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A329985
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a(1) = 1 and for n > 0, a(n+1) = a(k) - a(n) where k is the number of terms equal to a(n) among the first n terms.
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8
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1, 0, 1, -1, 2, -1, 1, 0, 0, 1, -2, 3, -2, 2, -2, 3, -3, 4, -3, 3, -2, 1, 1, -2, 4, -4, 5, -4, 4, -3, 4, -5, 6, -5, 5, -5, 6, -6, 7, -6, 6, -5, 4, -2, 1, 0, -1, 2, -1, 0, 2, -3, 2, 0, -1, 3, -4, 5, -4, 3, -1, 0, 1, -1, 2, -3, 5, -6, 7, -7, 8, -7, 7, -6, 5, -3
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OFFSET
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1,5
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COMMENTS
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In other words, for n > 0, a(n+1) = a(o(n)) - a(n) where o is the ordinal transform of the sequence.
The sequence has interesting graphical features (see plot in Links section).
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LINKS
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N. J. A. Sloane (in collaboration with Scott R. Shannon), Art and Sequences, Slides of guest lecture in Math 640, Rutgers Univ., Feb 8, 2020. Mentions this sequence.
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EXAMPLE
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The first terms, alongside their ordinal transform, are:
n a(n) o(n)
-- ---- ----
1 1 1
2 0 1
3 1 2
4 -1 1
5 2 1
6 -1 2
7 1 3
8 0 2
9 0 3
10 1 4
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MATHEMATICA
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A={1}; For[n=2, n<=76, n++, A=Append[A, Part[A, Count[Table[Part[A, i], {i, 1, n-1}], Part[A, n-1]]]-Part[A, n-1]]]; A (* Joshua Oliver, Nov 26 2019 *)
Nest[Append[#, #[[Count[#, #[[-1]] ] ]] - #[[-1]]] &, {1}, 75] (* Michael De Vlieger, Dec 01 2019 *)
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PROG
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(PARI) for (n=1, #(a=vector(76)), print1 (a[n]=if (n==1, 1, a[sum(k=1, n-1, a[k]==a[n-1])]-a[n-1])", "))
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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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