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A358402
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 minimum of n-1-k 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
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
Brady Haran and N. J. A. Sloane, Don't Know (the Van Eck Sequence), Numberphile video (2019).
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] == n-1 else min(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(), 96))) # Michael S. Branicky, Nov 14 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Scott R. Shannon, Nov 14 2022
STATUS
approved