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A274575
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For m=1,2,3,... write all the 2^m binary vectors of length m in increasing order, and replace each vector with (number of 1's) - (number of 0's). Start with an initial 0 for the empty vector.
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3
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0, -1, 1, -2, 0, 0, 2, -3, -1, -1, 1, -1, 1, 1, 3, -4, -2, -2, 0, -2, 0, 0, 2, -2, 0, 0, 2, 0, 2, 2, 4, -5, -3, -3, -1, -3, -1, -1, 1, -3, -1, -1, 1, -1, 1, 1, 3, -3, -1, -1, 1, -1, 1, 1, 3, -1, 1, 1, 3, 1, 3, 3, 5, -6, -4, -4, -2, -4, -2, -2, 0, -4, -2, -2, 0, -2, 0, 0, 2, -4, -2, -2, 0, -2, 0, 0, 2, -2, 0, 0, 2, 0, 2, 2, 4, -4, -2, -2, 0, -2, 0, 0, 2, -2, 0, 0, 2, 0, 2, 2, 4, -2, 0, 0, 2, 0, 2, 2, 4, 0
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OFFSET
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0,4
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COMMENTS
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This is the sequence of To-And-Fro positions: Positions of all backward-forward combinations in lexicographical order when assigning -1 to a backward move and +1 to a forward move and starting at 0.
-a(n) are the slopes of the different segments, from left to right, of the successive steps in the construction of the Takagi (a.k.a. Blancmange) function. - Javier Múgica, Dec 31 2017
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LINKS
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FORMULA
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a(2*n + 1) = a(n) - 1; a(2*n + 2) = a(n) + 1.
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EXAMPLE
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Terms a(3) to a(6) correspond to the binary vectors 00, 01, 10, 11, which get replaced by -2, 0, 0, 2, respectively. Terms a(7) to a(14) correspond to the binary vectors 000, 001, ..., 111 which get replaced by -3, -1, ..., 3. a(0) = 0
a(1) = a('backward') = -1
a(2) = a('forward') = +1
a(3) = a('backward and backward') = -2
a(4) = a('backward and forward') = 0
a(5) = a('forward and backward') = 0
a(6) = a('forward and forward') = +2
a(7) = a('backward, backward and backward') = -3
a(8) = a('backward, backward and forward') = -1
Arranged as a tree read by rows:
______0______
/ \
__-1__ __1__
/ \ / \
-2 0 0 2
/ \ / \ / \ / \
-3 -1 -1 1 -1 1 1 3
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PROG
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Basic
Dim a(2*k+2)
a(0) = 0
For n = 0 To k
a(2 * n + 1) = a(n) - 1
a(2 * n + 2) = a(n) + 1
Next n
(Python)
A = [0]
for n in range(0, nmax):
A.append(A[n//2]-(-1)**n)
return(A)
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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