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

0,6

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.

Links

Michael De Vlieger, Table of n, a(n) for n = 0..10000

Wikipedia, Fractal sequence

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.

Example

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.

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 *)

Crossrefs

Cf. A000217, A002260, A181391, A072638.

Sequence in context: A343938 A239397 A307884 * A177858 A166967 A136256

Adjacent sequences: A329218 A329219 A329220 * A329222 A329223 A329224

Keyword

nonn,easy

Author

David James Sycamore, Nov 22 2019

Status

approved