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A320037
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Write n in binary, then modify each run of 0's by appending one 1, and modify each run of 1's by appending one 0. a(n) is the decimal equivalent of the result.
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6
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2, 9, 6, 17, 38, 25, 14, 33, 70, 153, 78, 49, 102, 57, 30, 65, 134, 281, 142, 305, 614, 313, 158, 97, 198, 409, 206, 113, 230, 121, 62, 129, 262, 537, 270, 561, 1126, 569, 286, 609, 1222, 2457, 1230, 625, 1254, 633, 318, 193, 390, 793, 398, 817, 1638, 825, 414
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OFFSET
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1,1
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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) = 2 and f(1) = a(2) + a(3) = 15.
Then f(k) = 16*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) = 2a(2n)-1, a(4n+1) = 4a(2n)+2, a(4n+2) = 4a(2n+1)+1 and a(4n+3) = 2a(2n+1)+2. 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) = 15 shows that f(k) = 16*6^(k-1) - 2^(k-1) for k > 0.
(End)
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LINKS
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FORMULA
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a(4n) = 2a(2n)-1, a(4n+1) = 4a(2n)+2, a(4n+2) = 4a(2n+1)+1 and a(4n+3) = 2a(2n+1)+2. - Chai Wah Wu, Nov 21 2018
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EXAMPLE
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6 in binary is 110. Modify each run by appending the opposite digit to get 11001, which is 25 in decimal. So a(6) = 25.
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PROG
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(Python)
from re import split
return int(''.join(d+'0' if '1' in d else d+'1' 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,base
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AUTHOR
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STATUS
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approved
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