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a(n) = floor((a(n-1)+a(n-2)+n)/(a(n-3)+1), where a(1)=1, a(2)=2, a(3)=3.
4

%I #13 Jun 28 2024 23:57:35

%S 1,2,3,4,4,3,2,2,3,5,6,5,4,3,3,4,6,7,6,4,3,4,6,8,7,5,4,4,6,8,9,7,5,4,

%T 5,7,9,9,7,5,5,6,9,9,9,6,6,6,8,9,9,7,6,6,8,10,10,8,7,6,8,9,11,9,8,6,8,

%U 9,12,10,9,7,8,8,11,10,10,8,8,8,10,11,11

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

%H John Tyler Rascoe, <a href="/A372982/b372982.txt">Table of n, a(n) for n = 1..10000</a>

%t a[1] = 1; a[2] = 2; a[3] = 3;

%t a[n_] := a[n] = Floor[(a[n - 1] + a[n - 2] + n)/(a[n - 3] + 1)];

%t Table[a[n], {n, 1, 100}]

%o (Python)

%o from itertools import count, islice

%o def a_gen():

%o A = (1,2,3)

%o yield from A

%o for n in count(4):

%o A = A[1:]+((A[2]+A[1]+n)//(A[0]+1),)

%o yield A[-1]

%o A372982_list = list(islice(a_gen(), 100)) # _John Tyler Rascoe_, Jun 28 2024

%Y Cf. A372983, A372984, A372985.

%K nonn

%O 1,2

%A _Clark Kimberling_, Jun 27 2024