A384773 a(1) = 1, a(2) = 1. For n > 2 if a(n-1) = k is a novel term, a(n) = a(n-1-k). Otherwise if a(n-1) is a repeat term a(n) = number of m; 1 <= m <= n-2 such that a(m) = a(n-1).

1, 1, 1, 2, 1, 3, 1, 4, 2, 1, 5, 3, 1, 6, 4, 1, 7, 1, 8, 5, 1, 9, 1, 10, 6, 1, 11, 1, 12, 7, 1, 13, 8, 1, 14, 1, 15, 9, 1, 16, 10, 1, 17, 1, 18, 11, 1, 19, 12, 1, 20, 1, 21, 13, 1, 22, 1, 23, 14, 1, 24, 15, 1, 25, 1, 26, 16, 1, 27, 1, 28, 17, 1, 29, 18, 1, 30

Offset: 1

Author

David James Sycamore, Jun 09 2025

Keywords

• nonn
• easy

Comments

a(n) <= n for all n, with equality for n = 1. Same as A364749 until a(12).

The sequence of indices of terms a(n-1-k) following novel terms k (starting: 2,3,4,6,8,10,11,13,...) appears to be A335999.

Records subsequence is A000027, with records occurring at indices 1, A026278.

Examples

			a(1) = a(2) = 1 implies a(3) = 1 since 1 has been repeated once. Then a(4) = 2 because now 1 has been repeated twice. Since 2 is a novel term a(5) = a(4-2) = a(2) = 1. Since 1 has been repeated three times a(6) = 3, another novel term so a(7) = a(6-3) = a(3) = 1.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000027, A026278, A335999, A364749.

Programs

Mathematica

nn = 2^16; c[] := 0; a[1] = a[2] = 1; c[1]++; {1, 1}~Join~Do[(If[c[#] == 0, k = a[n - # - 1], k = c[#] ]; c[#]++) &[a[n - 1] ]; a[n] = k, {n, 3, nn}]; Array[a, nn] (* _Michael De Vlieger, Jun 10 2025 *)