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a(1)=1, a(2)=2, a(3)=2, a(4)=3; a(n) = a(n-a(n-1))+a(n-1-a(n-2)-a(n-2-a(n-2))) for n > 4.
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%I #11 Sep 08 2022 08:46:22

%S 1,2,2,3,3,3,4,5,5,5,5,6,7,7,8,8,8,8,8,9,10,10,11,11,11,12,13,13,13,

%T 13,13,13,14,15,15,16,16,16,17,18,18,18,18,19,20,20,21,21,21,21,21,21,

%U 21,22,23,23,24,24,24,25,26,26,26,26,27,28,28,29,29,29,29,29,30,31,31

%N a(1)=1, a(2)=2, a(3)=2, a(4)=3; a(n) = a(n-a(n-1))+a(n-1-a(n-2)-a(n-2-a(n-2))) for n > 4.

%C This sequence increases slowly.

%C k occurs A035612(k) times.

%C Each Fibonacci number occurs more times than any number before it.

%H Nathan Fox, <a href="/A316628/b316628.txt">Table of n, a(n) for n = 1..10000</a>

%H Nathan Fox, <a href="https://vimeo.com/322291024">Trees, Fibonacci Numbers, and Nested Recurrences</a>, Rutgers University Experimental Math Seminar, Mar 07, 2019

%F a(n+1)-a(n)=1 or 0.

%F a(n)/n -> C=(sqrt(5)-1)/(sqrt(5)+1).

%p A316628:=proc(n) option remember: if n <= 0 then 0: elif n = 1 then 1: elif n = 2 then 2: elif n = 3 then 2: elif n = 4 then 3: else A316628(n-A316628(n-1)) + A316628(n-1-A316628(n-2)-A316628(n-2-A316628(n-2))): fi: end:

%o (Magma) I:=[1,2,2,3]; [n le 4 select I[n] else Self(n-Self(n-1))+Self(n-1-Self(n-2)-Self(n-2-Self(n-2))): n in [1..100]]; // _Vincenzo Librandi_, Jul 09 2018

%o (GAP) a:=[1,2,2,3];; for n in [5..80] do a[n]:=a[n-a[n-1]]+a[n-1-a[n-2]-a[n-2-a[n-2]]]; od; a; # _Muniru A Asiru_, Jul 09 2018

%Y Cf. A000045, A005185, A035612, A046699.

%K nonn

%O 1,2

%A _Nathan Fox_, Jul 08 2018