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A320038
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Write n in binary, then modify each run of 0's by prepending one 1, and modify each run of 1's by prepending one 0. a(n) is the decimal equivalent of the result.
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6
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1, 6, 3, 12, 25, 14, 7, 24, 49, 102, 51, 28, 57, 30, 15, 48, 97, 198, 99, 204, 409, 206, 103, 56, 113, 230, 115, 60, 121, 62, 31, 96, 193, 390, 195, 396, 793, 398, 199, 408, 817, 1638, 819, 412, 825, 414, 207, 112, 225, 454, 227, 460, 921, 462, 231, 120, 241
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OFFSET
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1,2
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COMMENTS
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A variation of A175046. Indices of record values are given by A319423.
Let f(k) = Sum_{i=2^k..2^(k+1)-1} a(i), i.e., the sum ranges over all numbers with a (k+1)-bit binary expansion. Thus f(0) = a(1) = 1 and f(1) = a(2) + a(3) = 9.
Then f(k) = 10*6^(k-1) - 2^(k-1) for k > 0.
Proof: looking at the last 2 bits of n, it is easy to see that a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+1, a(4n+2) = 4*a(2n+1)+2 and a(4n+3) = 2*a(2n+1)+1. By summing over the recurrence relations for a(n), we get f(k+2) = Sum_{i=2^k..2^(k+1)-1} (f(4i) + f(4i+1) + f(4i+2) + f(4i+3)) = Sum_{i=2^k..2^(k+1)-1} (6a(2i) + 6a(2i+1) + 4) = 6*f(k+1) + 2^(k+2). Solving this first-order recurrence relation with the initial condition f(1) = 9 shows that f(k) = 10*6^(k-1) - 2^(k-1) for k > 0.
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LINKS
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FORMULA
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a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+1, a(4n+2) = 4*a(2n+1)+2 and a(4n+3) = 2*a(2n+1)+1. - Chai Wah Wu, Nov 25 2018
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EXAMPLE
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6 in binary is 110. Modify each run by prepending the opposite digit to get 01110, which is 14 in decimal. So a(6) = 14.
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MATHEMATICA
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a[n_] := Split[IntegerDigits[n, 2]] /. {a0:{(0)...} :> Prepend[a0, 1], a1:{(1)...} :> Prepend[a1, 0]} // Flatten // FromDigits[#, 2]&;
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PROG
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(Python)
from re import split
return int(''.join('0'+d if '1' in d else '1'+d for d in split('(0+)|(1+)', bin(n)[2:]) if d != '' and d != None), 2)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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