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a(0)=1; for n >= 1, a(n) = ceiling(Fibonacci(n)/a(n-1)).
1

%I #16 Mar 08 2019 07:40:19

%S 1,1,1,2,2,3,3,5,5,7,8,12,12,20,19,33,30,54,48,88,77,143,124,232,200,

%T 376,323,609,522,986,844,1596,1365,2583,2208,4180,3572,6764,5779,

%U 10945,9350,17710,15128,28656,24477,46367,39604,75024,64080,121392,103683

%N a(0)=1; for n >= 1, a(n) = ceiling(Fibonacci(n)/a(n-1)).

%F Conjectures from _Colin Barker_, Mar 08 2019: (Start)

%F G.f.: (1 + x + x^3)*(1 + x - x^9) / ((1 + x)*(1 - x^2 - x^4)).

%F a(n) = -a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5) for n>12.

%F (End)

%p with(combinat): a:=proc(n) if n=0 then 1 else ceil(fibonacci(n)/a(n-1)) end if end proc: seq(a(n),n=0..50); # _Emeric Deutsch_, Aug 09 2008

%t RecurrenceTable[{a[0]==1,a[n]==Ceiling[Fibonacci[n]/a[n-1]]},a,{n,50}] (* _Harvey P. Dale_, Dec 13 2013 *)

%Y Cf. A140828 (similar sequence).

%K nonn

%O 0,4

%A _Leroy Quet_, Jul 18 2008

%E More terms from _Emeric Deutsch_, Aug 09 2008