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A357943
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a(0) = 0; a(1) = 1, a(2) = 2; for n > 2, a(n) is the number of times the term a(n - 1 - a(n-1)) has appeared in the sequence.
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2
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0, 1, 2, 1, 1, 3, 1, 1, 5, 5, 5, 1, 3, 3, 3, 6, 3, 5, 5, 5, 5, 1, 7, 1, 1, 9, 5, 9, 8, 8, 9, 9, 1, 4, 2, 10, 4, 10, 4, 2, 2, 3, 3, 4, 4, 4, 7, 4, 7, 7, 7, 7, 7, 7, 8, 8, 7, 9, 9, 9, 9, 9, 9, 9, 4, 11, 4, 11, 9, 12, 12, 12, 12, 12, 12, 12, 12, 9, 13, 2, 13, 2, 6, 8, 8, 8, 13, 8, 6, 3, 3, 8, 9, 9
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OFFSET
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0,3
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COMMENTS
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In the first 100 million terms the longest run of consecutive equal terms is eight : a(69)..a(76) = 12. There is no other run of equal length in this range, and it is unknown if a longer run exists.
Other than the initial three terms, the first time a term exists that is one more than the previous term is a(29) = 8, a(30) = 9. Remarkably the first time two such consecutive terms exists is not until a(60917874) = 5394, a(60917875) = 5395, a(60917876) = 5396. It is unknown if three or more such terms exist.
Note that if the sequence starts with just a(0) = 0, a(1) = 1 then the resulting sequence is A003056.
The sequence is conjectured to contain all positive numbers. See A357944 for the index of where a given number first appears.
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LINKS
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EXAMPLE
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a(5) = 3 as the term at a(4 - a(4)) = a(4 - 1) = a(3) = 1, and 1 has appeared three times in the sequence.
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MATHEMATICA
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nn = 83; c[_] = 0; Array[Set[{a[#], c[#]}, {#, 1}] &, 3, 0]; Do[(a[n] = c[#]; c[c[#]]++) &[a[n - a[n - 1] - 1]], {n, 3, nn}]; Array[a, nn, 0] (* Michael De Vlieger, Oct 23 2022 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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