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a(n) = a(n-1) + a(floor(n/3)), a(1) = a(2) = 1.
1

%I #22 Dec 16 2019 20:05:20

%S 1,1,2,3,4,5,6,7,9,11,13,16,19,22,26,30,34,39,44,49,55,61,67,74,81,88,

%T 97,106,115,126,137,148,161,174,187,203,219,235,254,273,292,314,336,

%U 358,384,410,436,466,496,526,560,594,628,667,706,745,789,833,877

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

%C Also, the number of finite sequences b(1..r) satisfying b(1) = 1 and b(i+1) >= 3*b(i) and b(r) <= n.

%e For n = 10 the 11 sequences enumerated are (1), (1,3), (1,4), (1,5), (1,6), (1,7), (1,8), (1,9), (1,10), (1,3,9), (1,3,10).

%p a:= proc(n) option remember;

%p `if`(n<2, n, a(n-1)+a(iquo(n, 3)))

%p end:

%p seq(a(n), n=1..75); # _Alois P. Heinz_, Dec 16 2019

%t Nest[Append[#1, #1[[-1]] + #1[[Floor[#2/3] ]] ] & @@ {#, Length@ # + 1} &, {1, 1}, 57] (* _Michael De Vlieger_, Dec 16 2019 *)

%Y An analog of A033485.

%K nonn

%O 1,3

%A _Jeffrey Shallit_, Dec 16 2019