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A295235
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Numbers k such that the positions of the ones in the binary representation of k are in arithmetic progression.
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15
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 17, 18, 20, 21, 24, 28, 30, 31, 32, 33, 34, 36, 40, 42, 48, 56, 60, 62, 63, 64, 65, 66, 68, 72, 73, 80, 84, 85, 96, 112, 120, 124, 126, 127, 128, 129, 130, 132, 136, 144, 146, 160, 168, 170, 192, 224, 240, 248
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OFFSET
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1,3
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COMMENTS
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Also numbers k of the form Sum_{b=0..h-1} 2^(i+j*b) for some h >= 0, i >= 0, j > 0 (in fact, h = A000120(k), and if k > 0, i = A007814(k)).
There is a simple bijection between the finite sets of nonnegative integers in arithmetic progression and the terms of this sequence: s -> Sum_{i in s} 2^i; the term 0 corresponds to the empty set.
For any n > 0, A054519(n) gives the numbers of terms with n+1 digits in binary representation.
For any n >= 0, n is in the sequence iff 2*n is in the sequence.
For any n > 0, A000695(a(n)) is in the sequence.
The first prime numbers in the sequence are: 2, 3, 5, 7, 17, 31, 73, 127, 257, 8191, 65537, 131071, 262657, 524287, ...
For any k > 0, 2^k - 2, 2^k - 1, 2^k, 2^k + 1 and 2^k + 2 are in the sequence (e.g., 14, 15, 16, 17, and 18).
Every odd term is a binary palindrome (and thus belongs to A006995).
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LINKS
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EXAMPLE
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The binary representation of the number 42 is "101010" and has ones evenly spaced, hence 42 appears in the sequence.
The first terms, alongside their binary representations, are:
n a(n) a(n) in binary
-- ---- --------------
1 0 0
2 1 1
3 2 10
4 3 11
5 4 100
6 5 101
7 6 110
8 7 111
9 8 1000
10 9 1001
11 10 1010
12 12 1100
13 14 1110
14 15 1111
15 16 10000
16 17 10001
17 18 10010
18 20 10100
19 21 10101
20 24 11000
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MAPLE
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f:= proc(d) local i, j, k;
op(sort([seq(seq(add(2^(d-j*k), k=0..m), m=1..d/j), j=1..d), 2^(d+1)]))
end proc:
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MATHEMATICA
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bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n, 2]], 1];
Select[Range[100], SameQ@@Differences[bpe[#]]&] (* Gus Wiseman, Jul 22 2019 *)
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PROG
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(PARI) is(n) = my(h=hammingweight(n)); if(h<3, return(1), my(i=valuation(n, 2), w=#binary(n)); if((w-i-1)%(h-1)==0, my(j=(w-i-1)/(h-1)); return(sum(k=0, h-1, 2^(i+j*k))==n), return(0)))
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CROSSREFS
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Cf. A000051, A000079, A000120, A000225, A000668, A000695, A002450, A006995, A007814, A019434, A023001, A048645, A054519, A064896.
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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