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A384336 a(1) = 1, a(2) = 2. For n > 2, a(n) = number of a(k), k = 1..n-2 such that a(k) divides a(n-1).

%I A384336 #53 Jun 01 2025 16:36:25
%S A384336 1,2,1,1,2,4,5,3,3,4,6,7,3,5,4,7,4,8,9,6,9,7,5,5,6,10,9,8,10,10,11,3,
%T A384336 6,12,17,3,7,6,14,9,11,4,9,12,21,12,22,7,7,8,12,23,3,8,13,3,9,15,14,
%U A384336 12,26,6,17,4,10,12,29,3,10,13,4,11,5,7,9,17,5,8
%N A384336 a(1) = 1, a(2) = 2. For n > 2, a(n) = number of a(k), k = 1..n-2 such that a(k) divides a(n-1).
%C A384336 1 appears 3 times and 2 appears 2 times, all other numbers are conjectured to appear infinitely many times. The first occurrence of an odd prime is followed by 3, see Example, and the order of first appearances of primes starts: 2,5,3,7,11,19,23,13.
%H A384336 Alois P. Heinz, <a href="/A384336/b384336.txt">Table of n, a(n) for n = 1..16384</a>
%e A384336 Since a(1) = 1, and a(2) = 2, a(3) must be 1 since there is only one term (a(1) = 1) which is a divisor of 2. Then a(4) = 1 because a(1) = 1 is the only prior divisor of a(3) = 1.
%p A384336 a:= proc(n) option remember; `if`(n<3, n, add(
%p A384336      `if`(irem(a(n-1), a(j))=0, 1, 0), j=1..n-2))
%p A384336     end:
%p A384336 seq(a(n), n=1..100);  # _Alois P. Heinz_, Jun 01 2025
%t A384336 nn = 120; Array[Set[a[#], #] &, 2]; j = 2; Do[k = Count[Array[a, n - 2], _?(Divisible[j, #] &)]; Set[{a[n], j}, {k, k}], {n, 3, nn}]; Array[a, nn] (* _Michael De Vlieger_, Jun 01 2025 *)
%Y A384336 Cf. A000005.
%K A384336 nonn
%O A384336 1,2
%A A384336 _David James Sycamore_, May 28 2025
%E A384336 More terms from _Alois P. Heinz_, Jun 01 2025