

A358402


a(1) = 0; for n > 1, let a(n1) = m; if a(n1) is the first occurrence of m then a(n) = 0, but if there is a k < n1 with a(k) = m, a(n) is the minimum of n1k and j, where a(j) is the first occurrence of m in the sequence.


4



0, 0, 1, 0, 1, 2, 0, 1, 3, 0, 1, 3, 3, 1, 3, 2, 6, 0, 1, 3, 5, 0, 1, 3, 4, 0, 1, 3, 4, 4, 1, 3, 4, 3, 2, 6, 17, 0, 1, 3, 6, 5, 21, 0, 1, 3, 6, 6, 1, 3, 4, 18, 0, 1, 3, 5, 14, 0, 1, 3, 5, 5, 1, 3, 4, 14, 9, 0, 1, 3, 6, 17, 35, 0, 1, 3, 6, 6, 1, 3, 4, 16, 0, 1, 3, 5, 21, 43, 0, 1, 3, 6, 14, 27, 0, 1
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OFFSET

1,6


COMMENTS

This sequence can be considered a variation of Van Eck's sequence, see A181391, but where the sequence forms a loop of numbers, joined at its first and last term before each a(n) is calculated. When a previously unseen number first appears the following term is 0, and as 0 appears as the first term of the sequence, the next term after these 0's will always be 1.
See A358403 for the index where each number first appears.


LINKS



EXAMPLE

a(5) = 1 as a(4) = 0 and 0 appears as the first term of the sequence.
a(6) = 2 as a(5) = 1 and a(3) = 1, these being separated by two terms.
a(17) = 6 as a(16) = 2 and 2 appears as the sixth term of the sequence. Note that the number of terms between the two previous occurrences of 2 is 16  6 = 10 which is larger than 6, so 6 is chosen.


MATHEMATICA

nn = 120; q[_] = c[_] = 0; m = a[1] = 0; Do[If[c[#] == 0, k = 0; c[#] = q[#] = n  1, k = Min[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] (* Michael De Vlieger, Jan 21 2023 *)


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] == n1 else min(n1prev[an], first[an])
if an1 not in first: first[an1] = prev[an1] = n
prev[an] = n1
an = an1


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



