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%I #9 Oct 10 2019 11:37:23
%S 1,1,2,1,4,1,3,1,6,2,2,2,6,2,3,2,6,3,5,1,6,1,5,2,8,2,4,4,5,1,6,2,7,4,
%T 4,1,10,2,3,4,8,2,4,3,7,3,6,1,11,1,4,4,7,1,9,3,7,1,4,3,11,1,6,4,7,2,8,
%U 3,7,2,4,4,12,1,6,5,5,2,7,2,10,6,3,1,9,5,4,2,9,2,11,3,8,3,3,1,14,3,3,5,10,3
%N a(0)=1. a(n) = number of earlier terms a(k), 0<=k<=n-1, such that (k+a(k)) divides n.
%e (a(0)+0) divides 6; (a(1)+1) divides 6; and (a(5)+5) divides 6. These 3 cases are the only cases where (a(k)+k) divides 6, for 0<=k<=5. So a(6)=3.
%t f[l_List] := Block[{n = Length[l]},Append[l, Count[Table[Mod[n, k - 1 + l[[k]]], {k, n}], 0]]];Nest[f, {1}, 101] (* _Ray Chandler_, Jan 22 2007 *)
%Y Cf. A127460, A127463.
%K nonn
%O 0,3
%A _Leroy Quet_, Jan 15 2007
%E Extended by _Ray Chandler_, Jan 22 2007