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A178746
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Binary counter with intermittent bits. Starting at zero the counter attempts to increment by 1 at each step but each bit in the counter alternately accepts and rejects requests to toggle.
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3
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0, 1, 3, 6, 6, 7, 13, 12, 12, 13, 15, 26, 26, 27, 25, 24, 24, 25, 27, 30, 30, 31, 53, 52, 52, 53, 55, 50, 50, 51, 49, 48, 48, 49, 51, 54, 54, 55, 61, 60, 60, 61, 63, 106, 106, 107, 105, 104
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,3
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COMMENTS
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A simple scatter plot reveals a self-similar structure that resembles flying geese.
Ignoring the initial zero term, split the sequence into rows of increasing binary magnitude such that the terms in row m satisfy 2^m <= a(n) < 2^(m+1).
0: 1,
1: 3,
2: 6,6,7,
3: 13,12,12,13,15,
4: 26,26,27,25,24,24,25,27,30,30,31,
5: 53,52,52,53,55,50,50,51,49,48,48,49,51,54,54,55,61,60,60,61,63,
Then,
Row m starts at n = A005578(m+1) in the original sequence
The first term in row m is A081254(m)
The last term in row m is 2^(m+1)-1
The number of terms in row m is A001045(m+1)
The number of distinct terms in row m is A005578(m)
The number of ascending runs in row m is A005578(m)
The number of non-ascending runs in row m is A005578(m)
The number of descending runs in row m is A052950(m)
The number of non-descending runs in row m is A005578(m-1)
The sum of terms in row m is A178747(m)
The total number of '1' bits in the terms of row n is A178748(m)
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LINKS
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FORMULA
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If n is a power of 2, a(n) = n*3/2. Lim(a(n)/n) = 3/2.
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EXAMPLE
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0 -> low bit toggles -> 1 -> should be 2 but low bit does not toggle -> 3 -> should be 4 but 2nd-lowest bit does not toggle -> 6 -> should be 7 but low bit does not toggle -> 6 -> low bit toggles -> 7
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PROG
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(PARI) seq(n)={my(a=vector(n+1), f=0, p=0); for(i=2, #a, my(b=bitxor(p+1, p)); f=bitxor(f, b); p=bitxor(p, bitand(b, f)); a[i]=p); a} \\ Andrew Howroyd, Mar 03 2020
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CROSSREFS
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Cf. A178747 sum of terms in rows of a(n), A178748 total number of '1' bits in the terms of rows of a(n).
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KEYWORD
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AUTHOR
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STATUS
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approved
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