A329221 a(0)=0. If a(n)=k is the first occurrence of k then a(n+1)=a(k), otherwise a(n+1)=n-m where m is the index of the greatest prior term.

0, 0, 1, 0, 1, 2, 1, 1, 2, 3, 0, 1, 2, 3, 4, 1, 1, 2, 3, 4, 5, 2, 1, 2, 3, 4, 5, 6, 1, 1, 2, 3, 4, 5, 6, 7, 1, 1, 2, 3, 4, 5, 6, 7, 8, 2, 1, 2, 3, 4, 5, 6, 7, 8, 9, 3, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 1, 2, 3, 4, 5, 6, 7, 8

Offset: 0

Author

David James Sycamore, Nov 22 2019

Keywords

• nonn
• easy

Comments

Subsequence a(A000217(k+1)), k>=0 is an identical copy of the original. Erasure of the first occurrence of every k does not reproduce the original so this is not a fractal sequence. However, if a(0) and the copy subsequence are both erased, what remains is A002260. Hence this sequence contains both a copy identical to the original, and a fractal subsequence different from the original.

Examples

			a(0)=0 is the first occurrence of the term 0, therefore a(1)=a(0+1)=a(0)=0. a(1)=0 has been seen before, and 0 is the index of the greatest prior term (0), so a(2)=a(1+1)=1-0=1.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000
• Wikipedia, Fractal sequence

Crossrefs

Cf. A000217, A002260, A181391, A072638.

Programs

Mathematica

Block[{a, c, j, k, m, r, nn}, nn = 120; c[] := 0; a[0] = j = r = m = 0; Do[If[c[j] == 0, k = a[j], k = n - m - 1]; c[j]++; Set[{a[n], j}, {k, k}]; If[k > r, r = k; m = n], {n, nn}]; Array[a, nn + 1, 0] ] (* _Michael De Vlieger, Jun 30 2025 *)

Formula

a(k) = a(A000217(k+1)), k >= 0.

The n-th occurrence of k is a((k^2 + (2*n+1)*k + n*(n-1))/2), k >= 1.

The n-th occurrence of 0 is a(A072638(n)), n >= 0.