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A330261
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Start with an empty stack S; for n = 1, 2, 3, ..., interpret the binary representation of n from left to right as follows: in case of bit 1, push the number 1 on top of S, in case of bit 0, replace the two numbers on top of S, say u on top of v, with u-v; a(n) gives the number on top of S after processing n.
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3
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1, 0, 1, -1, 1, 0, 1, -2, 1, 1, 1, -1, 1, 0, 1, -3, 1, 4, 1, 0, 1, 0, 1, -2, 1, 1, 1, -1, 1, 0, 1, -4, 1, 5, 1, 7, 1, 0, 1, -1, 1, 0, 1, 0, 1, 0, 1, -3, 1, 2, 1, 0, 1, 0, 1, -2, 1, 1, 1, -1, 1, 0, 1, -5, 1, 5, 1, -4, 1, 0, 1, 3, 1, -3, 1, 1, 1, 0, 1, -2, 1, -2
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internal format)
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OFFSET
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1,8
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COMMENTS
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This sequence is a variant of A308551.
After processing n, S has A268289(n) elements.
Every integer appears infinitely many times in the sequence:
- the effect of the binary string b(0) = "110" is to leave 0 on top of S,
- the effect of the binary string b(1) = "1" is to leave 1 on top of S,
- the effect of the binary string b(-1) = "11100" is to leave -1 on top of S,
- let "|" denote the binary concatenation,
- for any k > 0:
- the effect of b(k+1) = b(-1)|b(k)|"0" is to leave k+1 on top of S,
- the effect of b(-k-1) = b(1)|b(-k)|"0" is to leave -k-1 on top of S,
- for any k, for any n > 0, if the binary representation of n ends with b(k), then a(n) = k, QED,
- see A330264 for the values in order of appearance.
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LINKS
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FORMULA
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a(2*k-1) = 1 for any k > 0.
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EXAMPLE
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The first terms, alongside the binary representation of n and the evolution of stack S, are:
n a(n) bin(n) S
-- ---- ------ ------------------------------------------------------------
1 1 1 () -> (1)
2 0 10 (1) -> (1,1) -> (0)
3 1 11 (0) -> (0,1) -> (0,1,1)
4 -1 100 (0,1,1) -> (0,1,1,1) -> (0,1,0) -> (0,-1)
5 1 101 (0,-1) -> (0,-1,1) -> (0,2) -> (0,2,1)
6 0 110 (0,2,1) -> (0,2,1,1) -> (0,2,1,1,1) -> (0,2,1,0)
7 1 111 (0,2,1,0) -> (0,2,1,0,1) -> (0,2,1,0,1,1) -> (0,2,1,0,1,1,1)
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PROG
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(PARI) See Links section.
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CROSSREFS
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KEYWORD
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sign,base
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AUTHOR
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STATUS
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approved
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