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a(0)=1. a(n) = floor(n/b(n)), where b(n) is the largest value among (a(0),a(1),...,a(n-1)).
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%I #11 Apr 28 2016 21:01:07

%S 1,1,2,1,2,2,3,2,2,3,3,3,4,3,3,3,4,4,4,4,5,4,4,4,4,5,5,5,5,5,6,5,5,5,

%T 5,5,6,6,6,6,6,6,7,6,6,6,6,6,6,7,7,7,7,7,7,7,8,7,7,7,7,7,7,7,8,8,8,8,

%U 8,8,8,8,9,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,10,9,9,9,9,9,9,9,9,9,10,10,10

%N a(0)=1. a(n) = floor(n/b(n)), where b(n) is the largest value among (a(0),a(1),...,a(n-1)).

%C Length of n-th run appears to be A103627(n + 2) for n > 1. - _Peter Kagey_, Apr 25 2016

%H Peter Kagey, <a href="/A137735/b137735.txt">Table of n, a(n) for n = 0..10000</a>

%F For all m>=1, a(k) = m if m^2 <= k <= m^2 +m-1, a(m^2 +m) = m+1, a(k) = m if m^2 +m+1 <= k <= m^2 +2m.

%e The largest value among terms a(0) through a(14) is 4. So a(15) = floor(15/4) = 3.

%t a = {1}; Do[AppendTo[a, Floor[n/Max@ a]], {n, 2, 120}]; {1}~Join~a (* _Michael De Vlieger_, Apr 25 2016 *)

%Y Cf. A103627, A137734.

%K easy,nonn

%O 0,3

%A _Leroy Quet_, Feb 09 2008