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a(0) = 1, a(n) = number of a(k), for 0 <= k <= n-1, that divide n.
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%I #12 May 26 2024 04:31:59

%S 1,1,2,2,4,2,5,2,7,2,8,2,9,2,10,3,11,2,12,2,14,4,12,2,18,3,12,5,16,2,

%T 18,2,18,5,14,6,26,2,15,4,23,2,22,2,22,9,18,2,29,3,22,5,22,2,29,7,27,

%U 5,21,2,37,2,21,11,26,7,31,2,25,6,33,2,41,2,25,13,27,7,32,2,36,9,26,2,45

%N a(0) = 1, a(n) = number of a(k), for 0 <= k <= n-1, that divide n.

%H Amiram Eldar, <a href="/A056148/b056148.txt">Table of n, a(n) for n = 0..10000</a>

%e a(6) = 5 because a(0), a(1), a(2), a(3) and a(5) divide 6.

%t a[0] = 1; a[n_] := a[n] = Count[Range[0, n - 1], _?(Divisible[n, a[#]] &)]; Array[a, 100, 0] (* _Amiram Eldar_, May 26 2024 *)

%K easy,nonn

%O 0,3

%A _Leroy Quet_, Aug 04 2000