Voozh

A358405

a(1) = 0; for n > 1, let a(n-1) = m; if a(n-1) is the first occurrence of m then a(n) = 0, but if there is a k < n-1 with a(k) = m, a(n) is the maximum of n-1-k and j, where a(j) is the first occurrence of m in the sequence.

0, 0, 1, 0, 2, 0, 2, 5, 0, 3, 0, 2, 5, 8, 0, 4, 0, 2, 6, 0, 3, 11, 0, 3, 10, 0, 3, 10, 25, 0, 4, 16, 0, 3, 10, 25, 29, 0, 5, 26, 0, 3, 10, 25, 29, 37, 0, 6, 29, 37, 46, 0, 5, 14, 0, 3, 14, 54, 0, 4, 29, 37, 46, 51, 0, 6, 19, 0, 3, 13, 0, 3, 10, 30, 0, 4, 16, 45, 0, 4, 16, 32, 0, 4, 16, 32, 82

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OFFSET

1,5

COMMENTS

This sequence uses the same selection rules as A358402 but here the maximum of the two offsets is chosen for a(n). The terms form an unusual pattern for larger values of n; see the linked images.

See A358406 for the index where each number first appears.

LINKS

Scott R. Shannon, Table of n, a(n) for n = 1..10000.

Scott R. Shannon, Image of the first 100000 terms.

Scott R. Shannon, Image of the first 1000000 terms.

Scott R. Shannon, Image of the first 10000000 terms.

EXAMPLE

a(5) = 2 as a(4) = 0 and a(2) = 0, these being separated by two terms.

a(8) = 5 as a(7) = 2 and 2 appears as the fifth term of the sequence. Note that the number of terms between the two previous occurrences of 2 is 7 - 5 = 2 which is smaller than 5, so 5 is chosen.

MATHEMATICA

nn = 120; q[_] = c[_] = 0; m = a[1] = 0; Do[If[c[#] == 0, k = 0; c[#] = q[#] = n - 1, k = Max[n - 1 - c[#], q[#]]; c[#] = n - 1] &[m]; a[n] = m = k; If[k == u, While[c[u] > 0, u++]], {n, 2, nn}]; Array[a, nn]

PROG

(Python)

from itertools import count, islice

def agen():

an, first, prev = 0, {0: 1}, {0: 1}

for n in count(2):

yield an

an1 = 0 if first[an] == n-1 else max(n-1-prev[an], first[an])

if an1 not in first: first[an1] = prev[an1] = n

prev[an] = n-1

an = an1

print(list(islice(agen(), 87))) # Michael S. Branicky, Nov 14 2022