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).

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, 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, 12, 26, 6, 17, 4, 10, 12, 29, 3, 10, 13, 4, 11, 5, 7, 9, 17, 5, 8

Offset: 1

Author

David James Sycamore, May 28 2025

Keywords

• nonn

Comments

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.

Examples

			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.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..16384

Crossrefs

Cf. A000005.

Programs

Maple

a:= proc(n) option remember; `if`(n<3, n, add(
     `if`(irem(a(n-1), a(j))=0, 1, 0), j=1..n-2))
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Jun 01 2025

Mathematica

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 *)

Extensions

More terms from Alois P. Heinz, Jun 01 2025