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A080637
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a(n) is the smallest positive integer which is consistent with the sequence being monotonically increasing and satisfying a(1)=2, a(a(n)) = 2n+1 for n > 1.
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9
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2, 3, 5, 6, 7, 9, 11, 12, 13, 14, 15, 17, 19, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 35, 37, 39, 41, 43, 45, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 96, 97, 98, 99, 100, 101, 102
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OFFSET
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1,1
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COMMENTS
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Sequence is the unique monotonic sequence satisfying a(a(n)) = 2n+1.
Except for the first term, numbers (greater than 2) whose binary representation starts with 11 or ends with 1. - Yifan Xie, May 26 2022
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LINKS
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FORMULA
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a(3*2^k - 1 + j) = 4*2^k - 1 + 3*j/2 + |j|/2 for k >= 0, -2^k <= j < 2^k.
a(2n+1) = 2*a(n) + 1, a(2n) = a(n) + a(n-1) + 1.
For n in the range 2*2^i <= n < 3*2^i, for i >= 0:
a(n) = n + 2^i.
a(n) = 1 + a(n-1).
Otherwise, for n in the range 3*2^i <= n < 4*2^i, for i >= 0:
a(n) = 2*(n - 2^i) + 1.
a(n) = 2 + a(n-1). (End)
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EXAMPLE
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a(8) = 12 because 2*2^2 <= 8 < 3*2^2, hence a(8) = 8 + 2^2 = 12;
a(13) = 19 because 3*2^2 <= 13 < 4*2^2, hence a(13) = 2*(13 - 2^2) + 1 = 19. (End)
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MAPLE
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t := []; for k from 0 to 6 do for j from -2^k to 2^k-1 do t := [op(t), 4*2^k - 1 + 3*j/2 + abs(j)/2]; od: od: t;
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MATHEMATICA
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b[n_] := b[n] = If[n<4, n+1, If[OddQ[n], b[(n-1)/2+1]+b[(n-1)/2], 2b[n/2]]];
a[n_] := b[n+1]-1;
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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