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A326280
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Let f(n) be a sequence of distinct Gaussian integers such that f(1) = 0 and for any n > 1, f(n) = f(floor(n/2)) + k(n)*g((1+i)^(A000120(n)-1) * (1-i)^A023416(n)) where k(n) > 0 is as small as possible and g(z) = z/gcd(Re(z), Im(z)); a(n) is the real part of f(n).
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2
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0, 1, 1, 1, 2, 2, 1, 0, 2, 3, 3, 4, 3, 2, 0, -1, 0, 2, 3, 3, 4, 4, 3, 4, 5, 4, 3, 4, 2, 0, -1, -2, -2, -1, 1, 0, 3, 4, 6, 2, 4, 5, 7, 6, 5, 5, 2, 1, 5, 7, 8, 7, 6, 4, 1, 5, 5, 3, 0, 2, -1, -2, -2, -2, -3, -3, -2, -3, -1, 1, 5, -2, 0, 3, 6, 4, 6, 7, 6, 0, 2, 4
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OFFSET
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1,5
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COMMENTS
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The idea underlying this sequence is to build an infinite binary tree of Gaussian integers:
- for any n > 0, f(n) has children f(2*n) and f(2*n+1),
- f(n), f(2*n) and f(2*n+1) form a right triangle,
- when u has child v and v has child w, then the angle between the vectors (u,v) and (v,w) is 45 degrees.
Among the first 2^20-1 terms, some values around the origin are missing: -2 - 3*i, -2, i, 2 - 2*i, 2, 4 + i, 5 - 2*i; will they ever appear?
Graphically, f has interesting features (see representations of f in Links section).
This sequence has similarities with A322574.
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LINKS
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EXAMPLE
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See representation of the first layers of the binary tree in links section.
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PROG
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(PARI) See Links section.
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CROSSREFS
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See A326281 for the imaginary part of f.
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KEYWORD
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AUTHOR
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STATUS
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approved
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