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A178746
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.
3
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
OFFSET
0,3
COMMENTS
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)
FORMULA
If n is a power of 2, a(n) = n*3/2. Lim(a(n)/n) = 3/2.
EXAMPLE
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
PROG
(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
CROSSREFS
Cf. A178747 sum of terms in rows of a(n), A178748 total number of '1' bits in the terms of rows of a(n).
Sequence in context: A177783 A228945 A348467 * A229986 A025500 A141218
KEYWORD
nonn,look
AUTHOR
David Scambler, Jun 08 2010
STATUS
approved