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Search: id:A091068
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| A091068 |
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a(0) = 0; for n>0, a(n) = a(n-1) - n if that is >= 0, else a(n) = a(n-1) + n - 1. |
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+0
3
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| 0, 0, 1, 3, 6, 1, 6, 12, 4, 12, 2, 12, 0, 12, 25, 10, 25, 8, 25, 6, 25, 4, 25, 2, 25, 0, 25, 51, 23, 51, 21, 51, 19, 51, 17, 51, 15, 51, 13, 51, 11, 51, 9, 51, 7, 51, 5, 51, 3, 51, 1, 51, 102, 49, 102, 47, 102, 45, 102, 43, 102, 41, 102, 39, 102, 37, 102, 35, 102, 33
(list; graph; listen)
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OFFSET
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0,4
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COMMENT
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A sequence equivalent to A091023. Let b(k) = A091023(k) for all k. Suppose we have just assigned b(x) = n. Then a(n-1) is the number of b(k) for 1 <= k < x that are not yet assigned.
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FORMULA
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This is a concatenation of blocks: b(-2) = [0], b(-1) = [0], b(0) = [1], b(1) = [3], b(2) = [6 1 6], b(3) = [12 4 12 2 12 0 12], b(4) = [25 10 25 8 25 6 25 4 25 2 25 0 25], ...
Let M(k) be the k-th term of A077854. Then block b(k) for k >= 2 is [M(k), x, M(k), x-2, M(k), x-4, M(k), ..., M(k), 0 or 1, M(k)] where x = M(k-1) - 2. The length of the block is M(k+1) - 2 M(k) + M(k-1) (the second difference of A077854, shifted one place).
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EXAMPLE
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a(4) = 6, 6-5 = 1 >= 0, so a(5) = 1. 1-6 < 0, so a(6) = 1 + 5 = 6.
When in A091023 we assign b(8) = 11, there are 2 unassigned b's to the left, namely b(3) and b(6) and indeed a(10) = 2.
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CROSSREFS
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Cf. A091023, A008344.
Sequence in context: A145389 A055263 A004157 this_sequence A065233 A072452 A155830
Adjacent sequences: A091065 A091066 A091067 this_sequence A091069 A091070 A091071
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KEYWORD
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nonn
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com), Feb 23 2004
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