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%I #9 Sep 16 2015 12:39:35
%S 1,1,2,2,4,4,2,2,6,7,6,6,2,2,10,8,2,3,2,2,12,15,14,11,14,15,5,17,16,
%T 11,3,2,14,16,2,3,17,15,4,2,16,17,3,2,19,4,22,19,6,2,16,24,5,2,40,33,
%U 22,21,26,17,13,26,3,6,22,17,2,2,23,7,25,2,29,3,27,8,23,26,9,20
%N a(1)=1, a(n) = number of earlier terms of the sequence that divide (the sum of positive integers which are <=n and missing from first (n-1) terms of sequence).
%e The positive integers which are <= 9 and are missing from the first 8 terms of the sequence are 3,5,6,7,8,9. The sum of these integers is 38, so a(9) is the number of earlier terms which divide 38. Therefore a(9) = 6.
%t a = {1}; Do[AppendTo[a, Length[Select[a, Mod[(Plus @@ Complement[Range[n], a]), # ] == 0 &]]], {n, 2, 80}]; a (* _Stefan Steinerberger_, Dec 26 2007 *)
%Y Cf. A130500.
%K nonn
%O 1,3
%A _Leroy Quet_, May 31 2007
%E More terms from _Stefan Steinerberger_, Dec 26 2007
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