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a(0)=1 and, for n>0, a(n)=a(Floor((n-1)/a[n-1]))+2.
4

%I #10 May 15 2017 22:56:04

%S 1,3,3,3,5,3,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,7,5,7,5,7,5,5,5,5,5,7,7,7,

%T 7,7,5,7,5,7,5,7,5,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,

%U 7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7

%N a(0)=1 and, for n>0, a(n)=a(Floor((n-1)/a[n-1]))+2.

%C Records are 1, 3, 5, 7, 9, ... and occur at {0,1,4,21,148,1333,14664,190633,2859496,48611433,...}, which appears to be A286286.

%H N. J. A. Sloane, <a href="/A132424/b132424.txt">Table of n, a(n) for n = 0..20000</a>

%p a[0]:=1: for n from 1 to 20000 do a[n]:=2+a[floor((n-1)/a[(n-1)])] end do: # _N. J. A. Sloane_, May 15 2017

%Y Cf. A286286.

%Y See A130147 for a related sequence.

%K nonn

%O 0,2

%A _John W. Layman_, Aug 20 2007