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%I #10 Oct 10 2019 11:37:44
%S 1,1,1,2,0,3,0,2,2,3,1,5,1,3,2,2,3,5,0,5,2,2,2,6,3,2,2,6,1,7,0,2,2,6,
%T 3,7,1,4,2,6,1,7,1,4,3,4,0,8,2,6,5,3,0,5,3,9,2,3,2,10,2,2,4,3,3,4,2,8,
%U 4,8,0,8,1,3,4,6,3,5,1,9,2,3,2,11,6,2,2,5,2,9,4,5,3,2,4,10,1,5,5,8,0,9,1,5
%N a(1)=1. a(n) = number of earlier terms a(k), 1<=k<=n-1, such that (k+a(k)) divides n.
%e (a(1)+1) divides 10; (a(5)+5) divides 10; and (a(8)+8) divides 10. These 3 cases are the only cases where (a(k)+k) divides 10, for 1<=k<=9. So a(10)=3.
%t f[l_List] := Block[{n = Length[l] + 1},Append[l, Count[Table[Mod[n, k + l[[k]]], {k, n - 1}], 0]]];Nest[f, {1}, 104] (* _Ray Chandler_, Jan 22 2007 *)
%Y Cf. A127461, A127462.
%K nonn
%O 1,4
%A _Leroy Quet_, Jan 15 2007
%E Extended by _Ray Chandler_, Jan 22 2007
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