/Min[a[n-1],a[Floor[n/2]]]]; Array[a,80] (* _Stefano Spezia_, May 18 2024 *)
%o (PARI) a(n)=if(n<2,1,floor(n/min(a(n-1),a(n\2))))
%Y Cf. A372970, A097051.
%K nonn,look
%O 1,2
%A _Benoit Cloitre_, May 18 2024
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