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A069013 a(1)=a(2)=a(3)=1; for n > 3, a(n) = floor(a(n-3) + a(n-2)/a(n-1)). 1

%I #16 Sep 08 2022 08:45:05

%S 1,1,1,2,1,3,2,2,4,2,4,4,3,5,4,4,6,4,5,6,4,6,6,5,7,6,6,8,6,7,8,6,8,8,

%T 7,9,8,8,10,8,9,10,8,10,10,9,11,10,10,12,10,11,12,10,12,12,11,13,12,

%U 12,14,12,13,14,12,14,14,13,15,14,14,16,14,15,16,14,16,16,15,17,16,16,18

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

%C To illustrate the formula: if k >= 2, then (2k, 2k, 2k-1, 2k+1, 2k, 2k, 2k+2, 2k, 2k+1, 2k+2, 2k) are next 11 terms.

%C E.g., if k=5, (10,10,9,11,10,10,12,10,11,12,10) form the terms from a(44) through a(54).

%H Harvey P. Dale, <a href="/A069013/b069013.txt">Table of n, a(n) for n = 1..1000</a>

%F For k >= 1:

%F a(11*k) = 2k+2,

%F a(11*k + 1) = 2k+2,

%F a(11*k + 2) = 2k+1,

%F a(11*k + 3) = 2k+3,

%F a(11*k + 4) = 2k+2,

%F a(11*k + 5) = 2k+2,

%F a(11*k + 6) = 2k+4,

%F a(11*k + 7) = 2k+2,

%F a(11*k + 8) = 2k+3,

%F a(11*k + 9) = 2k+4,

%F a(11*k + 10) = 2k+2.

%t nxt[{a_,b_,c_}]:={b,c,Floor[a+b/c]}; NestList[nxt,{1,1,1},90][[All,1]] (* _Harvey P. Dale_, Dec 12 2020 *)

%o (Magma) a:=[1,1,1]; [n le 3 select a[n] else Floor(Self(n-3)+Self(n-2)/Self(n-1)): n in [1..80]]; // _Marius A. Burtea_, Oct 06 2019

%K nonn

%O 1,4

%A _Benoit Cloitre_, Apr 02 2002

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