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a(0)=a(1)=1; a(n+2) = a(floor((n+1)/a(n+1))) + a(floor(n/a(n))).
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%I #16 Aug 08 2019 14:43:15

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

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

%U 8,8,8,8,8,8,8,8,8,8,8,8,8

%N a(0)=a(1)=1; a(n+2) = a(floor((n+1)/a(n+1))) + a(floor(n/a(n))).

%C The sequence is unbounded. - _Rémy Sigrist_, Aug 08 2019

%H Rémy Sigrist, <a href="/A130535/b130535.txt">Table of n, a(n) for n = 0..10000</a>

%H Rémy Sigrist, <a href="/A130535/a130535.txt">C program for A130535</a>

%p a[0]:=1: a[1]:=1: for n from 0 to 80 do a[n+2]:= a[floor((n+1)/a[n+1])]+a[floor(n/a[n])] end do: seq(a[n],n=0..80); # _Emeric Deutsch_, Aug 13 2007

%o (C) See Links section.

%Y Cf. A130147.

%K nonn,look,easy

%O 0,3

%A _Leroy Quet_, Aug 09 2007

%E More terms from _Emeric Deutsch_, Aug 13 2007