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%I #13 Jul 31 2024 22:11:57
%S 1,2,2,5,7,14,26,56,56,138,306,612,612,1224,3004,5758,11822,23476,
%T 45284,91792,184140,368224,735948,1472492,2944996,5889992,11411652,
%U 23191624,46290860,92672900,185346856,370693871,741375929,1479818680,2962582344,5925164688
%N a(n) = a(n-1) + a(n-2) + ... + a(n-k) where k = (a(n-1) mod (n-1)) + 1 for n >= 3, with a(1) = 1 and a(2) = 2.
%C It appears that the ratio a(n+1)/a(n) -> 2.
%F a(n) = Sum_{i=1 .. (a(n-1) mod (n-1)) + 1} a(n-i).
%e For n = 7, we add up the previous a(7-1) mod (7-1) + 1 = 3 terms to get a(7) = a(6) + a(5) + a(4) = 14 + 7 + 5 = 26.
%t Modanacci={1,2};Do[AppendTo[Modanacci,Sum[Modanacci[[-i]],{i,Mod[Modanacci[[-1]],Length[Modanacci]]+1}]],100]
%K nonn
%O 1,2
%A _Mehmet A. Ates_, Jul 29 2024
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